Answer

A continuous and closed path of an electric current is called an electric circuit. An electric circuit consists of electric devices, source of electricity and wires that are connected with the help of a switch.

Answer

The unit of electric current is ampere (A). $1 A$ is defined as the flow of $1 C$ of charge through a wire in $1 s$.

Answer

One electron possesses a charge of $1.6 \times 10^{-19} C$, i.e., $1.6 \times 10^{-19} C$ of charge is contained in 1 electron.

$\therefore 1 C$ of charge is contained in $1 / 1.6 \times 10^{-19}=6.25 \times 10^{18}=6 \times 10^{18}$

Therefore, $6 \times 10^{18}$ electrons constitute one coulomb of charge.

Answer

Any source of electricity like battery, cell, power supply, etc. helps to maintain a potential difference across a conductor.

Answer

If $1 J$ of work is required to move a charge of amount $1 C$ from one point to another, then it is said that the potential difference between the two points is $1 V$.

Answer

The energy given to each coulomb of charge is equal to the amount of work which is done in moving it.

Now we know that,

Potential difference $=$ Work Done/Charge

$\therefore$ Work done $=$ Potential difference $\times$ charge

Where, Charge $=1 C$ and Potential difference $=6 V$

$\therefore$ Work done $=6 \times 1$

$=6$ Joule.

11.3 CIRCUIT DIAGRAM

We know that an electric circuit, as shown in Fig. 11.1, comprises a cell (or a battery), a plug key, electrical component(s), and connecting wires. It is often convenient to draw a schematic diagram, in which different components of the circuit are represented by the symbols conveniently used. Conventional symbols used to represent some of the most commonly used electrical components are given in Table 11.1.

Table 11.1 Symbols of some commonly used components in circuit diagrams

11.4 OHM’S LAW

Is there a relationship between the potential difference across a conductor and the current through it? Let us explore with an Activity.

Activity 11.1

• Set up a circuit as shown in Fig. 11.2, consisting of a nichrome wire XY of length, say $0.5 m$, an ammeter, a voltmeter and four cells of $1.5 V$ each. (Nichrome is an alloy of nickel, chromium, manganese, and iron metals.)
• First use only one cell as the source in the circuit. Note the reading in the ammeter $I$, for the current and reading of the voltmeter $V$ for the potential difference across the nichrome wire $XY$ in the circuit. Tabulate them in the Table given.

• Next connect two cells in the circuit and note the respective readings of the ammeter and voltmeter for the values of current through the nichrome wire and potential difference across the nichrome wire.
• Repeat the above steps using three cells and then four cells in the circuit separately.
• Calculate the ratio of $V$ to $I$ for each pair of potential difference $V$ and current $I$.

Plot a graph between $V$ and $I$, and observe the nature of the graph.

Figure 11.3

$V-I$ graph for a nichrome wire. A straight line plot shows that as the current through a wire increases, the potential difference across the wire increases linearly - this is Ohm’s law. In this Activity, you will find that approximately the same value for $V / I$ is obtained in each case. Thus the $V-I$ graph is a straight line that passes through the origin of the graph, as shown in Fig. 11.3. Thus, $V / I$ is a constant ratio.

In 1827, a German physicist Georg Simon Ohm (1787-1854) found out the relationship between the current $I$, flowing in a metallic wire and the potential difference across its terminals. The potential difference, $V$, across the ends of a given metallic wire in an electric circuit is directly proportional to the current flowing through it, provided its temperature remains the same. This is called Ohm’s law. In other words -

$ \hspace{50 px}V \propto I \\ \text{or} \hspace{40 px} V/I = \text{ constant} \\ \hspace{70 px} = R \\ \text{or} \hspace{50 px} V = IR $

In Eq. (11.4), $R$ is a constant for the given metallic wire at a given temperature and is called its resistance. It is the property of a conductor to resist the flow of charges through it. Its SI unit is ohm, represented by the Greek letter $\Omega$. According to Ohm’s law,

$ R=V / I $

If the potential difference across the two ends of a conductor is $1 V$ and the current through it is $1 A$, then the resistance $R$, of the conductor is $1 \Omega$. That is, $1 ohm=\frac{1 \text{ volt }}{1 \text{ ampere }}$

Also from Eq. (11.5) we get

$ I=V / R $

It is obvious from Eq. (11.7) that the current through a resistor is inversely proportional to its resistance. If the resistance is doubled the current gets halved. In many practical cases it is necessary to increase or decrease the current in an electric circuit. A component used to regulate current without changing the voltage source is called variable resistance. In an electric circuit, a device called rheostat is often used to change the resistance in the circuit. We will now study about electrical resistance of a conductor with the help of following Activity.

