SATHEE: Chapter 05 Heating And Chemical Effects Of Current

Chapter 05 Heating And Chemical Effects Of Current

Introduction

Heating and chemical effects are the most important effects of electric current. Heating effect has many day-to-day applications for us. We use electric iron to press our clothes, room heater to get rid of cold, electrical lamps to lit our rooms, etc. We access these facilities due to heating effect of current. In these phenomena, electrical energy is converted into heat energy.

The other effect of electric current is to produce the chemical change. This effect is called the chemical effects of electric current. This chapter is based on the two effects of electric current.

ELECTRIC POWER

If a battery is used to establish an electric current in a conductor, these is a continuous transformation of chemical energy stored in the buttery to kinetic energy of the change carriers. This kinetic energy is quickly lost as a result of collisions between the charge carriers and the lattice ions, resulting in an increase in the temperature of the conductor. Therefore, we see that the chemical energy stored in the battery is continuously transformed into heat energy.

Consider a simple circuit vconsisting of a battery whose terminals are connected to a resistor R, as shown in figure 5.1.

The positive terminal of the battery (the longer plate) is at the higher potential, while the negative terminal (the shorter plate) is at the lower potential. Now imagine a positive quantity of charge $\Delta Q$ moving around the circuit from point $A$ through the battery and resistor and back to $A$. Point $A$ is a reference point that is grounded, and its potential is taken to be zero. As the charge moves from $A$ to $B$ through the battery, its electrical potential energy increases by an amount $V \Delta Q$ (where $V$ is the potential at $B$) while the chemical potential energy in the battery decreases by the same amount. However, as the charge moves from $C$ to $D$ through the resistor, it loses this electrical potential energy as it undergoes collisions with atoms in the resistor, thereby producing thermal energy. Note that if we neglect the resistance of the interconnecting wires there is no loss in energy for paths $B C$ and $D A$. When the charge returns to point $A$, it must have the same potential energy (zero) as it has at the start.

The rate at which the charge $\Delta Q$ loses potential energy in going through the resistor is given by

$$ \frac{\Delta U}{\Delta t}=\frac{\Delta Q}{\Delta t} V=I V $$

where $I$ is the current in the circuit. Of course, the charge regains this energy when it passes through the battery. Since the rate at which the charge loses energy equals the power $P$ lost in the resistor, we have

P = IV ………………… (1)

In this case, the power is supplied to a resistor by a battery. However, eq. (1) can be used to determine the power transferred to any device carrying a current $I$ and having a potential difference $V$ between its terminals.

Using eq. (1) and the fact that $V=I R$ for a resistor, we can express the power dissipated in the alternative forms

$$ P=I^{2} R=\frac{V^{2}}{R} $$

When $I$ is in amperes, $V$ in volts, and $R$ in ohms, the $S I$ unit of power is the watt (W). The power lost as heat in a conductor of resistance $R$ is called joule heating.

Watt is a small unit of power, its other bigger units are Kilowatt, Megawatt and Horsepower.

$ \begin{aligned} & 1 kW=1000 W=10^{3} W \\ & 1 {MW}=1000000 W=10^{6} W \\ & 1 hp=746 W \end{aligned} $

If $100 W-220 V$ is written on the bulb then it means that the bulb will consume 100 joule in one second if used at the potential difference of 220 volts.

The value of electricity consumed in houses is decided on the basis of the total electric energy used. Electric power tells us about the electric energy used per second not the total electric energy.

The total energy used in a circuit $=$ power of the electric circuit $\times$ time

If the power is in watt and time is in second, then the electric energy will be in joule or watt $\times$ second.

As the watt $\times sec$ is a small unit of energy, electric energy is measured in Kilowatt Hour (kWh)

1 Kilowatt hour $(kWh)=1000$ watt hour

$ =1000 \times 60 \times 60 \text{ watt sec }=36000000 \text{ joule }=3.6 \times 10^{6} \text{ joule } $

$1 kWh$ is called the 1 unit.

The electricity department uses this unit in the electric bills. If the 100 watt bulb or any other electric appliance is used for an hour then the total electric energy consumed is 1 kWh or 1 unit.

CHECK POINT:

The wattage marked on a light bulb is not an inherent property of the bulb; rather, it depends on the voltage to which it is connected, usually 110 or 120V. How many amperes flow through a 60-W bulb connected in a 120-V elrcult?

Show Answer

SOLUTION

Using the formula,

Watt $=$ volt $\times$ ampere

or, $60=120 \times$ ampere $\Rightarrow$ ampere $=\frac{60}{120}=0.5 A$

Thus, a 0.5 A current flows through a 60-W bulb connected in a 120-V circuit

ILLUSTRATION-5.1

An immersion rod of 1500 W is used everyday for 3 hours to boil water. If the rate of 1 unit of electric energy is $₹ 2.5$ then what will be the total cost of the electricity used in 30 days?

Show Answer

SOLUTION:

Electric energy used in a day $=1500 \times 3 Wh=4500 Wh$

Electric energy used in 30 days $=4500 \times 30 Wh=135000 Wh$

Total energy in $kWh$ unit $=\frac{135000}{1000} kWh=135 kWh$ or unit

Hence the total cost of the electricity $=135 \times ₹ 2.50=₹ 337.50$

ILLUSTRATION-5.2

An electric heater is rated $750 W-200 V$. Calculate (i) resistance of the heating element (ii) current flowing through it.

Show Answer

SOLUTION:

$P=750 W, p . d .=200 V, R=?, I=?$

$$ P=\frac{V^{2}}{R} \quad \therefore R=\frac{V^{2}}{P} $$

$\therefore \quad R=\frac{200 \times 200}{750}=53.33 \Omega$

$$ R=\frac{V}{I} \quad \therefore I=\frac{V}{R}=\frac{200}{53.33}=3.75 A $$

ILLUSTRATION-5.3

Calculate the monthly bill for a heater of resistance $50 \Omega$, which used $200 V$ mains, such that its daily use is $4 hr$. The energy costs ₹ 2.50 per kWh.

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SOLUTION:

Power of heater $P=\frac{V^{2}}{R}=\frac{(200)^{2}}{50}=800 W$

$\therefore$ Energy consumed by heat in a day $=P \times t=800 W \times 4 h=3200 Wh$

$\therefore$ Energy consumed by heater in 30 days (month) $=3200 Wh \times 30=96000 Wh=96 kWh$

$\therefore$ Monthly bill $=96 kWh \times \frac{₹ 2.50}{kWh}=₹ 240$

ILLUSTRATION-5.4

Rearrange the equation

Current $=$ voltage/ resistance

to express resistance in terms of current and voltage. Then solve the following: A certain device in a 120-V circuit has a current rating of $20 A$. What is the resistance of the device (how many ohms)?

Show Answer

SOLUTION:

Current $=$ voltage $/$ resistance

Rearranging, we get

Current $\times$ resistance $=$ voltage

or, Resistance $=$ voltage/current.

Now, Resistance $=\frac{\text{ voltage }}{\text{ current }}=\frac{120}{20}=6$ ohms.

Therefore, the resistance of the device is 6 ohms.

HEAT PRODUCED DUE TO ELECTRICITY

When a current passes through a conductor, heat is produced in it due to the collisions between the free electrons and the atoms of the conductor. The heat produced $(Q)$ depends on the following factors.

1. Resistance of conductor : More the resistance of conductor more is the number of collisions between free electrons and atoms. Hence if a conductor offers more resistance $Q \propto R$ 2. Current flowing through the conductor: Heat produced increases with the increase in the strength of the current flowing through conductor $Q \propto I^{2}$ 3. Time of flow of current : As the time of flow of current increases, the heat produced also increases.

$\quad$ $Q \propto t$

$\quad$Thus a conductor having larger resistance will generate more heat when a stronger current passes through it for a longer time. If $I$ is the current flowing a conductor of resistance $R$ for time $t$, the heat produced $(Q)$ is given by $Q \propto IRt$

PRACTICAL APPLICATION OF HEATING EFFECT

The heat generated when current passes through a resistive material is used in many common devices. The material through which the current passes is surrounded by an insulating substance in order to prevent the current from flowing through the cook to the earth when he or she touches the pan. A material that is a good conductor of heat surrounds the insulator

Hair dryer, in which a fan blows air past heating coils. In this case the warm air can be used to dry hair, but on a broader scale this same principle is used to dry clothes and to heat buildings.

A final example of a household uses the heating effect of electric current is the steam iron. A heating coil warms the bottom of the iron and simultaneously turns water to steam, which is sprayed from jets located in the bottom of the iron.

Other practical application includes heater, toaster, electric kettle.

CHEMICAL EFFECT OF CURRENT

Electric current can produce chemical change. This ability of current is called chemical effect. This effect is shown by D.C. only. A.C. current can’t produce chemical effect.

FARADAY’S LAWS OF ELECTROLYSIS

(i) 1st law

The mass of the substance liberated or deposited at an electrode during electrolysis is dircetly proportional to the quantity of charge passed through the electrolyte i.e.

mass or m = Zq = Z It, where Z = electrochemical equivalent (E.C.E.) of substance.

(ii) 2nd law

When the same amount of charge is passed through different electrolytes, the masses of the substance liberated or deposited at the various electrodes are proportional to their chemical equivalents i.e: $\frac{m_1}{m_2}=\frac{E_1}{E_2}$ where $m_1$ & $m_2$ are the masses of the substances liberated or deposited on electrodes during electrolysis and $E_1$ & $E_2$ are their chemical equivalents.

FARADAY CONSTANT

(i) Faraday constant is equal to the amount of charge required to liberate the mass of a substance at an clectrode during electrolysis, equal to its chemical equivalent in gram (i.e. one gram equivalent)

(ii) One faraday = 1F = 96500 C/gram equivalent.

ELECTROCHEMICAL CELL

An electrochemical cell is a device which by converting chemical energy into electrical energy maintains the flow of charge in a circuit. It usually consists of two electrodes of different materials and an electrolyte. The electrode at higher potential is called anode and the one at lower potential is cathode

PRIMARY CELLS

The cells which cannot be recharged electrically are called primary cells. Here the original state of cell cannot be brought back by passing electrical energy through cell from external source after cell is discharged.

e.g. Voltaic cell, Daniel cell, Leclanche cell, Manganese-alkaline cell, Mercury button cell etc.

SECONDARY CELLS

The cells in which chemical process is reversible are called secondary cells. Here original chemical state of cell can be brought back by passing electrical energy through cell from external source.

e.g. Lead acid accumulator, alkali cells etc.

ELECTROMOTIVE FORCE (EMF)

The emf of a cell is defined as work done by cell in moving a unit positive charge in the whole circuit including the cell once emf $E=W / q$;

Its SI unit is joule/coulomb or volt.

Emf is the maximum potential difference between the two electrodes of the cell when no current is drawn from the cell. It is the characteristic property of cell and depends on the nature of electrodes and electrolyte used in cell, and is independent of quantity of electrolyte, size of electrodes and distance between the electrodes.

INTERNAL RESISTANCE OF CELL

The opposition offered by the electrolyte of the cell to the flow of electric current through it is called the internal resistance of the cell. The internal resistance of cell depends on.

Distance between electrodes $(r \propto d)$, larger is the separation between electrodes more is the length of electrolyte through which ions have to move so more is internal resistance.

Conductivity or nature of electrolyte $(r \propto 1 / \sigma)$

Concentration of electrolyte $(r \propto c)$, Temperature of electrolyte $(r \propto 1 / T)$

Nature and area of electrodes dipped in electrolyte $(r \propto 1 / A)$

TERMINAL POTENTIAL DIFFERENCE

The potential difference between the two electrodes of a cell in a closed circuit i.e., when current is being drawn from the cell is called terminal potential difference.

(a) When cell is discharging

When cell is discharging current inside the cell is from anode to cathode .

Current $\quad I=\frac{E}{r+R}$ or $\quad E=I R+I r=V+I r \quad$ or $\quad V=E-I r$

When current is drawn from the cell, potential difference is less than emf of cell. Greater is the current drawn from the cell smaller is the terminal voltage. When a large current is drawn from a cell, its terminal voltage is reduced.

(b) When cell is charging

When cell is charging current inside the cell is from anode to cathode.

Fig. 5.4

Current $I=\frac{V-E}{r}$ $\quad$ or $\quad V=E+I r$

During charging terminal potential difference is greater than emf of cell.

(c) When cell is in open circuit

In open circuit $R=\infty \quad \therefore I=\frac{E}{R+r}=0 \quad$ So $\quad V=E$

In open circuit terminal potential difference is equal to emf and is the maximum potential difference which a cell can provide.

(d) When cell is short circuited

In short circuit $R=0$ so $I=\frac{E}{R+r}=\frac{E}{r}$ and $\quad V=I R=0$

In short circuit current from cell is maximum and terminal potential difference is zero.

(e) Power transferred to load by cell

$ \begin{aligned} & P=I^{2} R=\frac{E^{2} R}{(r+R)^{2}} \\ & P=P _{\max } \text{ if } r=R \end{aligned} $

Fig.5.5

Power transferred by cell to load is maximum when $r=R$ and $P _{\max }=\frac{E^{2}}{4 r}=\frac{E^{2}}{4 R}$.

DIFFERENCE BETWEEN EMF AND POTENTIAL DIFFERENCE

	EMFOFACELL		POTENTIAL DIFFERENCE
1.	The emf of a cell is the maximum potential difference between the two electrodes of a cell when the cell is in the open circuit.	1.	The potential difference between the two points is the difference of potential between these two points in a closed circuit.
2.	Emf is independent of resistance of circuit and depends upon the nature of electrodes and electrolyte.	2.	This depends upon the resistance between two points of the circuit and current flowing through the circuit.
3.	The term emf is used for source of current.	3.	Potential difference can be measured between any two points of circuit.
4.	This is a cause.	4.	It is an effect.

ILLUSTRATION-5.5

A dry cell of e.m.f. $1.5 V$ and internal resistance 0.10 $\Omega$ s connected across a resistor in series with a very low resistance ammeter.

When the circuit is switched on, the ammeter reading settles to a steady value of 2.0 A, What is the steady

(i) rate of chemical energy consumption of the cell.

(ii) rate of energy dissipation inside the cell.

(iii) rate of energy dissipation inside the resistor and

(iv) power output of the source?

Show Answer

SOLUTION:

Let $R$ be the external resistance, then

$(R+r)=\frac{V}{i}=\frac{1.5}{2.0}=0.75 \Omega$

$\therefore R=0.75-r=0.75-0.10=0.65 \Omega$

(i) The rate of consumption of chemical energy of the cell $=V \times i=1.5 \times 2.0=3.0 W$

(ii) The rate of energy dissipation inside the internal resistor $=i^{2} r=(2.0)^{2} \times 0.10=0.4 W$

(iii) The rate of energy dissipation inside the external resistor $=i^{2} R=(2.0)^{2} \times 0.65=2.6 W$

(iv) The power output of the source

$=$ total power

Rate of energy dissipation inside the cell

$=3.0-0.4=2.6 W$

ILLUSTRATION-5.6

Does more current flow out of a battery than into it? Does more current flow into a light bub than out of it? Explain.

Show Answer

SOLUTION:

The current that flow out of a battery is always equal to the current that flow into it.

Similarly, the current that flows into a light bulb is always equal to the current that flows out of it. Although there will be a voltage drop due to the resistance provided to the flow of current due to both the battery and the resistance, but the amount of current won’t be affected.

GROUPING OF CELLS

(i) Series Combination

The cells are said to be connected in series if negative terminal of first is connected to positive of second whose negative terminal is connected to positive of third cell. The external resistance is connected between free terminals of first and last cells.

Let $n$ identical cells each of emf $E$ and internal resistance $r$ be connected in series. The combination can be replaced by a single cell of emf $n E$ and internal resistance $n r$. The current flowing through load $I=\frac{n E}{R+n r}$

If $n r \ll R$ then $I=\frac{n E}{R}$. If equivalent internal resistance $n r$ is less than external resistance $R$ then current in circuit is equal to $n$ times circuit current due to single cell.

If $n r \gg R$ then $I=E / r$. If equivalent internal resistance $n r$ is greater than external resistance $R$ then current in circh short circuited current obtained from one cell.

Maximum current can be drawn from series combination of cells if external resistance is very large as compared to equivalent internal resistance.

If in series combination of $n$ cells, $p$ cells are reversed then equivalent emf

$ E _{e q}=(n-p) E-p E=(n-2 p) E \quad \text{ and } \quad r _{e q}=n r \text{ So current } I=\frac{(n-2 p) E}{n r+R} $

If unidentical cells are connected in series then

$ E _{e q}=E_1+E_2+\ldots .=\Sigma E_i \text{ and } \quad r _{e q}=r_1+r_2+\ldots .=\Sigma r_i $

$\therefore$ Current, $I=\frac{\Sigma E_l}{R+\Sigma r_i}$.

(ii) Parallel Combination

The cells are said to be connected in parallel if positive terminals of all the cells are connected tagether at one point and their negative terminals at another point. The external resistor is connected between these two points.

Let $m$ identical cells each of emf $E$ and internal resistance $r$ be connected in parallel. The combination can be reple cell of emf $E$ and internal resistance $r / m$. The current $I=\frac{E}{R+r / m}$.

If $r / m \ll R$ then $I=\frac{E}{R}$. If equivalent internal resistance $r / m$ is less than external resistance $R$ then current in circuit is equal to current produced by a single cell.

If $r / m \gg R$ then $I=\frac{m E}{r}$. If equivalent internal resistance $\frac{r}{m}$ is greater than external resistance then current in circuit is equal to $m$ times the current produced by a short circuited cell.

Thus maximum current can be drawn from parallel combination of cells if external resistance is small as compared to net internal resistance of cells.

(iii) Mixed Combination

If the $m n$ number of identical sources are combined, as shown, where we have the $m$ number of rows of sources in parallel and the $n$ number of sources in series in each row, the total emf of combination becomes $n e_s$ and the total internal resistance of combination becomes $n rl$ $m$. Then, the electric current through the load is

$$ I=\frac{n e_s}{(\frac{n r}{m}+R)}=\frac{e_s}{(\frac{r}{m}+\frac{R}{n})} $$

and the electrical power delivered to the load is

$$ P_2=R I^{2}=R \frac{n^{2} e_s^{2}}{(\frac{n r}{m}+R)^{2}}=R \frac{e_s^{2}}{(\frac{r}{m}+\frac{R}{n})^{2}} $$

which is maximum under the condition of total internal and external resistances being equal, i.e.,

$$ R=\frac{n r}{m} $$

ILLUSTRATION-5.7

Four identical cells, each of emf $e_s=2 V$, are joined is parallel, providing a current to the external circuit which consists of two resistances, each of $15 \Omega$, in parallel. The terminal voltage of the cells is $1.6 V$. Calculate the internal resistance $r$ of each cell.

Show Answer

SOLUTION:

The complete circuit diagram, consisting of the 4 identical cells and the two external resistances in parallel, is shown. The total emf in the circuit is $e_s^{\prime}=e_s=2 V$, the total internal resistance is $r / 4$ and the total external resistance is $15 / 2=7.5 \Omega$. Therefore, the current through the total external resistance is

Fig.5.11

$$ I=\frac{2}{(\frac{r}{4}+7.5)}=\frac{8}{(r+30)} $$

and the terminal voltage across the cells is

$ \begin{aligned} & V=e_s^{\prime}-I\bigg(\frac{r}{4}\bigg) \\ \Rightarrow \quad & 1.6=2-\frac{8}{(r+30)} \times \frac{r}{4}=\frac{60}{(r+30)} \\ \Rightarrow \quad r & =7.5 \Omega \end{aligned} $

ILLUSTRATION-5.8

If 64 cells of emf $e_s=2 V$ and internal resistance $r=2 \Omega$, are connected in $m$ rows, each consisting of $n$ cells in series, then find out these values of $m$ and $n$ for getting the maximum current through an external resistance $R=8 \Omega$.

Show Answer

SOLUTION:

The current through the external resistance $R$ is given by

$ \begin{aligned} I & =\frac{n e_s}{(\frac{n r}{m}+R)}=\frac{(m n) e_s}{(n r+m R)}=\frac{(64) \times 2}{(n \times 2+m \times 8)} \\ & =\frac{64}{(n+4 m)}=\frac{64}{(n+\frac{4 \times 64}{n})}=\frac{64}{(n+\frac{256}{n})} \end{aligned} $

where $m n=64$ is the total number of cells. The current will be maximum, when the denominator is minimum, i.e., when

$ \frac{d}{d n}(n+\frac{256}{n})=0 $

$ \Rightarrow 1-\frac{256}{n^{2}}=0 $

$ \Rightarrow n=16 $

$ \Rightarrow m=\frac{64}{n}=4 . $

DOMESTIC ELECTRIC CIRCUITS

Electricity from power stations is transmitted to homes through high voltage power lines. Electric potentials as high as 220000 V $AC$ are used. The electrical potential is decreased to 220 VAC with the help of step-down transformer before it enters the home. A cable containing three colour-coded wires enters the home. One wire is red, one is black, and one is green. The red wires are called live wires (or positive). Another wire with the black insulation is called neutral wires (or negative). There is an electrical potential difference of $220 V$ AC between a red and a black wire. The earth wire, which has insulation of green colour, is usually connected to a metal plate deep in the earth near the house. This is used as a safety measure, especially for those appliances that have a metallic body, for example, electric press, toaster, table fan, refrigerator, etc. The metallic body is connected to the earth wire, which provides a low-resistance conducting path for the current. Thus, it ensures that any leakage of current to the metallic body of the appliance keeps its potential to that of the earth, and the user may not get a severe electric shock.

Fuses and Circuit Breakers

A fuse is a safety device that acts as a switch. Fuses are connected into the circuit close to the source. As a result all the current flowing in the circuit passes through the fuse. When too much current flows, the fuse heats up sufficiently to bum out. This creates an air gap and no more current flows. The construction of a plug fuse is shown in Figure. Other kinds of fuses are cartridge fuses and knife blade fuses. The essential part is the fuse wire or a blade with a low melting point.

Fig. 5.12 : A plug fuse

Most homes built in recent years use circuit breakers instead of fuses in their electrical panels. A circuit breaker is connected in the same place as a fuse would be, and it performs the same function. Unlike a fuse, however, it does not bum out. When too much current flows, the circuit breaker simply opens like a switch. Then, when the problem that caused the excess current has been corrected, you can close the breaker again.

The Circuit Breaker Panel

At the home, the electricity first passes through an electric meter. The electric meter records the amount of electric energy used in the home. Then the electricity enters a circuit breaker panel. (In some older homes it may enter a fuse panel.) In this panel there is a master circuit breaker. The master circuit breaker can be operated manually to cut off the electricity to the entire home. It opens automatically if the current exceeds the -rated value. In homes that have a $100 A$ service, the master circuit breaker will allow up to $100 A$ to pass before it pops open.

There are three strips of conducting metal called bus bars in the breaker panel. The red wire is connected to one, the black wire to another, and the white wire to the third. As a result, two of the bus bars are live and one is neutral.

Electric Circuits in the Home

All the electric circuits within the home originate from the bus. Figure shows a $220 V$ AC cable going to a stove. Notice that the cable is connected to all three bus bars. This cable has two circuit breakers, one for the black wire and one for the red wire. Each major appliance has its own 220 V AC circuit.

A $110 V$ AC circuit is also shown. Notice that only two of the bus bars are used, the neutral bus bar and either of the live bus bars. This circuit also has its own circuit breaker. The $110 V$ AC circuit consists of the circuit breaker, two three-prong plugs, and two light fixtures and two switches. A ground wire connects all the bar. prevents injury if the boxes become live.

Loads in the same circuit, such as wall outlets and lights, are always connected in parallel. If one load bums out the others still operate. Also, as more and more loads are turned on, the electrical potential remains the same. However, the electric current increases. But one load does not affect another unless the circuit becomes overloaded.

Causes of Overloaded Circuits

There are two main causes of overloaded circuits. First, an appliance or a conducting wire can develop a short circuit. Now the current is no longer limited by the load. In this case, current passes from the live wire to the neutral wire or from the live wire to the ground wire without passing through the load. The current becomes very large very fast. This heats up the circuit. Second, too many appliances can be plugged into the same circuit. Each appliance draws a certain current. The combination can draw too much current and the circuit again heats up.

A circuit breaker is designed to prevent overloading. It trips and opens the circuit as soon as the circuit becomes too hot.

CHECK POINT:

Why are devices in household circuits almost never connected in series?

Show Answer

SOLUTION:

The devices in household circuits are never connected in series because if they are connected so they can’t be turned on of separately as required. If one of the device will be turned off then all of them will also be turned off. Moreover, if the devices are lighting one, their brightness will decrease if more devices will be added.

ILLUSTRATION-5.9

An immersion heater having $50 \Omega$ resistance is connected to $250 V$ mains. Find the time required to heat $2 kg$ of water from $34^{\circ} C$ to $100^{\circ} C$, assuming there is no loss of heat energy (Sp. heat cap. of water $=4 \times 10^{3} J / kg^{\circ} C$)

Show Answer

SOLUTION:

Current passing through the heater

$ I=\frac{V}{R}=\frac{250}{50}=5 A $

Heat produced $=I^{2} R t$ and Heat required $=m C \theta$.

Heat produced $=5^{2} \times 50 \times t$

$ m C \theta=2 \times 4 \times 10^{3} \times(100-35)=8 \times 10^{3} \times 65 $

Since

$ 5^{2} \times 50 \times t=8 \times 10^{3} \times 65 $

$\therefore \quad t=\frac{520 \times 1000}{1250}=416 s=6 \min 56 s$

ILLUSTRATION-5.10

An electric refrigerator rated $400 W$ operates 8 hours/day. What is the cost of the energy to operate it for 30 days at ₹ $3.00 per$ $kWh$ ?

Show Answer

SOLUTION:

The total energy consumed by the refrigerator in 30 days would be

$400 W \times 8.0$ hours/day $\times 30$ days $=96999 Wh=96 k Wh$

Thus the cost of energy to operate the refrigerator for 30 days is

$96 kWh \times ₹ 3.00$ per $kWh=₹ 288.00$

ILLUSTRATION-5.11

What is the operating resistance of a 100-watt household light bulb ? The operating line voltage of household electricity is $120 V$.

Show Answer

SOLUTION:

Given : $P=100 W, V=120 V$, Find R

From $P=\frac{V^{2}}{R}$, We have $R=\frac{V^{2}}{P}=\frac{(120 V)^{2}}{100 W}=144 \Omega$

ILLUSTRATION-5.12

On a $220 V$ power line the powers of two electric bulbs are $60 W$ and $30 W$ separately. Find the powers of their parallel and series combinations when connected on the same line.

Show Answer

SOLUTION:

The resistance of the $220 V-60 W$ bulb is $\quad R_1=\frac{V^{2}}{P_1}=\frac{(220)^{2}}{60} \Omega$

The resistance of the $220 V-30 W$ bulb is

$ R_2=\frac{V^{2}}{P_2}=\frac{(220)^{2}}{30} \Omega $

When the bulbs are connected in parallel, their equivalent resistance is

$R=\frac{R_1 \times R_2}{R_1+R_2}=\frac{\frac{(220)^{2}}{60} \times \frac{(220)^{2}}{30}}{\frac{(220)^{2}}{60}+\frac{(220)^{2}}{30}}=\frac{(220)^{2}}{90} \Omega$

The power of the parallel combination is $\quad P=\frac{V^{2}}{R}=\frac{(220 V)^{2}}{{(220)^{2} / 90} \Omega}=90 W$

When the bulbs are connected in series, their equivalent resistance is

$R^{\prime}=R_1+R_2=\frac{(220)^{2}}{60}+\frac{(220)^{2}}{30}=(220)^{2} \times \frac{1}{20} \Omega$

The power is $\quad P^{\prime}=\frac{V^{2}}{R^{\prime}}=\frac{(220 V)^{2}}{{(220)^{2} \times 1 / 20} \Omega}=20 W$

ILLUSTRATION-5.13

An electric kettle has coils A and B. When coil A is switched on, the water boils in 10 minutes, and when coil $B$ is switched on, the water boils in 20 minutes. Calculate the time taken by the water to boil if the coils connected in (a) series and in (b) parallel are switched on.