Activity 11.2

• Take a nichrome wire, a torch bulb, a $10 W$ bulb and an ammeter ( 0 - 5 A range), a plug key and some connecting wires.
• Set up the circuit by connecting four dry cells of $1.5 V$ each in series with the ammeter leaving a gap XY in the circuit, as shown in Fig. 11.4.

Nichrome wire

Torch bulb

Figure 11.4

• Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. Take out the key from the plug. [Note: Always take out the key from the plug after measuring the current through the circuit.]
• Replace the nichrome wire with the torch bulb in the circuit and find the current through it by measuring the reading of the ammeter.
• Now repeat the above step with the $10 W$ bulb in the gap XY.
• Are the ammeter readings different for different components connected in the gap XY? What do the above observations indicate?
• You may repeat this Activity by keeping any material component in the gap. Observe the ammeter readings in each case. Analyse the observations.

In this Activity we observe that the current is different for different components. Why do they differ? Certain components offer an easy path for the flow of electric current while the others resist the flow. We know that motion of electrons in an electric circuit constitutes an electric current. The electrons, however, are not completely free to move within a conductor. They are restrained by the attraction of the atoms among which they move. Thus, motion of electrons through a conductor is retarded by its resistance. A component of a given size that offers a low resistance is a good conductor. A conductor having some appreciable resistance is called a resistor. A component of identical size that offers a higher resistance is a poor conductor. An insulator of the same size offers even higher resistance.

11.5 FACTORS ON WHICH THE RESISTANCE OF A CONDUCTOR DEPENDS

Activity 11.3

• Complete an electric circuit consisting of a cell, an ammeter, a nichrome wire of length $l$ [say, marked (1)] and a plug key, as shown in Fig. 11.5.

Figure 11.5 Electric circuit to study the factors on which the resistance of conducting wires depends

• Now, plug the key. Note the current in the ammeter.
• Replace the nichrome wire by another nichrome wire of same thickness but twice the length, that is $2 l$ [marked (2) in the Fig. 11.5].
• Note the ammeter reading.
• Now replace the wire by a thicker nichrome wire, of the same length $l$ [marked (3)]. A thicker wire has a larger cross-sectional area. Again note down the current through the circuit.
• Instead of taking a nichrome wire, connect a copper wire [marked (4) in Fig. 11.5] in the circuit. Let the wire be of the same length and same area of cross-section as that of the first nichrome wire [marked (1)]. Note the value of the current.
• Notice the difference in the current in all cases.
• Does the current depend on the length of the conductor?
• Does the current depend on the area of cross-section of the wire used?

It is observed that the ammeter reading decreases to one-half when the length of the wire is doubled. The ammeter reading is increased when a thicker wire of the same material and of the same length is used in the circuit. A change in ammeter reading is observed when a wire of different material of the same length and the same area of cross-section is used. On applying Ohm’s law [Eqs. (11.5) - (11.7)], we observe that the resistance of the conductor depends (i) on its length, (ii) on its area of cross-section, and (iii) on the nature of its material. Precise measurements have shown that resistance of a uniform metallic conductor is directly proportional to its length $(l)$ and inversely proportional to the area of cross-section $(A)$. That is,

$ \hspace{60 px} R \propto l \\ \text{ and } \hspace{40 px} R \propto 1 / A $

Combining Eqs. (11.8) and (11.9) we get

$ \begin{aligned} R & \propto \frac{l}{A} \\ \text{ or, } \quad R & =\rho \frac{l}{A} \end{aligned} $

where $\rho$ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor. The SI unit of resistivity is $\Omega m$. It is a characteristic property of the material. The metals and alloys have very low resistivity in the range of $10^{-8} \Omega m$ to $10^{-6} \Omega m$. They are good conductors of electricity. Insulators like rubber and glass have resistivity of the order of $10^{12}$ to $10^{17} \Omega m$. Both the resistance and resistivity of a material vary with temperature.