Show Answer

SOLUTION:

Let $R_1$ and $R_2$ be the resistances of the coils $A$ and $B$ respectively, and $t_1$ and $t_2$ the time-intervals taken by the water to boil when $A$ and $B$ are switched on turn by turn. If $W$ be the heat energy required to boil the water, then

$ W=\frac{V^{2} t_1}{R_1}=\frac{V^{2} t_2}{R_2} $

where $V$ is supply voltage. This gives

$ \frac{R_2}{R_1}=\frac{t_2}{t_1} $

Here $t_1=10$ min and $t_2=20 min . \quad \therefore \quad \frac{R_2}{R_1}=\frac{20}{10}=2$

$ \text{ or } \quad R_2=2 R_1 $

(a) When the coils are connected in series, the equivalent resistance is

$ R=R_1+R_2=R_1+2 R_1=3 R_1 . $

If $t$ be the time taken by the water to boil, then from eq. (1), we have

$ \frac{R}{R_1}=\frac{t}{t_1} ; t=3 t_1=30 min . \quad[\because R\rangle _{.R_1=3]} $

(b) When the coils are connected in parallel, the equivalent resistance is

$R^{\prime}=\frac{R_1 R_2}{R_1+R_2}=\frac{R_1(2 R_1)}{R_1+2 R_1}=\frac{2}{3} R_1$ If $t^{\prime}$ be the time taken by the water to boil, then from eq. (1), we have

$ \frac{R}{R_1^{\prime}}=\frac{t^{\prime}}{t_1} \quad \text{ or } \quad t^{\prime}=\frac{2}{3} t_1=\frac{20}{3} \min \quad[\because R^{\prime} / R_1=2 / 3] $

ILLUSTRATION-5.14

Two bulbs A and B are rated $100 W, 120 V$ and $10 W, 120 V$ respectively. They are connected across a $120 V$ source in series. Calculate the current through each bulb. Which bulb will consume more energy?

Show Answer

SOLUTION:

Resistance of bulb A (rating 100W, 120V) is $R_1=\frac{V^{2}}{P_1}=\frac{(120)^{2}}{100}=144 \Omega$

Resistance of bulb B (rating 10W, 120V) is $R_2=\frac{V^{2}}{P_2}=\frac{(120)^{2}}{10}=1440 \Omega$

When they are connected in series,

Total resistance, $R=R_1+R_2=144+1440=1584 \Omega$

Current in the circuit $I=\frac{V}{R}=\frac{120}{1584}=0.076 A$

Same current passes through each bulb.

Power consumed in the bulb $A$ is $P_1=R^{2} R_1$.

Power consumed in the bulb B is $P_2=R^{2} R_2$.

Since, $\quad R_1<R_2$

Since, $\quad R_1<R_2 \quad \therefore \quad P_1<P_2$. Hence, the bulb B (rated 10W, 120V) consumes more energy than the bulb A (rated 100W, 120V) when they are connected in series.

ILLUSTRATION-5.15

There are 5 rooms in a house. Each room has a $100 W$ bulb and a $40 W$ tube light. If every day the bulb is used for 1 hour and tube light is used for 5 hours in each room then what will be the cost of total electric energy consumed in 30 days when 1 unit of electric energy costs ₹ 2.5 .

Show Answer

SOLUTION:

Energy used every day in 5 bulbs $=5 \times 100 \times 1 Wh=500 Wh$

Energy used everyday in 5 tube lights $=5 \times 40 \times 5 Wh=1000 Wh$

Total energy used everyday by bulb and tube light $=(500+1000) Wh=1500 Wh$

Total energy used in 30 day $=1500 \times 30 Wh=45000 Wh=\frac{45000}{1000} KWh=45 KWh$ (unit)

Cost of the total energy used $=45 \times 2.50 ₹=₹ 112.50$

ILLUSTRATION-5.16

Compute the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of $50 V$.

Show Answer

SOLUTION:

Here, $Q=96,000 C, t=1$ hour $=60 \times 60 s, V=50 V$

Current, $I=\frac{Q}{t}=\frac{96000}{60 \times 60}=\frac{80}{3} A$

Resistance, $R=\frac{V}{I}=\frac{50 \times 3}{80}=\frac{15}{8} \Omega$

Heat produced, $H=P^{2} R t=\frac{80}{3} \times \frac{80}{3} \times \frac{15}{8} \times 60 \times 60=4800000 J=4800 kJ$

ILLUSTRATION-5.17

When a $12 V$ battery is connected across an unknown resistor, there is a current of $2.5 mA$ in the circuit. Find the value of the resistance of the resistor.

Show Answer

SOLUTION:

Potential difference, $V=12 V$

Current, $I=2.5 mA=2.5 \times 10^{-3} A$

We know that, $\quad V=I R ; R=\frac{V}{I}=\frac{12}{2.5 \times 10^{-3}}=4.8 \times 10^{3} \Omega=4.8 k \Omega$

ILLUSTRATION-5.18

Two lamps, one rated $100 W$ at $220 V$, and the other $60 W$ at $220 V$, are connected in parallel to electric mains supply, What current is drawn from the line if the supply voltage is $220 V$ ?

Show Answer

SOLUTION:

We know that, $P=\frac{V^{2}}{R} \quad \therefore \quad R=\frac{V^{2}}{P}$

Resistance of 1st lamp, $R_1=\frac{V^{2}}{P}=\frac{220 \times 220}{100}=484 \Omega$

Resistance of 2nd lamp, $R_2=\frac{220 \times 220}{60}=\frac{2420}{3} \Omega$

Since, two lamps are connected in parallel, so its equivalent resistance is given by

$ \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{484}+\frac{3}{2420}=\frac{5+3}{2420}=\frac{8}{2420} \Rightarrow R=\frac{2420}{8} \Omega $

Current drawn from the line, $I=\frac{V}{R}=\frac{220 \times 8}{2420}=0.73 A$

ILLUSTRATION-5.19

A current of $0.5 A$ is passing through a $CuSO_4$ solution. How many $Cu^{++}$ions will be deposited on catlode in 10 second?

Show Answer

SOLUTION:

If $n$ is number of copper ions then $q=n \times 2 e$

$q=It \quad$ or $\quad n \times 2 e=It \quad$ so $n=\frac{It}{2 e}=\frac{0.5 \times 10}{2 \times 1.6 \times 10^{-19}}=1.5625 \times 10^{19}$.

ILLUSTRATION-5.20

Calculate the current flowing through a nichrome wire, immersed in $100 g$ of oil at $20^{\circ} C$, such that final temperature of oil ries to $80^{\circ} C$ in five minutes, when the supply of emf is $10 V$. [$Sp$. heat capacity of oil is $0.85 Jg^{-10} C^{-1}$]

Show Answer

SOLUTION:

$I=$ ?, p.d. $=10 V, t=5 min=300 s$

$m=100 g$, S.H.C. $=0.85 Jg^{-10} C^{-1}, \theta_R=(80-20)=60^{\circ} C$

Energy produced in wire $=I \times V \times t=I \times 10 \times 300 Vs$

Energy absorbed by oil $=m c \theta_R=100 \times 0.85 \times 60 J$

$\therefore$ By the law of conservation of energy $I \times 10 \times 300=100 \times 0.85 \times 60 \Rightarrow I=\frac{100 \times 0.85 \times 60}{10 \times 300}=1.7 A$

ILLUSTRATION-5.21

Two coils of resistance $3 \Omega$ and $6 \Omega$ are connected across battery of emf $12 V$. Find the electrical energy consumed in 1 minute in each resistance when they are connected in series. Draw the circuit diagram.

Show Answer

SOLUTION:

The circuit is shown in figure.

Total resistance of the circuit $R=\beta+6=9 \Omega$.

Current in the circuit

$ I=\frac{V}{R}=\frac{12}{9}=\frac{4}{3} A $

Since the resistance are in series, same current flows in each resistance.

Electrical energy spent in $R_1=R^{2} R_1 t=(\frac{4}{3})^{2} \times 3 \times 60=320 J$

Electrical energy spent in $R_2=I 2 R_2 t=(\frac{4}{3})^{2} \times 6 \times 60=640 J$

MISCELLANEOUS

SOLVED EXAMPLES

1. A bulb has voltage rating of $220 V$ and power rating of $40 W$. How can this bulb be made to glow with normal brightness if a voltage source of e.m.f. $330 V$ is available?

Show Answer

Solution:

Here $V=220 V ; P=40 W$

Resistance of the bulb, $R=\frac{V^{2}}{P}=\frac{(220)^{2}}{40}=1210 \Omega$

Current in the circuit, $I=\frac{330}{1210+S}$

Potential difference across the bulb,

$$ I R=\left(\frac{330}{1210+S}\right) 1210 $$

As per question, $\quad 220=\frac{330 \times 1210}{1210+S}$

On solving, we get $S=605 \Omega$

2. If two bulbs of wattage 25 and $30 W$, each rated at 220 volts are connected in series with a 440 volt supply, which bulb will fuse?

Show Answer

Solution: Resistance of $25 W$ bulb, $R_I=\frac{(220)^{2}}{25} \Omega$;

$ \text{ Current } I_1=\frac{25}{220} A $

Resistance of $30 W$ bulb, $R_2=\frac{(220)^{2}}{30} \Omega$,

$ \text{ Current } I_2=\frac{30}{220} A $

When bulbs are connected in series, effective resistance is

$R=R_1+R_2=(220)^{2}[\frac{1}{25}+\frac{1}{30}]=220 \times 220 \times \frac{11}{150} \Omega$

When supply voltage is $440 V$, then current is

$I^{\prime}=\frac{440}{R}=\frac{440 \times 150}{220 \times 220 \times 11}=\frac{27.27}{220} A$

As $I^{\prime}>I_1$ but less than $I_2$, hence the bulb of 25 watt will fuse.

3. Fig. shows three similar lamps $L_1, L_2$ and $L_3$ connected across a power supply.

If the lamp $L_3$ fuse, how will the light emitted by $L_1$ and $L_2$ change?

Show Answer

Solution: Let $R$ be the resistance of each lamp. If $E$ be the applied e.m.f., then the current in the circuit, $I_1$, is given by

$ I_1=\frac{E}{R+(R / 2)}=\frac{2 E}{3 R} $

Current flowing through $L_2$ or $L_3=\frac{1}{2}[\frac{2 E}{3 R}]=\frac{E}{3 R}$

When $L_3$ is fused, the whole current flows through $L_1$ and $L_2$.

Thus, $\quad I_2=\frac{E}{R+R}=\frac{E}{2 R}$

So, current through $L_1$ decreases and through $L_2$ increases.

4. 1kW heat is to be used with $220 V$ d.c. supply. (i) What is the current in the heater? (ii) What is its resistance? (iii) What is the power dissipated in the heater (iv) How much heat in calories is produced per second? (v) How many grams of water at $100^{\circ} C$ will be converted per minute into steam at $100^{\circ} C$ with the heater?

Assume that heat loses due to radiation are negligible. Latent heat of steam $=540 cal g^{-1}$.

Show Answer

Solution: (i) $I=P / V=1000 / 220=4.5 A$

(ii) $R=V^{2} / P=(220)^{2} / 1000=48.4 \Omega$

(iii) $P=1 kW=1000 W$

(iv) $H=1000 / 4.2=238.1 cal s^{-1}$

(v) Let $m$ gram water be converted into steam at $100^{\circ} C$ in one minute. Then

$ m \times 540=\frac{1000 \times 60}{4.2} \text{ or } m=\frac{1000 \times 60}{4.2 \times 540}=26.46 g . $

5. A fuse wire with a circular cross section and having diameter of 0.4 mm blows with a current of 3 amp. The value of the current for which another fuse wire made of the same material but having circular cross-section with diameter of 0.06 mm will blow is

Show Answer

Solution: For a fuse wire $I^{2} \propto r^{3}$

$ \therefore \quad \frac{I_1^{2}}{I_2^{2}}=\frac{r_1^{3}}{r_2^{3}} $

where $r_1=\frac{0.04}{2}=0.02 cm$ and $r_2=\frac{0.06}{2}=0.03 cm$.

$\therefore \frac{(3)^{2}}{I_2^{2}}=(\frac{0.02}{0.03})^{3}=(\frac{2}{3})^{3}$

or $\quad I_2^{2}=(3)^{2}(\frac{3}{2})^{3}$ or $I_2=3 \times(\frac{3}{2})^{3 / 2}$

6. An accumulator is connected first to an external resistance $R_1$ and then to another external resistance $R_2$ for the same time. At what value of the internal resistance of the accumulator will the amount of heat dissipated in the external resistances be the same in the two cases?

Show Answer

Solution: When an accumulator of emf $E$ and internal resistance $r$ is connected across a load resistance $R$, the heat dissipated in the external circuit

$ H=I^{2} R t=\frac{E^{2} R t}{(R+r)^{2}} \quad[$ as $I=\frac{E}{(R+r)}]$

According to given problem :

$\frac{E^{2} R_1 t}{(R_1+r)^{2}}=\frac{E^{2} R_2 t}{(R_2+r)^{2}}$ i.e., $(R_2-R_1)(r^{2}-R_1 R_2)=0$

And as $R_2 \neq R_1$ (given) $\quad r^{2}-R_1 R_2=0$

7. How will you connect 24 cells each of internal resistance $1 \Omega$ so as to get maximum power output across a load of $10 \Omega$ ?

Show Answer

Solution: For maximum power output $\frac{R}{n}=\frac{r}{m}$

and $n \times m=p=24$

So $R=\frac{n}{m} r=\frac{p r}{m^{2}}$

or $m^{2}=\frac{p r}{R}=\frac{24 \times 1}{10}$ or $m=\sqrt{2.4}=1.56$

(a) If $m=1$ then $n=\frac{p}{m}=24$

$ P_1=\frac{R E^{2}}{(\frac{R}{n}+\frac{r}{m})^{2}}=\frac{10 E^{2}}{(\frac{10}{24}+1)^{2}}=5 E^{2} $

(b) If $m=2$ then $n=\frac{p}{m}=\frac{24}{2}=12$

$ P_2=\frac{10 E^{2}}{(\frac{10}{12}+\frac{1}{2})^{2}}=5.6 E^{2} $

So to get maximum output power cells must be arranged in two rows having 12 cells in each row.