Table 11.2 reveals that the resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidise (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices, like electric iron, toasters etc. Tungsten is used almost exclusively for filaments of electric bulbs, whereas copper and aluminium are generally used for electrical transmission lines.

Table 11.2 Electrical resistivity* of some substances at $20^{\circ} C$

• You need not memorise these values. You can use these values for solving numerical problems.

Example 11.3

(a) How much current will an electric bulb draw from a $220 V$ source, if the resistance of the bulb filament is $1200 \Omega$ ? (b) How much current will an electric heater coil draw from a $220 V$ source, if the resistance of the heater coil is $100 \Omega$ ?

Solution

(a) We are given $V=220 V ; R=1200 \Omega$.

From Eq. (12.6), we have the current $I=220 V / 1200 \Omega=0.18 A$.

(b) We are given, $V=220 V, R=100 \Omega$.

From Eq. (11.6), we have the current $I=220 V / 100 \Omega=2.2 A$.

Note the difference of current drawn by an electric bulb and electric heater from the same $220 V$ source!

Example 11.4

The potential difference between the terminals of an electric heater is $60 V$ when it draws a current of $4 A$ from the source. What current will the heater draw if the potential difference is increased to $120 V$ ? Solution

We are given, potential difference $V=60 V$, current $I=4$ A.

According to Ohm’s law, $R=\frac{V}{I}=\frac{60 V}{4 A}=15 \Omega$.

When the potential difference is increased to $120 V$ the current is given by

current $=\frac{V}{R}=\frac{120 V}{15 \Omega}=8 A$.

The current through the heater becomes $8 A$.

Example 11.5

Resistance of a metal wire of length $1 m$ is $26 \Omega$ at $20^{\circ} C$. If the diameter of the wire is $0.3 mm$, what will be the resistivity of the metal at that temperature? Using Table 11.2, predict the material of the wire.

Solution

We are given the resistance $R$ of the wire $=26 \Omega$, the diameter $d=0.3 mm=3 \times 10^{-4} m$, and the length $l$ of the wire $=1 m$.

Therefore, from Eq. (11.10), the resistivity of the given metallic wire is

$\rho=(R A / l)=(R \pi d^{2} / 4 l)$

Substitution of values in this gives

$\rho=1.84 \times 10^{-6} \Omega m$

The resistivity of the metal at $20^{\circ} C$ is $1.84 \times 10^{-6} \Omega m$. From Table 11.2, we see that this is the resistivity of manganese.

Example 11.6

A wire of given material having length $l$ and area of cross-section $A$ has a resistance of $4 \Omega$. What would be the resistance of another wire of the same material having length $l / 2$ and area of cross-section $2 A$ ?

Solution

For first wire

$ R_1=\rho \frac{l}{A}=4 \Omega $

Now for second wire

$R_2=\rho \frac{l / 2}{2 A}=\frac{1}{4} \rho \frac{l}{A}$

$R_2=\frac{1}{4} \quad R_1$

$R_2=1 \Omega$

The resistance of the new wire is $1 \Omega$.

QUESTIONS

Answer

The resistance of a conductor depends upon the following factors:

$\rightarrow$ Length of the conductor

$\rightarrow$ Cross-sectional area of the conductor

$\rightarrow$ Material of the conductor

$\rightarrow$ Temperature of the conductor

Answer

The current will flow more easily through thick wire. It is because the resistance of a conductor is inversely proportional to its area of cross - section. If thicker the wire, less is resistance and hence more easily the current flows.

Answer

According to Ohm’s law

$V=I R$ $\Rightarrow I=V / R \ldots$

Now Potential difference is decreased to half

$\therefore$ New potential difference $V^{\prime}=V / 2$

Resistance remains constant

So the new current $I^{\prime}=V^{\prime} / R$

$=(V / 2) / R$

$=(1 / 2)(V / R)$

$=(1 / 2) I=I / 2$

Therefore, the amount of current flowing through the electrical component is reduced by half.

Answer

The resistivity of an alloy is higher than the pure metal. Moreover, at high temperatures, the alloys do not melt readily. Hence, the coils of heating appliances such as electric toasters and electric irons are made of an alloy rather than a pure metal.

Answer

(a) Resistivity of iron $=10.0 \times 10^{-8} \Omega$

Resistivity of mercury $=94.0 \times 10^{-8} \Omega$

Resistivity of mercury is more than that of iron. This implies that iron is a better conductor than mercury.