8. If electrochemical equivalent of hydrogen is $Z_H$ kg/coul-equivalent and chemical equivalent of copper is W, then the electrochemical equivalent of copper is

Show Answer

Solution: $\frac{Z _{C u}}{Z_H}=\frac{W}{W_H}$

We know that, $W_H=1$

$\therefore \quad Z _{C u}=W . Z_H$

9. Find the mass of silver liberated in a silver voltameter carrying a current of $1.5 A$, during 15 minutes. The electro chemical equivalent of silver is $1.12 \times 10^{-6} kg / C$.

Show Answer

Solution: Here, $m=?, i=1.5 A, Z=1.12 \times 10^{-6} kg / c$

and $t=15 \times 30 s$

Using $m=$ Zit yields

mass of silver liberated is

$m=(1.12 \times 10^{-6} \frac{kg}{c})(\frac{1.5 c}{s})(450 s)=7.6 \times 10^{-4} kg=0.76 g$

10. In a water voltameter, the act of passing a certain amount of current for a certain time produces of $1.2 g$ of $H_2$ at STP. Find the amount of $O_2$ liberated during that period.

Show Answer

Solution: Since the same current flows through both the electrodes of a water voltameter, so the amount of oxygen and hydrogen liberated (for the same charge) will be in direct proportional to their respective equivalent weights by Faraday’s second law of electrolysis,

i.e., $\frac{m_0}{m_H}=\frac{8}{1} \quad$ or $\quad m_0=\left(\frac{8}{1}\right) m_H=8 \times 1.2 \mathrm{~g}=9.6 \mathrm{~g}$

1 EXERCISE

Fill in the Blanks

DIRECTIONS : Complete the following statements with an appropriate word / term to be filled in the blank space(s).

1. Kilowatt is the unit of electrical …………… but kilowatt-hour is the unit of electrical ……………

Show Answer

Answer: power, energy

2. Energy spent in kilowatt-hour

$$=\frac{\text{ volt } \times \ldots \ldots \times \ldots…}{1000}$$

Show Answer

Answer: ampere, hour

3. A fuse is a short piece of wire of high …………… and low ……………

Show Answer

Answer: resistance, melting point

4. Fuse wire has a …………… melting point and is made of an alloy of …………… and …………… If the current in a circuit rises too high, the fuse wire ……………

Show Answer

Answer: low, lead, tin, melts

5. A fuse is connected in …………… to the …………… wire.

Show Answer

Answer: series, live

6. Electric energy is produced by the …………… of charges.

Show Answer

Answer: separation

7. Energy converted per unit charge is measured with an instrument called a ($n$) ……………

Show Answer

Answer: voltmeter

8. The electrinal energy dissipated in a resistor is given by $W=$ ……………

Show Answer

Answer: $V \times I \times t$

9. The unit of power is ……………

Show Answer

Answer: watt (W)

10. One watt of power is consumed when $1 A$ of current flows at a potential difference of ……………

Show Answer

Answer: 1V

11. 1 kWh = ……………

Show Answer

Answer: 3,600,000 J

12. The alloy which is used for making the filament of bulbs is ……………

Show Answer

Answer: Tungsten

13. Power transmission is carried out at high …………… and low ……………

Show Answer

Answer: voltage, current

14. Rate at which electric work is done is called ……………

Show Answer

Answer: electric power

15. The process of depositing a thin layer of desired metal over another metal by passing an electric current through some electrolyte is called ……………

Show Answer

Answer: electroplating

16. For a given cell, its terminal voltage depends on …………… and ……………

Show Answer

Answer: external resistance, internal resistance

True / False

DIRECTIONS : Read the following statements and write your answer as true or false.

1. The filament resistance of glowing bulb is greater, to its resistance when it is not glowing?

Show Answer

Answer: True

2. The commercial unit of electrical energy is kilowatt-hour (kWh).

Show Answer

Answer: True

3. Pure tungsten has high resistivity and a high melting point (nearly $3000^{\circ} C$).

Show Answer

Answer: True

4. When a metallic conductor is heated the atoms in the metal vibrate with greater amplitude and frequency.

Show Answer

Answer: True

5. The used up chemicals in a dry cell can be reobtained by charging it.

Show Answer

Answer: False

6. In an electrolyte, the movement of ions is responsible for electric current.

Show Answer

Answer: True

7. Electrotyping is an application of electrolysis.

Show Answer

Answer: True

8. One kilowatt is equal to 10 horse power.

Show Answer

Answer: False

9. The mass of the element deposited at the electrode when one coulomb of charge is passed through the electrolyte is called electrochemical equivalent of the element.

Show Answer

Answer: True

10. The terminal voltage of a cell in open circuit condition is less than its emf

Show Answer

Answer: False

11. Fuse should be connected to phase wire of the circuit.

Show Answer

Answer: True

12. Fuse is a thin wire which melts and breaks the electric circuit due to only high voltage.

Show Answer

Answer: False

13. The difference between EMF (E) of the cell and terminal voltage $(V)$ of a cell is called lost voltage.

Show Answer

Answer: True

14. The filament of bulb offers high resistance to electric current.

Show Answer

Answer: True

Match the Following

DIRECTIONS : Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in column I have to be matched with statements (p, q, r, s) in column II.

1. Column II gives name of material use for device given in column I

	Column I		Column II
(A)	Resistance of resistance box	(p)	tungsten
(B)	Fuse wire	(q)	maganin
(C)	Bulb	(r)	tin-lead alloy
(D)	Insulator	(s)	glass

Show Answer

Answer: $(A) \to q ;(B) \to r ; \quad(C) \to p ; \quad(D) \to s$

2. For the circuit shown in the adjoining figure, match the entries of column I with the entries of column II.

	Column I		Column II
(A)	Potential difference across battery A	(p)	A
(B)	Potential difference across battery B	(q)	B
(C)	Power is supplied by battery	(r)	14 V
(D)	Power is consumed by battery	(s)	None

Show Answer

Answer: $(A) \to r ;(B) \to s ; \quad(C) \to p ; \quad(D) \to q$

3. For a potentiometer circuit match the column

	Column I		Column II
(A)	Sensitivity	(p)	increases with length of wire
(B)	Accuracy	(q)	decreases with length of wire
(C)	Range	(r)	increases with emf of primary cell
(D)	Potential gradient	(s)	maximum if internal resistance & rheostat resistance is zero.

Show Answer

Answer: $(A) \to q ;(B) \to p ; \quad(C) \to r,s ; \quad(D) \to r,s$

Very Short Answer Questions:

DIRECTIONS: Give answer in one word or one sentence.

1. What is a fuse and specify its uses?

Show Answer

Answer: Fuse is a thin wire which melts and breaks the electric circuit due to (i) overloading (ii) short circuit (iii) fluctuation in power supply.

2. Find the total internal resistance of the cells when they are connected in

$\quad$ (a) series and $\quad$ (b) parallel connections

Show Answer

Answer: (a) In series connection, total internal resistance ($r$) is equal to the sum of the internal resistances of the individual cells; $r=r_1+r_2+r_3$

(b) In parallel connection, total internal resistance is equal to the sum of the reciprocals of individual internal resistances.

3. What is the use of kilowatt-hour meter?

Show Answer

Answer: Gives the reading of energy consumed.

4. What is the value of mechanical equivalent of heat (J)?

Show Answer

Answer: $J=\frac{W}{Q}=4.18 J cal^{-1}$

5. Define terminal voltage, internal resistance and lost voltage.

Show Answer

Answer: (i) The P.D. between two terminals of the cell, when current is drawn from it, i.e., the cell is in closed circuit, is called its terminal voltage (V).

(ii) The electrolyte offers certain resistance to the flow of current. This is called the internal resistance of cell $(r)$.

(iii) The difference between $EMF(E)$ of the cell and terminal voltage $(V)$ of a cell is called lost voltage.

6. What is watt-hour?

Show Answer

Answer: Watt-hour is a unit in which the amount of electric energy consumed by a device is measured.

7. Define electric energy or work.

Show Answer

Answer: The energy spent or work done an moving on electric charge through conductor is called electric energy or electric work

8. Define electric power and give its units in S.I.

Show Answer

Answer: The rate at which electric work is done is called electric power. In S.I., power $=$ watt $=1 Js^{-1}$.

9. What is electro plating?

Show Answer

Answer: The process of depositing a thin layer of desired metal over another metal by passing an electric current through some electrolyte is called electro plating.

10. How is the electric current passing though a metallic wire produces heat?

Show Answer

Answer: Colliding electrons lose their K.E. as heat.

11. What is the operating resistance of a 100-watt household light bulb? The operating line voltage of household electricity is 120V.

Show Answer

Answer: Given : $P=100 W, V=120 V$, Find R.

From $P=\frac{V^{2}}{R}$, We have $R=\frac{V^{2}}{P}=\frac{(120 V)^{2}}{100 W}=144 \Omega$

12. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal ?

Show Answer

Answer: Coils of electric toasters and electric irons are made of an alloy rather than a pure metal because (i) the resistivity of an alloy is much higher than that of pure metal, and (ii) an alloy does not undergo oxidation easily even at high temperature.

13. Why does the cord of an electric heater not glow while the heating element does ?

Show Answer

Answer: The cord of the electric heater made of copper does not glow because negligible heat is produced in it by passing current. It is due to its extremely low resistance.

14. How does the amount of current in a household circuit differ from the amount of current in a reading lamp?

Show Answer

Answer: The amount of current in a household circuit is more than the amount of current in a reading lamp as the current of the circuit is divided among all the appliances in a parallel household circuit.

15. Would you expect to find DC or AC in the filament of a light bulb in your home? How about in the headlight of an automobile?

Show Answer

Answer: There is $A C$ flowing in the filament of the light bulb in our homes but DC flows in the bulbs of automobiles.

16. A 220 volt 100 watts bulb is connected to 110 volts source, calculate the power consumed by the bulb.

Show Answer

Answer: Resistance of the bulb

$R=\frac{V}{I}=\frac{V^{2}}{I V}=\frac{V^{2}}{W}$ or $R=\frac{220 \times 220}{100}=484 ohm$.

Power consumed when connected to 110 volt source.

$=\frac{V^{2}}{R}=\frac{110 \times 110}{484}=25$ watt.

17. A current of 0.5 A is passing through a $CuSO_4$, solution. How many $Cu^{++}$ ions will be deposited on cathode in 10 second?

Show Answer

Answer: If $n$ is the number of copper ions then $q=n \times 2 e$

$q=It \quad$ or $n \times 2 e=It \quad n=\frac{It}{2 e}=\frac{0.5 \times 10}{2 \times 1.6 \times 10^{-19}}$ $=1.5625 \times 10^{19}$.

18. An electric bulb is connected to a 220V generator. The current is 0.50V. What is the power of the bulb?

Show Answer

Answer: $P=VI=220 V \times 0.50 A=110 J / s=110 W$

19. What uses more energy, a 250 W TV set in 1 hr. or a 1200W toaster in 10 minutes?

Show Answer

Answer: $250 W$ TV set in 1 hour.

20. Name two devices which use the heating effect of current.

21. Why is the filament of an electric bulb not made of carbon?

22. 60W-220V is written on a bulb. What does it mean?

23. Name the term used to represent the values of the voltage and wattage (power) of an electrical appliance taken together.

Short Answer Questions

DIRECTIONS: Give answer in 2-3 sentences.

1. Obtain an expression for the (i) electrical energy (ii) electrical power, spent in flow of current through a conductor.

2. A 60W auto lamp allows 5 amps to pass through it. Find

(i) The p.d. across its terminal.

(ii) the resistance of the filament of the lamp

(iii) energy consumed in 2 hours

Show Answer

Answer: (i) Power, P = V $\times$ I

$60=V \times I$

$ V=\frac{60}{5}=12 \text{ volts } $

(ii) Resistance $=R=\frac{V}{I}=\frac{12}{5}=2.4 \Omega$

(iii) Energy consumed in 2 hours

$ =\frac{\text{ watts }}{1000} \times hr=\frac{60}{1000} \times 2=0.12 kWh $

3. What is the electric power? Derive a formula for it?

4. An electric iron consumes energy at a rate of 840W when heating is at the maximum rate and 360W when the heating is at the minimum. The voltage is 220V. What are the current and the resistance in each case?

Show Answer

Answer: We know that the power input is, $P=V I$

Thus the current $I=P / V$

(a) When heating is at the maximum rate,

$I=840 W / 220 V=3.82 A$

and the resistance of the electric iron is

$R=V / I=220 V / 3.82 A=57.60 \Omega$

(b) When heating is at the minimum rate,

$I=360 W / 220 V=1.64 A$

and the resistance of the electric iron is

$R=V / I=220 V / 1.64 A=134.15 \Omega$.

5. A geyser is rated 1500W, 250V. This geyser is connected to 250V mains. Calculate: (i) the current drawn, (ii) the energy consumed in 50 hours. (iii) the cost of energy consumed at ‘220 per kWh.

Show Answer

Answer: $\begin{matrix} \text{(i) } 6 A & \text{ (ii) } 75 kWh & \text{ (iii) } 165\end{matrix} $

6. An electric iron of resistance $20\Omega$ takes a current of 5A. Calculate the heat developed in 30s.

Show Answer

Answer: Here, $R=20 \Omega, I=5 A, t=30 s$

Heat developed $=I^{2} R t=5 \times 5 \times 20 \times 30=15,000 J=15 kJ$

7. Compare the power used in the $2\Omega$ resistor in each of the following circuits:

(i) a 6V battery in series with $1\Omega$ and $2\Omega$ resistors and (ii) a 4V battery in parallel with $12\Omega$ and $2\Omega$ resistors.

Show Answer

Answer: (i) Equivalent resistance of $1 \Omega$ and $2 \Omega$ in series,

$R=1 \Omega+1 \Omega=3 \Omega$

Potential difference, $V=6 V$, Current, $I=\frac{V}{R}=\frac{6}{3}=2 A$

Current in series circuit is same.

$\therefore$ Current in $2 \Omega$ resistor $=2 A$

Power in $2 \Omega$ resistor, $P=I^{2} R=2^{2} \times 2=8 W$

(ii) Potential difference across $2 \Omega$ resistor $=4 V$

Power, $P^{\prime}=\frac{V^{2}}{R}=\frac{4^{2}}{2}=8 W$

$P: P^{\prime}=8 W: 8 W=1: 1$

8. An electric heater of resistance $8\Omega$ draws 15A from the service mains 2 hours. Calculate the rate at which heat is developed in the heater.

Show Answer

Answer: Here, $R=8 \Omega, \quad I=15 A, t=2 h=2 \times 60 \times 60 s$

Heat developed, $H=I^{2} R t=(15)^{2} \times 8 \times 2 \times 60 \times 60 J$

The rate at which heat is developed is power.

$ \begin{aligned} \text{ Power } & =\frac{\text{ Heat developed }(I^{2} Rt)}{\text{ Time taken }(t)} \\ & =\frac{15 \times 15 \times 8 \times 2 \times 60 \times 60}{2 \times 60 \times 60}=1800 J / s \end{aligned} $

9. 100J of heat are produced each second in a $4\Omega$ resistance. Find the potential difference across the resistor.

Show Answer

Answer: $H=100 J, R=4, t=1 s, V=$ ?

We have the current through the resistor as

$ I=\sqrt{H / Rt}=\sqrt{[100 J /(4 \Omega \times 1 s)]}=5 A $

Thus the potential difference across the resistor $V$ is given by $V=I R=5 A \times 4 \Omega=20 V$

10. In the home electrical energy is converted into other energy forms. What is the energy conversion when

(a) an electric bell is used?

(b) an electric toaster is used ?

(d) an electric fan is used?

Show Answer

Answer: (a) electrical $\rarr$ sound

(b) electrical sound

(d) electrical $\rarr$mechanical energy

11. An immersion heater having $50 \Omega$ resistance is connected to $250 V$ mains. Find the time required to heat 2 kg of water from $34^{\circ} C$ to $100^{\circ} C$, assuming there is no loss of heat energy (Sp. heat cap. of water $=4 \times 10^{3} J / g^{\circ} C$ ).

Show Answer

Answer: Current passing through the heater

$ I=\frac{V}{R}=\frac{250}{50}=5 A $

Heat produced $=I^{2} R t$ and Heat required $=m C \theta$.

Heat produced $=5^{2} \times 50 \times t$

$ m C \theta=2 \times 4 \times 10^{3} \times(100-35)=8 \times 10^{3} \times 65 $

Since, $\quad 5^{2} \times 50 \times t=8 \times 10^{3} \times 65$

$\therefore \quad t=\frac{520 \times 1000}{1250}=416 s=6 min 56 s$

12. Two bulbs are marked $60 W, 220 V$ and $100 W, 220 V$. These are connected in parallel to $220 V$ main. Which one of the two will glow brighter?

Show Answer

Answer: $100 W$ bulb as $P \propto \frac{1}{R}, i \propto R(V$ is constant $)$

13. A toaster produces more heat than a light bulb when connected in a parallel to the $220 V$ mains. Which of the has greater resistance?

Show Answer

Answer: Since, $R \propto \frac{I}{P}(=\frac{V^{2}}{P})$ toaster will have low resistance and bulb will have high resistance

14. An electric heater is used everyday for 120 minutes. The electricity bill for 30 days is 60 units. Calculate the power of the electric heater.

Show Answer

Answer: Energy consumed in 30 days $=60$ units

(i.e), $\frac{\text{ days } \times \text{ watt } \times \text{ hours }}{1000}=60$

$\Rightarrow \frac{30 \times P \times 2}{1000}=60 ; P=1000$ watt.

15. In a household 5 tubelights of $40 W$ each are used for 5 hours and an electric press of $500 W$ for 4 hours everyday. Calculate the total electrical energy consumed by the tubelights and press in a month of 30 days.

Show Answer

Answer: Energy consumed by five tubelights of $40 W$ for 5 hours in 1 day $=\frac{40}{1000} \times 5=1 kWh$

Energy consumed by an electric press of $500 W$ for 4 hours in 1 day $=\frac{500 \times 4}{1000}=2 kWh$

Energy consumed in 1 day $=1+2=3 kwh$

$\therefore$ Total electrical energy consumed in 30 days

$ =3 \times 30=90 kWh=90 \text{ units } $

16. A $100 W$ electrical bulb is lighted for 2 hours everyday and five $40 W$ tubes are lighted for 4 hours everyday. Calculate

(i) The energy consumed for 60 days and

(ii) Cost of electricity consumed at the rate of ’ $1.50 per$ $kWh$.

Show Answer

Answer: (i) Energy consumed in days

$ \begin{aligned} & =100 \times 2 \times 60+5 \times 40 \times 4 \times 60 \\ & =12000+48000=60000 wh=60 kWh \end{aligned} $

(ii) Cost of electricity $=₹ 1.50 \times 60=₹ 90$

17. What is meant by rating of electrical appliance?

Show Answer

Answer: It is the product of the voltage and the current which can flow through the device.

18. What are the factors on which heat dissipated by a conductor depends?

Show Answer

Answer: Heat dissipated depends upon (i) the square of current through the conductor (ii) the resistance of the conductor and (iii) the time for which current is passed.

19. Why is heat produced when a current is passed through a conductor?

Show Answer

Answer: During their motion the electrons collide with one another and hence lose some kinetic energy. This loss in kinetic energy is dissipated as heat across the conductor.

20. Define one watt.

Show Answer

Answer: The power expended by a source is said to be I watt if one ampere of current flows through it under a potential difference of 1 volt.

21. Why should the melting point of a fuse wire be low?

Show Answer

Answer: So that it melts and breaks the circuit as soon as the safe limit of current is exceeded.

22. The filament of electric bulb is made of tungsten. Why?

Show Answer

Answer: Because it has a high melting point and a high resistivity and does not easily react with air.

Long Answer Questions :

DIRECTIONS: Give answer in four to five sentences.

1. What are the safety measures to be taken while dealing with household electricity?

Show Answer

Answer: It goes beyond doubt that electricity plays a very important role in our lives. It is one of the most important and convenient sources of energy at homes and industries. However, we daily come across reports regarding hazards of electricity. Use of electricity can prove to be very dangerous if certain precautions and safety measures are not observed in the design of electrical devices and in handling them. The various safety measures to be taken are:

(i) Use wires of high quality, proper amperage and good insulating material.

(ii) Cover all naked wires and joints with insulating tape.

(iii) All connections at plugs, switches, sockets must be tight.

(iv) Replace any defective plugs, switches and sockets.

(v) Never touch any part of the circuit without putting on rubber shoes or rubber gloves.

(vi) Use fuse (now a days MCB) of proper rating and material. (vii) All electrical appliances must be properly earthed.

(viii) Connect switches and fuse to live wires.

(ix) Put out the main switch in case of short circuiting and fire.

(x) Do not use water as fire extinguishers in case of fire due to electricity.

2. (i) What do you understand by earthing?

(ii) What are the advantages of earthing in a household electric circuit?

(iii) Explain how it is done.

Show Answer

Answer: (i) By earthing we mean that the metallic body of an electric appliance is connected to thick, copper wire, which is buried deep in the earth and at its end is a copper plate surrounded by a mixture of charcoal and common salt.

(ii) It is a kind of safety device which saves us from an electric shock, in case when the metal casing of the appliance happens to touch the live wire or due to short circuiting or leakage of electric current.

Whenever an appliance which is earthed, gets short circuited, the current from the metal casing of the appliance flows into the earth which acts as an electric sink, i.e., its potential always remains zero. Due to the flow of heavy current the fuse in that circuit melts and disconnects the appliance from the circuit. So the user who happens to touch the appliance is protected from receiving any electric shock. Another advantage is that due to overheating the house wiring system is saved from being damaged and same time from being burnt out.

(iii) A three core, cord having three wires coated with insulation of red, brown and green colour, in used for connecting the appliance to the mains for drawing current from the mains. At one end of the cord, red is connected to the pin marked L (live), the brown to the pin marked $N$ (nentral) and green to thick pin of the plug. The three wires at the other end of the cord are connected to the appliance such that live and neutral wires are connected to the element and the earth wire is connected to the metal body of the appliance so as to earth it. Once the plug is put in the socket, the current through the appliance passes as soon as the switch is pressed. The live wire gets connected to the live wire of the mains. Neutral wire is connected to the neutral of the mains and the earth wire gets connected to the earth in the mains.

3. Three 250 watt heaters are connected in parallel to a 100 volt supply. Calculate :

(i) the total current taken from the supply,

(ii) the resistance of each heater.

(iii) the energy supplied in $kWh$ to the three heaters in 5 hours.

Show Answer

Answer: Given, power of one heater (P) = 250 watt, potential (V) = 100 volt, time (t) = 5 hours.

$(i) P=\frac{V^2}{R} ; R=\frac{V^2}{P}=\frac{100 \times 100}{250}=40 \Omega$

The three 250 watt heaters are connected in parallel then total current, $I = I_{1} + I_{2} + I_{3}$

$=\frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}=\frac{100}{40}+\frac{100}{40}+\frac{100}{40}$

= 2.5 + 2.5 + 2.5 = 7.5 A

(ii) The resistance of each heater is $40 \Omega$

(iii) The energy supplied to the first heater is

$H = I^2 \times Rt = (2.5)^2 \times 40 \times 5$ Wh, As current flowing each heater is 2.5 A

$\therefore $ H = 6.25 $\times$ 200 Wh = 1250 Wh = $\frac{1250}{1000}$ kWh = 1.25 kWh

Then three heaters in connection = 3 $\times$ 1.25 kWh

= 3.75 kWh energy is used in the circuit.

4. A heating coil is immersed in a calorimeter of heat capacity $50 J^{\circ} C^{-1}$ containing $1.0 kg$ of a liquid of specific heat capacity $450 J kg^{-1}{ }^{\circ} C^{-1}$. The temperature of liquid rises by $10^{\circ} C$ when $2.0 A$ current is passed for 10 minutes.

Find (i) the resistance of the coil (ii) the potential difference across the coil. State the assumption used in your calculations.

Show Answer

Answer: (i) 2.08 $\Omega$ (ii) 4.16 V

5. State Joule’s law. Explain an experimental method to determine the value of $J$.

Show Answer

Answer: The same work done in different forms produces the same quantity of heat in all cases.

W = JQ

$\mathrm{J}=\frac{\mathrm{W}}{\mathrm{Q}}=4.18 \mathrm{~J} \mathrm{cal}^{-1}$

6. Sate Farady’s first law and second law of electrolysis and verify it experimentally.

Show Answer

Answer: 1st Law The mass of the substance deposited at the cathode is directly proportional to the quantity of electricity.

$m \propto I$

$m \propto$ It

$m \propto s t$

$m=Zit=ZQ$

$m_1: m_2: m_3=z_1: z_2: z_3$

2nd Law

The same quantity of electricity passes through different electrolytes, the masses of ions liberated at the respective electrodes are proportional to the chemical equivalents.

$m_1: m_2: m_3=E_1: E_2: E_3$

$E_1: E_2: E_3=z_1: z_2: z_3$

7. Explain the working of the following

(i) Filament lamp

(ii) Flourescent lamp

(iii) Discharge lamp

(iv) Electric are lamp

(v) Drycells

Show Answer

Answer: (i) Filament offers high resistance to electric current.

Electric energy is converted into light and heat energy.

(ii) Discharge lamp is coated with fluorescent material from inside.

High voltages are applied to coils, heat them which results in emission of electrons. Collision of electrons

(iii) Glass mercury vapours produces lights.

high voltage is two electrodes fused to its. When ionized. Accelerated ions collide with neutral gas atoms that produce glow.

(iv) Produces high intensity light.

Carbon rods having pointed tips at one end and separated by small gap.

When $40 V$ to $60 V$ is applied across the carbon rods, glow is produced.

(v) Form of a primary cell.

Consists of zinc container (cathode) at the centre of which a carbon rod (anode) is fixed. Dry paste of $NH_4 Cl+ZnCl_2$ serves as an electrolyte.

Electrolyte is mixed with powdered coke.

$NH_4 Cl \to NH_4^{+}+Cl$

$2 NH_4{ }^{+}+2 Cl^{-} \to 2 NH_3+H_2$

$H_2+2 MnO_4 \to H_2 O+MNO_3$

$Zn^{2+}+2 Cl^{-} \to ZnCl_2$

Here, chemical energy is converted in electrical energy.