(b) It can be observed from Table 12.2 that the resistivity of silver is the lowest among the listed materials. Hence, it is the best conductor.

Answer

Three cells of potential $2 V$, each connected in series therefore the potential difference of the battery will be $2 V+2 V+2 V=6 V$. The following circuit diagram shows three resistors of resistances $5 \Omega, 8 \Omega$ and $12 \Omega$ respectively connected in series and a battery of potential $6 V$ and a plug key which is closed means the current is flowing in the circuit.

Answer

An ammeter should be connected in the circuit in series with the resistors. To measure the potential difference across the resistor it should be connected in parallel, as shown in the following figure.

The resistances are connected in series.

Ohm’s law can be used to obtain the readings of ammeter and voltmeter. According to Ohm’s law,

$V=I R$,

Where,

Potential difference, $V=6 V$

Current flowing through the circuit/resistors $=I$

Resistance of the circuit, $R=5+8+12=25 \Omega$

$I=V / R=6 / 25=0.24 A$

Potential difference across $12 \Omega$ resistor $=V_1$

Current flowing through the $12 \Omega$ resistor, $I=0.24 A$

Therefore, using Ohm’s law, we obtain

$V_1=I R=0.24 \times 12=2.88 V$

Therefore, the reading of the ammeter will be $0.24 A$.

The reading of the voltmeter will be $2.88 V$.

Answer

(a) When $1 \Omega$ and $10^{6} \Omega$ are connected in parallel:

Let $R$ be the equivalent resistance.

$ \begin{aligned} & \therefore \frac{1}{R}=\frac{1}{1}+\frac{1}{10^{6}} \\ & R=\frac{10^{6}}{1+10^{6}} \approx \frac{10^{6}}{10^{6}}=1 \Omega \end{aligned} $

Therefore, equivalent resistance $\approx 1 \Omega$

(b) When $1 \Omega, 103 \Omega$ and $106 \Omega$ are connected in parallel:

Let $R$ be the equivalent resistance.

$ \begin{aligned} & \frac{1}{R}=\frac{1}{1}+\frac{1}{10^{3}}+\frac{1}{10^{6}} \frac{10^{6}+10^{3}+1}{10^{6}} \\ & R=\frac{1000000}{1001001}=0.999 \Omega \end{aligned} $

Therefore, equivalent resistance $=0.999 \Omega$

Answer

Resistance of electric lamp, $R_1=100 \Omega$

Resistance of toaster, $R_2=50 \Omega$

Resistance of water filter, $R_3=500 \Omega$

Potential difference of the source, $V=220 V$

These are connected in parallel, as shown in the following figure. $R_1=100 \Omega$

Let $R$ be the equivalent resistance of the circuit. $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{100}+\frac{1}{50}+\frac{1}{500}$

According to Ohm’s law,

$V=I R$

$I=\frac{V}{R}$

Where,

Current flowing through the circuit $=I$

$I=\frac{220}{\frac{500}{16}}=\frac{220 / 16}{50}=7.04 A$

7.04 A of current is drawn by all the three given appliances.

Therefore, current drawn by an electric iron connected to the same source of potential $220 V=7.04 A$

Let $R^{\prime}$ be the resistance of the electric iron. According to Ohm’s law,

$V=I R^{\prime}$

$R^{\prime}=\frac{V}{I}=\frac{220}{7.04}=31.25 \Omega$

Therefore, the resistance of the electric iron is $31.25 \Omega$ and the current flowing through it is $7.04 A$.

Answer

There is no division of voltage among the appliances when connected in parallel. The potential difference across each appliance is equal to the supplied voltage.

The total effective resistance of the circuit can be reduced by connecting electrical appliances in parallel.

Answer

There are three resistors of resistances $2 \Omega, 3 \Omega$, and $6 \Omega$ respectively.

(a) The following circuit diagram shows the connection of the three resistors.

Here, $6 \Omega$ and $3 \Omega$ resistors are connected in parallel.

Therefore, their equivalent resistance will be given by

$ \frac{1}{\frac{1}{6}+\frac{1}{3}}=\frac{6 \times 3}{6+3}=2 \Omega $

This equivalent resistor of resistance $2 \Omega$ is connected to a $2 \Omega$ resistor in series.