2 EXERCISE

Multiple Choice Questions:

DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

1. Two electric lamps each of 100 watts $220 V$ are connected in series to a supply of 220 volts. The power consumed would be -

(a) 100 watts

(b) 200 watts

(d) 50 watts

Show Answer

Answer: (d)

2. An electric bulb is rated $220 V$ and $100 W$. When it is operated on $110 V$, the power consumed will be

(a) $100 W$

(b) $75 W$

(d) $25 W$

Show Answer

Answer: (d)

3. If it takes 8 minutes to boil a quantity of water electrically, how long will it take to boil the same quantity of water using the same heating coil but with the current doubled

(a) 32 minutes

(b) 16 minutes

(d) 2 minutes

Show Answer

Answer: (d)

4. Kilowatt-hour is the unit of

(a) potential difference

(b) electric power

(d) charge

Show Answer

Answer: (c)

5. An electric bulb is filled with

(a) hydrogen

(b) oxygen and hydrogen

(d) nitrogen and argon

Show Answer

Answer: (d)

6. When current is passed thought and electric bulb, its filament glows, but the wire leading current to the bulb does not glow because

(a) less current flows in the leading wire as compared to that in the filament

(b) the leading wire has more resistance than the filament

(d) filament has coating of fluorescent material over it

Show Answer

Answer: (c)

7. From a power station, the power is transmitted at a very high voltage because -

(a) it is generated only at high voltage

(b) it is cheaper to produce electricity at high voltage

(d) there is less loss of energy in transmission at high voltage

Show Answer

Answer: (b)

8. When a fuse is rated $8 A$, it means

(a) it will not work if current is less than 8A

(b) it has a resistance of 8 ohm

(d) it will burn if current exceeds 8A

Show Answer

Answer: (a)

9. Fuse wire is made of

(a) platinum

(b) copper

(d) alloy in tin and lead

Show Answer

Answer: (c)

10. Which is not a device based on the heating effect of electricity

(a) heater

(b) toaster

(d) press

Show Answer

Answer: (b)

11. Which of the following terms does not represent electrical power in a circuit?

(a) $I^{2} R$

(b) $I R^{2}$

(d) $V^{2} / R$

Show Answer

Answer: (d)

12. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be

(a) $1: 2$

(b) $2: 1$

(d) $4: 1$

Show Answer

Answer: (d)

13. On which one of the following the emf of a cell does not depend

(a) The nature of the metal of electrodes

(b) The size of the plates

(d) The nature of electrodes

Show Answer

Answer: (b)

14. The primary cell which is used in daily life is -

(a) Leclanche cell

(b) Dry cell

(d) Simple voitaic cell

Show Answer

Answer: (c)

15. Which one of the following primary cells has emf 1.08 volts and which remains fairly constant?

(a) Daniel cell

(b) Simple voltaic cell

(d) Dry cell

Show Answer

Answer: (d)

16. Primary cell are connected in parallel to

(a) Increase voltage

(b) decrease capacity

(d) make electric current constant

Show Answer

Answer: (d)

17. In a closed circuit drawing current from cell, the emf of a cell is always

(a) Less than potential difference

(b) More than potential difference

(d) Double of the potential difference

Show Answer

Answer: (b)

18. The filament of an electric bulb is of tungsten because

(a) Its resistance is negligible

(b) It is cheaper

(d) Filament is easily made

Show Answer

Answer: (c)

19. When the current passes through the filament, it gets heated to incandescence and give light while the connecting wires are not heated because

(a) The connecting wires are good conductor of heat while the filament is bad conductor

(b) The connecting wires are of low resistance while the filament is of high resistance

(d) The connecting wires are bad conductor of heat while the filament is good conductor

Show Answer

Answer: (d)

20. The bulbs which emit a bluish light, are

(a) Filled with argon

(b) Filled with nitrogen

(d) coated from inside with a light blue colour

Show Answer

Answer: (a)

21. Which one of the following heater element is used in electric press

(a) copper wire

(b) nichrome wire

(d) iron wire

Show Answer

Answer: (a)

22. If 10 cells each of 1.4 volts are connected in parallel, their equivalent potential difference will be

(a) 1.4 volts

(b) 0.14 volts

(d) 10/1.4 volts

Show Answer

Answer: (b)

23. What should be the characteristic of fuse wire?

(a) High melting point, high specific resistance

(b) Low melting point, low specific resistance

(d) Low melting point, high specific resistance

Show Answer

Answer: (d) Fuse wire should be such that it melts immediatley when strong current flows through the circuit. The same is possible if its melting point is low and resistivity is high.

24. The heating element of an electric heater should be made with a material, which should have

(a) high specific resistance and high melting point

(b) high specific resistance and low melting point

(d) low specific resistance and high melting point

Show Answer

Answer: (a) A heating wire should be such that it produces more heat when current is passed through it and also does not melt. It will be so if it has high specific resistance and high melting point.

25. Resistance of conductor is doubled keeping the potential difference across it constant. The rate of generation of heat will

(a) become one fourth

(b) be halved

(d) become four times

Show Answer

Answer: (b) The rate of generation of heat, for a given potential difference is, $P=V^{2} / R$

26. A current I passes through a wire of length $l$, radius $r$ and resistivity $r$. The rate of heat generated is

(a) $\frac{l^{2} \rho \ell}{r}$

(b) $\frac{l^{2} \rho \ell}{\pi r^{2}}$

(d) none of these

Show Answer

Answer: (b) The rate of heat generation $=I^{2} R=I^{2}(\rho \ell / \pi r^{2})$.

27. The resistance $R_1$ and $R_2$ are joined in parallel and a current is passed so that the amount of heat liberated is $H_1$ and $H_2$ respectively. The ratio $H_1 / H_2$ has the value

(a) $R_2 / R_1$

(b) $R_1 / R_2$

(d) $R_2^{2} / R_1^{2}$

Show Answer

Answer: (a) Heat produced, $H=V^{2} t / R$ i.e. $H \propto 1 / R$ so $H_1 / H_2=R_2 / R_1$.

28. Two electric bulbs rated $P_1$ watt $V$ volts and $P_2$ watt $V$ volts are connected in parallel and applied across $V$ volts. The total power (in watts) will be

(a) $P_1+P_2$

(b) $\sqrt{P_1 P_2}$

(d) $\frac{P_1+P_2}{P_1 P_2}$

Show Answer

Answer: (a) In parallel combination, total power $P=P_1+P_2$.

29. First electro chemical cell was designed by

(a) Leclanche

(b) Faraday

(d) Some one other than those mentioned above

Show Answer

Answer: (d) First electro-chemical cell was designed by Allexandro de-Volta.

30. The electric cell is a device to obtain

(a) electric charge

(b) electric force

(d) electric energy from chemical energy

Show Answer

Answer: (d)

31. In which of the following cells polarisation is the major defect?

(a) Voltaic cell

(b) Daniel cell

(d) Fuse cell

Show Answer

Answer: (a)

32. Which of the following is NOT the name of a secondary cell?

(a) Storage cell

(b) Fuel cell

(d) Acid cell

Show Answer

Answer: (b)

33. The current capacity of the charged secondary cell does not depend on

(a) rate of charging

(b) rate of discharging

(d) amount of active material

Show Answer

Answer: (c)

34. Which of the following is not the storage cell?

(a) Acid cell

(b) NIFE cell

(d) Dry cell

Show Answer

Answer: (d) Dry cell being primary cell, cannot be recharged.

35. In charging a battery of motor car, the following effect of electric current is used

(a) magnetic

(b) heating

(d) induction

Show Answer

Answer: (c) Chemical effect of current is used in charging a car battery.

36. Same current is being passed through a copper voltameter and silver voltameter. The rate of increase in weight of the cathodes in the two voltameters will be proportional to

(a) relative densities

(b) atomic masses

(d) none of the above

Show Answer

Answer: (d) The masses of the copper and silver deposited on cathodes of copper voltameter and silver voltameter are in the ratio of their chemical equivalent of copper and silver respectively.

37. If $E$ be chemical equivalent of an element and $Z$ is its electrochemical equivalent, then $E / Z$ is measured in

(a) farad

(b) newton

(d) faraday

Show Answer

Answer: (d)

38. A silver and a zinc voltameter are connected in series and a current $I$ is passed through them for a time $t$, liberating W grams of zinc. The weight of silver deposited is nearly

(a) 1.7 W grams

(b) 2.4 W grams

(d) 1.2 W grams

Show Answer

Answer: (c) $m_{A g} / m _{Z n}=E _{A g} / E _{Z n}=108 / 31$

or $m _{A g}=m _{Z n} \times 108 / 31=W \times 108 / 31$

39. What determines the e.m.f. between the two metals placed in an electrolyte?

(a) relative position of metals in the electro chemical series

(b) distance between them

(d) nature of electrolyte

Show Answer

Answer: (d)

40. For electroplating a spoon, it is placed in the voltameter at

(a) the position of anode

(b) the position of cathode

(d) anywhere in the electrolyte

Show Answer

Answer: (b) In electroplating, the metallic ions are positive, w/hich are deposited on cathode.

41. Electroplating does not help in

(a) fine finish to the surface

(b) shining appearance

(d) protect metals against corrosion

Show Answer

Answer: (c) Electroplating does not help in making the metals become hard.

42. Faraday’s laws are consequence of conservation of

(a) energy

(b) energy and magnetic field

(d) magnetic field

Show Answer

Answer: (a) Faraday’s laws are based on the conversion of electrical energy into mechanical energy; which is in accordance with the law of conservation of energy.

43. According to Faraday’s law of electrolysis, the amount of decomposition is proportional to

(a) 1/time for which current passes

(b) electrochemical equivalent of the substance

(d) 1/electrochemical equivalent

Show Answer

Answer: (b) The amount of decomposition (i.e., mass of the substance liberated during electrolysis) is proportional to electro chemical equivalent of the substance.

44. The current inside a copper voltameter

(a) is half the outside value

(b) is the same as the outside value

(d) depends on the concentration of $CuSO_4$

Show Answer

Answer: (b) Current inside a copper voltameter is same as that of outside.

45. In electrolysis, the amount of mass deposited or liberated at an electrode is directly proportional to

(a) amount of charge

(b) square of current

(d) square of electric charge

Show Answer

Answer: (a) $m=Z t=Z q$ where $I$ is the current and $q$ is the amount of charge.

46. Of the following, the one that does not make use of chemical effect of electric current is

(a) electroplating

(b) electrotyping

(d) voltaic cells

Show Answer

Answer: (c)

More Than One Correct

DIRECTIONS: This section contains multiple choice questions, Each question has 4 choices (a), (b), (c) and (d) out of which ONE OR MORE may be correct.

1. A constant voltage is applied between the two ends of a uniform metallic wire. If both the length and radius of wire are doubled, then:

(a) the heat developed in the wire will be doubled

(b) the electric field in the wire will be doubled

(d) the electric field in the wire will be halved

Show Answer

Answer: (a,d)

2. The emf of a cell is:

(a) the potential difference across its terminals

(b) the potential difference across its terminals when no current is passing through it

(d) the total work done per coulomb of electricity taken in a circuit in which the cell is connected

Show Answer

Answer: (b,d)

3. In the circuit shown in the figure:

(a) the voltage at the point $A$ is $6 V$

(b) the voltage at the point $D$ is $-0.5 V$

(d) if the voltmeter is connected across a $6 V$ battery, it will read $5 V$

Show Answer

Answer: (b,c)

4. When the terminals of a cell of e.m.f. $1.5 V$ are connected to an ammeter of resistance $4 \Omega$, the ammeter reads 0.30 amp.

Which of the following statements is/are correct?

(a) The cell is non-ideal

(b) If a $4 \Omega$ resistor is also connected across terminals of the cell, the ammeter will read 0.50 amp.

(c) If a $4 \Omega$ resistor is also connected across the terminals of the cell, one-third of the electrical powe generated will dissipate as heat within the cell.

(d) If a voltmeter of resistance $4 \Omega$ is used to measure potential difference between terminals of the cell, it will read $1.2 V$.

Show Answer

Answer: (a,c,d)

Potential difference across the accumulator $=0.3 \times 4=1.2 V$ $<1.5 V$; hence cell is non-ideal. Internal resistance of the cell

$=\frac{1.5-1.2}{0.3}=1.0 \Omega$.

5. Which of the following statements is/are correct?

(a) Potential difference between terminals of a non-ideal battery can never be greater than its e.m.f.

(b) If a non-ideal battery is short circuited by a wire, heat generater in the wire is less than electric energy developed in the battery.

(c) e.m.f. of an ideal battery is first measured by a potentiometer and then by a voltmeter. Both the measurements are equally correct.

(d) Ncne of the above.

Show Answer

Answer: (a, b, c, d)

In case of non-ideal battery, an amount of energy is dissipated in the form of heat due to its internal resistance. If the battery is ideal then there is no internal drop hence reading of potentiometer and voltmeter will be equally correct.

6. Two cells of unequal e.m.fs. $E_1$ and $E_2$ and internal resistances $r_1$ and $r_2$ are joined as shown. $V_A$ and $V_B$ are the potentials at $A$ and $B$ respectively:

(a) one cell will continuously supply energy to the other

(b) the potential difference across both the cells will be equal

(d) $V_A-V_B=\frac{(E_1 r_2+E_2 r_1)}{(r_1+r_2)}$

Show Answer

Answer: (a, b, c)

Let $E_1<E_2$.

Current in the circuit $=I=\frac{E_1-E_2}{r_1+r_2}$

$V_A-V_B=E_2+I r_2=P D$ across each cell

Hence, $\quad V_A-V_B>E_2$

Current flows in $E_2$ from the positive plate to the negative plate inside the cell and hence it absorbs energy.

7. Which of the following statement(s) is/are correct?

(a) If $n$ identical cells are connected in series and-then the battery thus formed is short circuited by a conduction ${ }^{-}$ wire, current through the wire will be independent of $n$.

(b) If $n$ identical cells are connected in parallel and then the battery thus formed is short circuited by conducting wire, current through the wire will be directly proportional to $n$.

(c) If $n$ identical cell are connected in parallel and then the battery thus formed is short circuited by a wire having a constant resistance, current through the wire will increase as $n$ increases.

(d) None of the above.

Show Answer

Answer: (a, b, d)

In series, $I=\frac{n E}{n r}=\frac{E}{r}$ (i.e., $I$ is independent of $n$)

In parallel, $I=\frac{E}{r / n}=\frac{n E}{r}$, i.e., $I \propto n$

After shorting by a wire having a constant resistance

$I=E(R+\frac{r}{n})$

8. For the battery shown in the figure:

(a) equivalent internal resistance $R$ is given by:

$\frac{1}{R}=(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3})$

(b) if $E_3=\frac{(E_1 R_2+E_2 R_1)}{(R_1+R_2)}$, equivalent e.m.f. of the battery will be equal to $E_3$

(d) equivalent e.m.f. of the battery not only depends upon values of $E_1, E_2$ and $E_3$ but depends upon values of $R_1, R_2$ and $R_3$ also

Show Answer

Answer: $(a, d)$

Equivalent e.m.f. of the battery is obtained by applying Kirchhoff’s 2nd law to the two loops.

$E_3=\frac{E_1 R_2+E_2 R_1}{R_1+R_2}$

9. A cell of e.m.f. $5 V$ and internal resistance $1 \Omega$ will give maximum power output to:

(a) a single resistor of $1 \Omega$

(b) two $1 \Omega$ resistors connected in series

(d) two $2 \Omega$ resistors connected in parallel

Show Answer

Answer: (a,d)

10. Two heaters designed for the same voltage $V$ have different power ratings. When connected individually across a source of voltage $V$, they produce $H$ amount of heat each in times $t_1$ and $t_2$ respectively. When used together across the same source, they produce $H$ amount of heat in time $t$ :

(a) if they are in series, $t=t_1+t_2$

(b) if they are in series, $t=2(t_1+t_2)$

(d) if they are in parallel, $t=\frac{t_1 t_2}{2(t_1+t_2)}$

Show Answer

Answer: (a,c)

Let $R_1+R_2$ be the resistances of the two heaters and $H$ be the heat produced.