Therefore, the equivalent resistance of the circuit $=2 \Omega+2 \Omega=4 \Omega$

Hence the total resistance of the circuit is $4 \Omega$.

(b) The following circuit diagram shows the connection of the three resistors.

All the resistors are connected in series. Therefore, their equivalent resistance will be given as

$ \frac{1}{\frac{1}{2}+\frac{1}{3}+\frac{1}{6}}=\frac{1}{\frac{3+2+1}{6}}=\frac{6}{6}=1 \Omega $

Therefore, the total resistance of the circuit is $1 \Omega$.

Answer

There are four coils of resistances $4 \Omega, 8 \Omega, 12 \Omega$ and $24 \Omega$ respectively.

(a) If these coils are connected in series, then the equivalent resistance will be the highest, given by the sum $4+8+12+24=48 \Omega$

(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by

$ \frac{1}{\frac{1}{4}+\frac{1}{8}+\frac{1}{12}+\frac{1}{24}}=\frac{1}{\frac{6+3+2+1}{24}}=\frac{24}{12}=2 \Omega $

Therefore, $2 \Omega$ is the lowest total resistance.

Answer

The heating element of the heater is made up of alloy which has very high resistance so when current flows through the heating element, it becomes too hot and glows red. But the resistance of cord which is usually of copper or aluminium is very law so it does not glow.

Answer

Given Charge, $Q=96000 C$

Time, $t=1 hr=60 \times 60=3600 s$

Potential difference, $V=50$ volts

Now we know that $H=$ VIt

So we have to calculate / first

As $l=Q / t$

$\therefore I=96000 / 3600=80 / 3 A$

$ H=50 \times \frac{80}{3} \times 60 \times 60=4.8 \times 10^{6} J $

Therefore, the heat generated is $4.8 \times 10^{6} J$.

Answer

The amount of heat $(H)$ produced is given by the joule’s law of heating as $H=Vlt$ Where, Current, $I=5 A$

Time, $t=30 s$

Voltage, $V=$ Current $\times$ Resistance $=5 \times 20=100 V$

$H=100 \times 5 \times 30=1.5 \times 10^{4} J$.

Therefore, the amount of heat developed in the electric iron is $1.5 \times 10^{4} J$.

Answer

The rate of consumption of electric energy in an electric appliance is called electric power. Hence, the rate at which energy is delivered by a current is the power of the appliance.

Answer

Power $(P)$ is given by the expression, $P=V I$

Where,

Voltage, $V=220 V$

Current, $I=5 A$

$P=220 \times 5=1100 W$

Energy consumed by the motor $=Pt$

Where,

Time, $t=2 h=2 \times 60 \times 60=7200 s$

$\therefore P=1100 \times 7200=7.92 \times 10^{6} J$

Therefore, power of the motor $=1100 W$

Energy consumed by the motor $=7.92 \times 10^{6} J$

Answer

(d) 25

Answer

(b) $I R^{2}$

Answer

(d) $25 W$

Answer

(c) $1: 4$

Answer

To measure the potential difference between two points, a voltmeter should be connected in parallel to the points.

Answer

Area of cross-section of the wire, $A=\pi(d / 2) 2$

Diameter $=0.5 mm=0.0005 m$

Resistance, $R=10 \Omega$

We know that $R=\rho_A^{l}$

$l=\frac{R A}{\rho}$

$=\frac{10 \times 3.14 \times(\frac{0.0005}{2})^{2}}{1.6 \times 10^{-8}}$

$=\frac{10 \times 3.14 \times 25}{4 \times 1.6}=122.72 m$

$\therefore$ length of the wire $=122.72 m$

If the diameter of the wire is doubled, new diameter $=2 \times 0.5=1 mm=0.001 m$ Let new resistance be $R^{\prime}$

$ \begin{aligned} & R^{\prime}=\rho_A^{l} \\ & =\frac{1.6 \times 10^{-8} \times 122.72}{\pi(\frac{1}{2} \times 10^{-3})^{2}} \\ & =\frac{1.6 \times 10^{-8} \times 122.72 \times 4}{3.14 \times 10^{-6}} \\ & =250.2 \times 10^{-2}=2.5 \Omega \end{aligned} $

Therefore, the length of the wire is $122.7 m$ and the new resistance is $2.5 \Omega$.