$\therefore \quad H=\frac{V^2}{R_1} t_1=\frac{V^2}{R_2} t_2$

When used in series, $H=\frac{V^{2} t}{R_1+R_2}$

When used in parallel, $H=\left(\frac{V^2}{R_1}+\frac{V^2}{R_2}\right) \mathrm{t}$

11. In a household electric circuit:

(a) all electric appliances drawing power are joined in parallel

(b) a switch may be either in series or in parallel with the appliance which it controls

(d) if switch is in parallel with an appliance, the fuse will blow (burn out) when the switch is put on (closed)

Show Answer

Answer: (a, c, d)

12. A cell drives a current through a circuit. The e.m.f. of the cell is equal to the work done in moving unit charge:

(a) from the positive to negative plate of the cell

(b) from the positive plate back to the positive plate

(d) from any point in the circuit back to the same point

Show Answer

Answer: (b, c, d)

13. A cell of e.m.f. $E$ and internal resistance $r$ drives a current $I$ through an external resistance $R$ :

(a) the cell supplies $E I$ power

(b) heat is produced in $R$ at the rate $E I$

(d) heat is produced in the cell at the rate $E I(\frac{r}{R+r})$

Show Answer

Answer: (a,c,d)

14. In the circuit shown in the fiugre:

(a) power supplied by the battery is 200 watt

(b) current flowing in the circuit is $5 A$

(d) current in wire $A B$ is zero

Show Answer

Answer: (a, c)

$4 \Omega$ and $6 \Omega$ resistances are short circuited. Therefore, no current will flow through these two resistances. Current passing through the battery is $I=(20 / 2)=10 A$.

This is also the current passing in wire $A B$ from $B$ to $A$.

Power supplied by the battery.

$P=E I=(20)(10)=200$ Watt

Potential difference across $4 \Omega$ resistance

$=$ Potential difference across $6 \Omega$ resistance $=0$

Fill in the Passage:

DIRECTIONS : Complete the following passage(s) with an appropriate word/term to be filled in the blank spaces.

$\text{I.}$

melt $\quad$ $\quad$ fuse $\quad$ $\quad$ live wire $\quad$ $\quad$ exceeds

zero $\quad$ $\quad$ eliminated

Different electric circuits can withstand certain maximum amount of current without getting heated up and damaging themselves. For example, for a lighting circuit this limit is $5 A$ and for a heating circuit, it is $15 A$. If current approaches this value, fuse wire will ………1………. Thus, the current ratings of ………2………. should be less than the maximum current that can safely flow through the circuit.

Fuse should be connected in ………3………. of circuit. If it is connected to neutral wire, the fuse will melt when current ………4………. its rating, but the appliance will still be connected to high potential of supply through live wire. If a person touches the appliance, his body provides a path for current to flow to earth which is at ………5………. potential. Thus, person may get severe shock which may prove fatal. Hence, fuse is connected to live wire so that if it melts, the appliance is no longer connected to the high potential of the supply and possibility of electric shock is ………6……….

Show Answer

Answer: 1. melt

2. fuse

3. live wire

4. exceeds

5. zero

eliminated

$\text{II.}$

insulation $\quad$ $\quad$ zero $\quad$ $\quad$ earth $\quad$ $\quad$ body

conducting $\quad$ $\quad$ touched $\quad$ $\quad$ $\quad$ conducting

When the ………1………. of live wire in a circuit inside a electric appliance melts due to some reasons, the live wire may touch the metal casing of the appliance. Thus, metal casing is connected to high potential. This metal casing of appliances such as mixer, fan, refrigerator, toaster etc., is continuously ………2………. by us by our bare hands. Hence, we may get a severe shock as current will flow through our ………3………. to ………4………. which is considered as ………5………. potential. To prevent this, metal casings of all appliances are connected to earth by a ………6………. wire. This is called earthing.

Thus, ………7………. involves connecting metal body of electric appliance to thick copper wire which is connected to a copper plate buried deep in the earth. The copper plate is surrounded by a mixuture of charcoal and common salt.

Show Answer

Answer:

1. insulation

touched
body
earth
zero
conducting
earthing

$\text{III.}$

conductor $\quad$ $\quad$ $\quad$ free electrons $\quad$ $\quad$ higher potential

lower potential $\quad$ $\quad$ colliding $\quad$ $\quad$ $\quad$ vibrate

temperature $\quad$ $\quad$ kinetic energy $\quad$ $\quad$ $\quad$ insulator

When a potential difference is applied between the ends of a ………1………, current begins to flow through it. The current flows as conductor contains large number of ………2……… which begin to drift from the end at ………3……… to the end at ………4……… In the process they constantly keep on ………5……… with atoms of conductor. The atoms gain energy and beging to ………6……… more vigorously about their mean position. The average ………7……… of atoms of conductor increases.

The results in a rise in the ………8……… of the conductor and we say that the conductor has been heated. Thus, flow of current has a heating effect on the conductor.

Show Answer

Answer:

1. conductor

free electrons
lower potential
higher potential
colliding
vibrate
kinetic energy

8 . temperature.

Passage Based Questions:

DIRECTIONS : Study the given paragraph(s) and answer the following questions.

Passage-I

Two tungston lamps with resistances $R_1$ and $R_2$ respectively at full incandescence are connected first in parallel and then in series, in a lighting circuit of negaligible internal resistance. It is given that: $R_1>R_2$.

1. Which lamp will glow more brightly when they are connected in parallel?

(a) Bulb having lower resistance

(b) Bulb having higher resistance

(d) None of the two bulbs

Show Answer

Answer: (a) When the lamps are connected in parallel, then potential difference $V$ across each lamp will be same and will be equal to potential necessary for full brightness of each bulb. Because illumination produced by a lamp is proportional to electric power consumed in it, and power consumed,

$ P_1=\frac{V^{2}}{R_1}<\frac{V^{2}}{R_2}=P_2 $

Hence, illumination produced by 2 nd bulb will be higher than produced by lst bulb, i.e., bulb having lower resistance will shine more brightly.

2. If the lamp of resistance $R_1$ now burns out, how will the illumination produced change?

(a) Net illumination will increase

(b) Net illumination will decrease

(d) Net illumination will reduced to zero

Show Answer

Answer: (b) When $R_1$ burns out, then power is dissipated in $R_2$ only. Because internal resistance is quite low in lighting circuit, potential difference is still equal to $V$, hence. power dissipated in $2^{nd}$ lamp, i.e..

$ \frac{v^{2}}{R_2}<(\frac{v^{2}}{R_1}+\frac{v^{2}}{R_2}) \text{. } $

i.e., net power consumed initially. In other words, net illumination will now decrease.

3. Which lamp will glow more brightly when they are connected in series?

(a) Bulb having lower resistance

(b) Bulb having higher resistance

(d) None of the two bulbs

Show Answer

Answer: (b) When two lamps are connected in series, the potential difference across each lamp will be different but current $I$ flowing through each lamp will be same.

Hence. $ \quad P_1=I^{2} R_1>I^{2} R_2=P_2 $

i.e., illumination produced by Ist lamp will be higher as compared to that produced by 2nd lamp, i.e.. lamp having higher resistance will glow more brightly.

4. If the lampe of resistance $R_2$ now burns out and the lamp of resistance $R_1$ alone is plugged in, will the illumination increase or decrease?

(a) Illumination will remain same

(b) Illumination will increase

(d) Illumination will decrease

Show Answer

Answer: (b) When lamp of resistance $R_2$ bums out and only lamp of resistance $R_1$ is connected in the circuit then current flowing the circuit will change. La new current be $I^{\prime}$. Because potential difference still remains same (due to low intemal resistance), hence

$I^{\prime} R_1=I(R_1+R_2)$

or $I^{\prime}=\frac{I(R_1+R_2)}{R_1}$

If $P^{\prime}$ is the power consumed, then

$P^{\prime}=I^{\prime} 2 R_1=I^2 \frac{\left(R_1+R_2\right)\left(R_1+R_2\right)}{R_1}$

When both the lamps were present then total power consumed was given by:

$P_S=P_1+P_2=I^{2}(R_1+R_2)$, ie.. $P^{\prime}>P_S$

i.e., illumination gets increased when only one bulb is used.

5. Would physically bending a supply wire cause any change in the illumination?

(a) Illumination will remain same

(b) Illumination will increase

(d) It is not possible to predict from the given data

Show Answer

Answer: (a) If a water pipe is given bend at some points, then it definitely reduces the flow of water in the pipe but this is not true in case of an electric current flowing in a conductor because electric current is established in a conductor due to drift motion of electrons in it along the line of the potential gradient. Hence, illumination is not affected due to bending along the length of supply wires.

Passage-II

Figure shows the circuit of a potentiometer. The length of the potentiometer wire $A B$ is $50 cm$. The emf of the battery resistances $R_1$ and $R_2$ are $15 ohm$ and $5 ohm$ respectively. When both the keys are open, the null point is obtained at a distance of $31.25 cm$ from end $A$ but when both the keys are closed, the balance length reduces to $5 cm$ only. Given $R _{AB}=10 \Omega$.

1. The emf of the cell $E_2$ is:

(a) 1 volt

(b) 2 volt

(d) 4 volt

Show Answer

Answer: (a) Potential gradient.

$K=\frac{V _{A B}}{L}=\frac{I \times R}{L}=\frac{4 \text{ volt }}{(15+10) ohm} \times \frac{10 ohm}{50 cm}$

$=\frac{40}{25 \times 50}$ Volt $/ cm$

$\therefore E_2=K l=\frac{40}{25 \times 50} \times 31.25=1$ volt

2. The internal resistance of the cell $E_2$ is:

(a) $4.5 \Omega$

(b) $5.5 \Omega$

(d) $7.5 \Omega$

Show Answer

Answer: (d) When both the keys are closed: In the situation. $R_1$ is short circuited and terminal potential difference of $E_2$ is balanced by the potentiometer, i.e..

$E_2 \times R_2=K l={ }_{10 \times 50}^{4 \times 10} \times 5$

$ \begin{aligned} & \frac{1 \times 5}{5+r}=\begin{matrix} 4 \\ 50 \end{matrix} \times 5 \text{ or } r=7.5 \Omega \end{aligned} $

3. The balance length when key $K_2$ is open and $K_1$ is closed, is given by:

(a) 10.5 cm

(b) 11.5 cm

(d) 13.5 cm

Show Answer

Answer: (c) When key $K_2$ is open and $K_1$ is closed: In this case $R_1$ is short circuited and e.m.f. $E_2$ is balanced against potentiometer,

i.e., $E_2=\left[\begin{array}{cc}4 \\ 10\end{array} \times \begin{array}{c}10 \\ 50\end{array}\right] \times l=\frac{1 \times 10 \times 50}{4 \times 10}=12.5 \mathrm{~cm}$

4. The blance length when key $K_1$ is open and $K_2$ is closed, is given by:

(a) 10.5 cm

(b) 11.5 cm

(d) 13.5 cm

Show Answer

Answer: (c) When key $K_1$ is open and $K_2$ is closed: In this case, $R_1$ is included in the potentiometer circuit and terminal potential difference of $E_2$ is balanced by the potentiometer.

i.e., $\frac{E_2 R_2}{R_2+r}=[\frac{4}{25} \times \frac{10}{50}] \times l$

or $\frac{1 \times 5}{5+7.5}=\frac{4}{1.25} \times l$

or $l=12.5 cm$

Passage-III

A 12 volt battery is connected to two light bulbs, as shown in figure. Light bulb 1 has resistance 3 ohm, while light bulb 2 has resistance 6 ohm. The battery has essentially no internal resistance and all the wires are essentially resistanceless too. When a light bulb is unscrewed, no current flows through that branch of the circuit. For instance, if light bulb 2 is unscrewed, current flows only around the lower loop of the circuit, which consists of the battery and light bulb 1 . When two resistances are joined in series, their equivalent resistances $R_{eq}=R_1+R_2$ but when two resistances are wired in parallel, their net resistance is given by:

$\frac{1}{R_{e q .}}=\frac{1}{R_1}+\frac{1}{R_2}$

1. When bulb 1 is screwed in, but bulb 2 is unscrewed, the power generated in bulb 1 is:

(a) 4 watt

(b) 12 watt

(d) 48 watt

Show Answer

Answer: (d) Since, the power generated in the bulb equals the voltage across the bulb times the current through the bulb $(P=I V)$, we need to know the current and voltage. Because the lower circuit loop contains only the battery and bulb 1 , the bulb feels the battery’s full voltage, 12 volts. So, by Ohm’s law, a current,

$ I=\frac{V}{K}=\frac{12 V}{3 \Omega}=4 amp $

flows through that bulb. Therefore, the bulb generates power,

$ P=I V=(4 A)(12 V)=48 W $

2. Which of the following statements is false?

(a) Some of the energy produced by the light bulb takes the form of heat

(b) The battery is the source of all the electrons flowing around the circuit

(d) The potential in the wire to the left of the light bulb differs from the potential in the wire to the right of that bulb

Show Answer

Answer: (b) Most of the charges flowing around the circuit are valence electrons stripped off the metal atoms in the wire and light bulbs. A battery does not supply all of the charges. It merely pushes around charges already present in the circuit.

Statements (c) and (d) are both true. All charges flowing into the light bulb also flow back out; no curent gest used up. But inside the bulb, those charges lose energy. This lost electrical energy converts into light and heat. So, the current has lower potential after flowing through the bulb.

3. Bulb 2 is now screwed in, as a result, bulb 1

(a) turns off

(b) becomes dimmer

(d) becomes brighter

Show Answer

Answer: (c) If the battery has internal resistance, bulb 1 would dim. But here, screwing in bulb 2 does not prevent bulb 1 from feeling the full 12 volts produced by the battery. Both bulbs get a full does of that voltage. So, the bulb 2 turns on without lessening the brightness ofbulb 1. If you picked (b), you probably thought that the battery can supply only a certain amount of current. According to this reasoning, bulb 2 draws current at the expense of bulb 1 . But the battery does not supply all of the electrons flowing around the circuit. It merely pushes around electrons already present in the metal wires and bulb filaments, as mentioned in earlier question. Screwing in bulb 2 allows the valence electrons in that branch of the circuit to join the current.