Answer

The plot between voltage and current is called $I V$ characteristic. The voltage is plotted on $x$-axis and current is plotted on $y$-axis. The values of the current for different values of the voltage are shown in the given table.

The IV characteristic of the given resistor is plotted in the following figure.

The slope of the line gives the value of resistance $(R)$ as,

Slope $=1 / R=B C / A C=2 / 6.8$

$R=6.8 / 2=3.4 \Omega$

Therefore, the resistance of the resistor is $3.4 \Omega$.

Answer

Resistance $(R)$ of a resistor is given by Ohm’s law as, $V=I R$

$R=V / I$

Where,

Potential difference, $V=12 V$

Current in the circuit, $l=2.5 mA=2.5 \times 10^{-3} A$

$ R=\frac{12}{2.5 \times 10^{-3}}=4.8 \times 10^{3} \Omega=4.8 k \Omega $

Therefore, the resistance of the resistor is $4.8 k \Omega$

Answer

There is no current division occurring in a series circuit. Current flow through the component is the same, given by Ohm’s law as

$V=I R$

$I=V / R$

Where,

$R$ is the equivalent resistance of resistances $0.2 \Omega, 0.3 \Omega, 0.4 \Omega, 0.5 \Omega$ and $12 \Omega$. These are connected in series. Hence, the sum of the resistances will give the value of $R$.

$R=0.2+0.3+0.4+0.5+12=13.4 \Omega$

Potential difference, $V=9 V$

$l=9 / 13.4=0.671 A$

Therefore, the current that would flow through the $12 \Omega$ resistor is $0.671 A$.

Answer

For $x$ number of resistors of resistance $176 \Omega$, the equivalent resistance of the resistors connected in parallel is given by Ohm’s law as $V=I R$

$R=V / I$

Where,

Supply voltage, $V=220 V$

Current, $I=5 A$

Equivalent resistance of the combination $=R$, given as

$ \begin{aligned} & \frac{1}{R}=x \times(\frac{1}{176}) \\ & R=\frac{176}{x} \end{aligned} $

From Ohm’s law,

$ \begin{aligned} & \frac{V}{I}=\frac{176}{x} \\ & x=\frac{176 \times I}{V}=\frac{176 \times 5}{220}=4 \end{aligned} $

Therefore, four resistors of $176 \Omega$ are required to draw the given amount of current.

Answer

If we connect the resistors in series, then the equivalent resistance will be the sum of the resistors, i.e., $6 \Omega+6 \Omega+6 \Omega=18 \Omega$, which is not desired. If we connect the resistors in parallel, then the equivalent resistance will be $6 / 2=3 \Omega$ is also not desired. Hence, we should either connect the two resistors in series or parallel.

(a) Two resistor in parallel

Two $6 \Omega$ resistors are connected in parallel. Their equivalent resistance will be

$ \frac{1}{\frac{1}{6}+\frac{1}{6}}=\frac{6 \times 6}{6+6}=3 \Omega $

The third $6 \Omega$ resistor is in series with $3 \Omega$. Hence, the equivalent resistance of the circuit is $6 \Omega+$ $3 \Omega=9 \Omega$.

(b) Two resistor in series

Two $6 \Omega$ resistors are in series. Their equivalent resistance will be the sum $6+6=12 \Omega$. The third $6 \Omega$ resistor is in parallel with $12 \Omega$. Hence, equivalent resistance will be

$ \frac{1}{\frac{1}{12}+\frac{1}{6}}=\frac{12 \times 6}{12+6}=4 \Omega $

Therefore, the total resistance is $4 \Omega$.

Answer

Resistance $R_1$ of the bulb is given by the expression,

Supply voltage, $V=220 V$

Maximum allowable current, $I=5 A$

Rating of an electric bulb $P=10$ watts

Because $R=V^{2} / P$

$R 1=\frac{(220)^{2}}{10}=4840 \Omega$

According to Ohm’s law,

$V=I R$

Let $R$ is the total resistance of the circuit for $x$ number of electric bulbs

$R=V / I$

$=\frac{220}{5}=44 \Omega$

Resistance of each electric bulb, $R_1=4840 \Omega$

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\ldots . .$. upto $x$ times.

$\frac{1}{R}=\frac{1}{R_1} \times x$

$x=\frac{R_1}{R}=\frac{4840}{44}=110$

$\therefore$ Number of electric bulbs connected in parallel are 110 .