4. When both light bulbs are screwed in, the current through the battery is

(a) 1.2 ampere

(b) 2 ampere

(d) 6 ampere

Show Answer

Answer: (d) The total current through the battery is simply,

$ I=\frac{V}{R _{e q}} $

where, $R _{e q}$ denotes the equivalent resistance of the whole circuit. These two light bulbs are wired in parallel; any given electron flowing around the circuit passes through one bulb or the other, but not both. So, we must use the inparallel formula,

$$\frac{1}{R _{e q} .}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2 \Omega}$$

and hence, $R _{\text{eq. }}=2 \Omega$

So, $I=\frac{V}{R _{e q}}=\frac{12 V}{2 \Omega}=6 amp$

Intuitively, the equivalent resistance of two in-parallel resistors is less than either individual resistance because the current can split itself between two paths.

5. With only light bulb 1 screwed in, a never quit 12 volt battery goes dead in 24 days. With both light bulbs screwed in a never quit 12 volt battery goes dead in:

(a) 12 days

(b) 14 days

(d) 18 days

Show Answer

Answer: (c)

Assertion & Reason :

DIRECTIONS : Each of these questions contains an Assertion followed by reason. Read them carefully and answer the question on the basis of following options. You have to select the one that best describes the two statements.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(d) If Assertion is incorrect but Reason is correct.

1. Assertion : When current through a bulb decreases by $0.5 %$, the glow of bulb decreases by $1 %$.

Reason : Glow (Power) which is directly proportional to square of current.

Show Answer

Answer: (b) $Glow=Power(P)=I^{2} R$

$\therefore \frac{d P}{P}=2(\frac{d I}{I})=2 \times 0.5=1 \%$

2. Assertion : Long distance power transmission is done at high voltage.

Reason : At high voltage supply power losses are less.

Show Answer

Answer: (b) Power loss $=i^{2} R=(\frac{P}{V})^{2} R$

$[P=$ Transmitted power $]$

3. Assertion : Resistance of $50 W$ bulb is greater than that of $100 W$.

Reason : Resistance of bulb is inversely proportional to rated power.

Show Answer

Answer: (b) $P=\frac{V^{2}}{R} ; \quad R \propto \frac{1}{P}$ (same rated voltage)

4. Assertion : $40 W$ tube light give more light in comparison to $40 w$ bulb.

Reason : Light produced is same from same power.

Show Answer

Answer: (d) In tube light majority portion of radiation comes under visible region while bulb radiation consists of visible, ultraviolet, infrared radiation giving less visible part.

5. Assertion : A resistor of resistance $R$ is connected to an ideal battery. If the value of $R$ is decreased, the power dissipated in the circuit will increase.

Reason : The power dissipated in the circuit is directly proportional to the resistance of the circuit.

Show Answer

Answer: (c) Here, $P=\frac{E^{2}}{R}$, so $P \propto R$ only when $I$ is constant. Here $I$ increases as $R$ is decreased. Hence the reason is wrong.

6. Assertion : A voltmeter and ammeter can be used together to measure resistance and power.

Reason : Resistance and power both are proportional to voltage and current.

Show Answer

Answer: (c) As $R=V / I$ and $P=VI$, by .measuring $V$ and $I$ simultaneously in circuit we can measure both resistance and power, using the given relation.

7. Assertion : A torch bulb give light if operated on $AC$ of same voltage and current as DC.

Reason : Heating effect is common to both $AC$ and $DC$.

Show Answer

Answer: (a)

8. Assertion : A tube light emits white light.

Reason : Emission of light in a tube takes place at a very high temperature.

Multiple Matching Questions :

DIRECTIONS : Following question has four statements (A, B, C and D) given in Column I and four statements (p, q, r and s) in Column II. Any given Assertion Column I can have correct matching with one or more statement(s) given in Column II. Match the entries in column I with entries in column II.

1. For the circuit shown in the adjoining figure, match the entries of column I with the entries of column II.

		Column II
(a)	(p)	Current drawn from the battery is maximum
(b)	(q)	Current drawn from the battery is the least
(c)	(r)	Bulbs will lit the brightest
(d)	(s)	Bulbs will lit with brightness lying between maximum and minimum value

Show Answer

Answer: (A) $\rarr$ q ; (b) $\rarr$ r ; (C) $\rarr$ p,s (D) $\rarr$ s

2. In the shown circuit diagram, all the electric bulbs are identical. Then, match the entries of column I with the entries of column II

	Column I		Column II
(A)	Current drawn by $L_1$ is,	(p)	maximum
(B)	Intensity of $L_1$ and $L_2$ is	(q)	minimum
(C)	Intensity of $L_3$ and $L_4^{2}$ is	(r)	same
(D)	Intensity of $L_1$ and $L_4$ is	(s)	different

Show Answer

Answer: (A) $\rarr$ p ; (B) $\rarr$ p ; r (C) $\rarr$ q,r ; (D) $\rarr$ s

3. Six batteries of increasing e.m.f. and increasing internal resistance are connected in a circuit as shown in the given figure. Match the entries of column I with the entries of column II.

	Column I		Column II
(A)	Potential of point $A$	(p)	Zero
(B)	Potential of point $B$	(q)	$2 V$
(C)	Potential of point $C$	(r)	$4 V$
(D)	Potential of point $D$	(s)	$6 V$
		(t)	None

Show Answer

Answer: (A) $\rarr$ p ; (B) $\rarr$ p ; (C) $\rarr$ p ;(D) $\rarr$ p

HOTS Subjective Questions :

DIRECTIONS : Answer the following questions.

1. How is a fuse put in an electric circuit? State the purpose of using a fuse in a circuit.

Show Answer

Answer: A fuse is safety device having a short length of a thin wire of tin-lead alloy, that melts and breaks the circuit if the current exceeds a safe value.

2. A $60-W$ bulb is switched on in a room. A $240-W$ heater is also turned on in the same room. The voltage of the mains is $120 V$ and the resistance of the connecting leads is $6 \Omega$. What is the change in the voltage at the bulb when the heater is turned on.

Show Answer

Answer: The resistance of bulb and heater we respectively $240 \Omega$ and $60 \Omega$, and these are connected in parallel. Their equivalent resistance is $48 \Omega$. When only bulb is switched on, current from the mains is

$120 V$

$(240+6) \Omega=0.488 A$. When heater is also turned on,

current becomes $\frac{120 V}{(48+6) \Omega}=2.22 A$

3. Explain why an electric bulb becomes dim when an electric heater in parallel circuit is made on. Why dimness decrease after sometime?

Show Answer

Answer: The resistance of a heater coil is less than that of electric bulb filament. When heater is switched on in parallel, more current start flowing through the heater coil and current through the bulb filament decreases making it dim.

After sometime, when heater coil becomes hot its resistance increases. As a result current through the heater coil decreases and the current through the bulb filament increases and thus dimness of the bulb decreases.

4. Why is tungsten metal selected for making filaments of incandescent lamp bulbs?

Show Answer

Answer: Tungsten metal has high resistivity and high melting point.

5. A heater joined in parallel with a $60 W$ bulb is replaced by a more or less or remains the same?

Show Answer

Answer: Resistance of $60 W$ bulb is more than in $100 W$ bulb. So equivalent resistance of the network drops on placing $100 W$ bulb and so the power or heat $(\frac{V^{2}}{R} t)$ is more

6. Nichrome and copper.wires of same length and same radius are connected in series. Current 1 is passed through then. Why does the nichrome wire get heated first?

Show Answer

Answer: Nichrome has greater resistivity.

7. Two bulbs whose resistances are in the ratio $1: 2$, are connected in parallel to a source of constant voltage. What will be the ratio of power dissipation of these?

Show Answer

Answer: Since, $P=\frac{V^{2}}{R}, P_1: P_2=R_2: R_1=2: 1$

8. In a factory, an electric bulb of 500 W is used for 2 hours and electric bulb of $500 W$ is used for 2 hours and electric motor of 0. 5 horse power is used for 5 hours everydary. Calculate the cost of using the bulb and motor for 30 days if cost of electrical energy is three rupees per unit.

Show Answer

Answer: Electrical energy consumed in 30 days

$=500 \times 2 \times 30+0.5 \times 746 \times 5 \times 30(1 hp=746 W)$

$=30000+55950=85950 Wh=85.95 kWh$

Cost of energy $=3 \times 85.95=₹ 258$

9. An electric lamp is marked $100 W, 220 V$. It is used for 5 hours daily. Calculate

(i) its resistance while glowing

(ii) energy consumed in $kWh$ per day,

Show Answer

Answer: (i) $R=\frac{V^{2}}{P}=\frac{220^{2}}{100}=484 \Omega$

(ii) Energy consumed per day

$ =P \times t=100 \times 5=500 Wh=0.5 kWh . $

10. A torch bulb is rated $3 V$ and $600 mA$. Calculate its resistance if it is lighted for 4 hours.

Show Answer

Answer: Given : $V=3$ Volt,

$ I=600 mA=600 \times 10^{-3} A=0.6 A $

From Ohm’s law

$V=IR \Rightarrow R=\frac{V}{I}=\frac{3}{0.6}=\frac{30}{6}=5 \Omega$

11. A heater coil is rated $100 W, 200 V$. It is cut into two identical parts. Both parts are connected together in parallel to the same source of $200 V$. Calculate the energy liberated per second in the new combination.

Show Answer

Answer: Resistance of heater coil

$V^{2} =\frac{200^{2}}{100}=400 \Omega$

Resistance of each cut part $=200 \Omega$

When connected in parallel, the net resistance will be $100 \Omega$

Energy liberated per second

$ =\frac{V^{2}}{R}=\frac{200^{2}}{100}=400 \Omega \text{ joule } $

12. A $60 W$ electric lamp gives off energy in the form of light at a rate of 7.5 joule per second. What percentage of energy 13. does the lamp transform into light energy?

Show Answer

Answer: Power of lamp $=60 W$

Light liberated $=7.5 W$

$\%$ of transfer $=\frac{7.5}{60} \times 100=12.5 %$

13.

Refrigerator, $\quad$ electrical $\quad$ chip-pan, $\quad$ battery $\quad$ charger,

loudspeaker, $\quad$ air-conditioner, $\quad$ electric drill, $\quad$ electroplating-yat

Choose only from the above electrical appliances those that make use of the

(a) chemical effect of an electric current.

(b) heating effect of an electric current.

Show Answer

Answer: (a) Battery charger, electroplating-vat

(b) electric chip-pan

14. This calculation is aimed at finding out how many joules there are in 1 kilowatt hour :

1 watt second = 1 joule

1 kilowatt second = _______________joule

1 kilowatt minute = _______________$\times 60$ joules

1 kilowatt hour = _______________$\times 60 \times 60$ joules

1 kilowatt hour = _______________joules

Show Answer

Answer: (a) 1 kilowatt second $=1000$ joules

1 kilowatt minute $=1000 \times 60 \times 60$ joules

1 kilowatt hour $=1000 \times 60 \times 60$ joules

$=3600000$ joules

15. A man sits in a room for an hour watching T.V. There are four $100 W$ light bulbs switched on an the T.V. has a power rating of $150 W$. The only electrical appliance in the room is an air conditioner which has a power rating of $450 W$. Calculate the cost of electricity used in that room for the hour if 1 unit (kilowatt hour) costs $25 c$. It might help to divide the calculation as follows :

Number of kilowatt hours for

light bulbs = ___________________

T.V. = ___________________

air conditioner = ___________________

Total kilowatt hours = ___________________

Total cost is ___________________

Show Answer

Answer: Light bulbs = 1000

T.V. = 0.15 kWh

air conditioner = 0.45 kWh

Total = 1.0 kWh = 25e

16. What does it mean to say that your car has a 12-volt tautery?

Show Answer

Answer: It means that one of the battery terminals is $12 V$ higher in potential than the other one. Soon we will see that it also means that, when a circuit is connected between these terminals, each coulomb of charge in the resulting current will be given $12 J$ of energy as it passes through the battery (and $12 J$ of energy “spent” in the circuit).

17. A balloon may easily be charged to several thousind volts Does that mean it has several thousand joules of energy? Explain.

Show Answer

Answer: No, it doesn’t mean that the balloon has several thousand joules of energy always. It can have several thousand joules of energy only if the balloon acquires high charge.

18. Calculate the cost of electric bill of a house for the month of March. The following appliances were used in the house for the duration shown respectively. The cost of electrical energy is 50 paise per unit.

$6-100 W$ lamps 4h each

$5-60 W$ lamps 5h each

$1-750 W$ iron 2h

$1-2 kW$ geyser 2h

Show Answer

Answer: Energy consumed by different appliances is

$100 W$ lamps, $6 \times 100 \times 4=2400 Wh$

$60 W$ lamps, $5 \times 60 \times 5 \quad=1500 Wh$

$750 W$ electric iron, $750 \times 2=1500 Wh$

$2 kW$ geyser, $2000 \times 2=4000 Wh$

$ 9400 Wh $

No. of units consumed per day $=\frac{9400}{1000} Wh=9.4 kWh$

Total consumption for 31 days $=9.4 \times 31=291.4 kWh$

Total cost of the bill $=291.4 \times \frac{1}{2}=₹ 145.70$

Chapter 05 Heating And Chemical Effects Of Current

Introduction

ELECTRIC POWER

HEAT PRODUCED DUE TO ELECTRICITY

PRACTICAL APPLICATION OF HEATING EFFECT

CHEMICAL EFFECT OF CURRENT

FARADAY’S LAWS OF ELECTROLYSIS

(i) 1st law

(ii) 2nd law

FARADAY CONSTANT

ELECTROCHEMICAL CELL

PRIMARY CELLS

SECONDARY CELLS

ELECTROMOTIVE FORCE (EMF)

INTERNAL RESISTANCE OF CELL

TERMINAL POTENTIAL DIFFERENCE

DIFFERENCE BETWEEN EMF AND POTENTIAL DIFFERENCE

GROUPING OF CELLS

(i) Series Combination

(ii) Parallel Combination

(iii) Mixed Combination

DOMESTIC ELECTRIC CIRCUITS

Fuses and Circuit Breakers

The Circuit Breaker Panel

Electric Circuits in the Home

Causes of Overloaded Circuits

MISCELLANEOUS

SOLVED EXAMPLES

1 EXERCISE

Fill in the Blanks

True / False

Match the Following

Very Short Answer Questions:

Short Answer Questions

Long Answer Questions :

2 EXERCISE

Multiple Choice Questions:

More Than One Correct

Fill in the Passage:

Passage Based Questions:

Assertion & Reason :

Multiple Matching Questions :

HOTS Subjective Questions :

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

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