Answer

Supply voltage, $V=220 V$

Resistance of one coil, $R=24 \Omega$

(i) Coils are used separately

According to Ohm’s law,

$V=I_1 R_1$

Where,

$I_1$ is the current flowing through the coil

$I_1=V / R_1=220 / 24=9.166 A$

Therefore, 9.16 A current will flow through the coil when used separately.

(ii) Coils are connected in series

Total resistance, $R_2=24 \Omega+24 \Omega=48 \Omega$

According to Ohm’s law, $V=I_2 R_2$

Where,

$I_2$ is the current flowing through the series circuit

$I_2=V / R_2=220 / 48=4.58 A$

Therefore, $4.58 A$ current will flow through the circuit when the coils are connected in series.

(iii) Coils are connected in parallel

Total resistance, $R_3$ is given as $=$

$ \frac{1}{\frac{1}{24}+\frac{1}{24}}=\frac{24}{2}=12 \Omega $

According to Ohm’s law,

$V=I_3 R_3$

Where,

$I_3$ is the current flowing through the circuit $I_3=V / R_3=220 / 12=18.33 A$

Therefore, 18.33 A current will flow through the circuit when coils are connected in parallel.

Answer

(i) Potential difference, $V=6 V$ $1 \Omega$ and $2 \Omega$ resistors are connected in series. Therefore, equivalent resistance of the circuit, $R=$ $1+2=3 \Omega$

According to Ohm’s law,

$V=I R$

Where,

$I$ is the current through the circuit

$1=6 / 3=2 A$

This current will flow through each component of the circuit because there is no division of current in series circuits. Hence, current flowing through the $2 \Omega$ resistor is 2 A. Power is given by the expression, $P=(I)^{2} R=(2)^{2} \times 2=8 W$

(ii) Potential difference, $V=4 V$

$12 \Omega$ and $2 \Omega$ resistors are connected in parallel. The voltage across each component of a parallel circuit remains the same. Hence, the voltage across $2 \Omega$ resistor will be $4 V$.

Power consumed by $2 \Omega$ resistor is given by

$P=V^{2} / R=4^{2} / 2=8 W$

Therefore, the power used by $2 \Omega$ resistor is $8 W$.

Answer

Both the bulbs are connected in parallel. Therefore, potential difference across each of them will be $220 V$, because no division of voltage occurs in a parallel circuit.

Current drawn by the bulb of rating $100 W$ is given by,Power $=$ Voltage $x$ Current

Current $=$ Power/Voltage $=60 / 220 A$

Hence, current drawn from the line $=100 / 220+60 / 220=0.727$ A

Answer

Energy consumed by an electrical appliance is given by the expression, $H=Pt$

Where,

Power of the appliance $=P$

Time $=t$

Energy consumed by a TV set of power $250 W$ in $1 h=250 \times 3600=9 \times 10^{5} J$

Energy consumed by a toaster of power $1200 W$ in 10 minutes $=1200 \times 600$

Energy consumed by a toaster of power $1200 W$ in 10 minutes $=1200 \times 600$

$=7.2 \times 10^{5} J$

Therefore, the energy consumed by a $250 W$ TV set in $1 h$ is more than the energy consumed by a toaster of power $1200 W$ in 10 minutes.

Answer

Rate of heat produced by a device is given by the expression for power as, $P=I^{2} R$ Where, Resistance of the electric heater, $R=8 \Omega$

Current drawn, $I=15 A$

$P=(15)^{2} \times 8=1800 J / s$

Therefore, heat is produced by the heater at the rate of $1800 J / s$.

Answer

(a) The melting point and of Tungsten is an alloy which has very high melting point and very high resistivity so does not burn easily at a high temperature.

(b) The conductors of electric heating devices such as bread toasters and electric irons are made of alloy because resistivity of an alloy is more than that of metals which produces large amount of heat.

(c) In series circuits voltage is divided. Each component of a series circuit receives a small voltage so the amount of current decreases and the device becomes hot and does not work properly. Hence, series arrangement is not used in domestic circuits.

(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A), i.e. when area of cross section increases the resistance decreases or vice versa.

(e) Copper and aluminium are good conductors of electricity also they have low resistivity. So they are usually used for electricity transmission.