Answer

(d) the same in all the cases

Explanation

There are no changes in any of the circuits, hence current will be same in all the circuits.

Answer

(c) maximum in case(ii)

Explanation

Here two transistors are in series. In figure (iii) total resistance will be less than individual resistances as they are connected parallel. Higher resistance produces more heat hence option c) is the right answer.

Answer

(d) nature of the material

Answer

(a) ${10}^{20}$

Explanation

$I=Q/t$

$Q=It$

$Q=1\times 16$

$Q=16q$

$Q=ne$

$n=Q/e$

$n=16/1.6\times {10}^{-19}$ $n=10\times {10}^{19}$

$n={10}^{20}$ electrons

The number of electrons flowing is ${10}^{20}$ electrons

Answer

(b) (ii)

Answer

(d) $1\mathrm{\Omega }$

Explanation

Maximum resistance is obtained when resistors are connected in series.

$R={\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{5}}$

$=5/5$

$=1\mathrm{\Omega }$

Answer

(b) $1/25\mathrm{\Omega }$

Explanation

Minimum resistance is obtained when resistors are connected parallel

${\displaystyle \frac{1}{R}}=5+5+5+5+5=25\mathrm{\Omega }$

$R={\displaystyle \frac{1}{25\mathrm{\Omega }}}$

Answer

(a) (i)

Explanation

Here positive terminal of next cell is adjacent to negative terminal of previous cell.

Answer

(a) Work done/(Current×Time)

Answer

(c) $2A$

Explanation

$P={\displaystyle \frac{RA}{l}}$

When Length doubles

$P={\displaystyle \frac{RA}{2l}}$

${\displaystyle \frac{RA}{l}}={\displaystyle \frac{RA}{2l}}$

$A=2A$

Answer

(c) R3 $>$ R2 $>$ R1

Explanation

Current flow is inversely proportional to resistance. Highest resistance will show less flow of current hence answer is c).

Answer

(c) $300\mathrm{\%}$

Explanation

Heat generated by a resistor is directly proportional to square of current. Hence, when current becomes double, dissipation of heat will multiply by $2=4$. This means there will be an increase of $300\mathrm{\%}$.

Answer

(c) the shape of the resistor is changed

Answer

(c) Brightness of bulb B will be more than that of A

Explanation

Bulbs are connected in parallel so resistance of combination would be less than arithmetic sum of resistance of all the bulbs. So. there will be no negative effect on flow of current. As a result, bulbs would glow according to their wattage.

Answer

(c) $20J$

Explanation

Equivalent resistance of the circuit is $R=4+2=6\mathrm{\Omega }$

current, $I=V/R)$ 6/6=1A

the heat dissipated by $4ohm$ resistor is, $H=I^{Rt}=20J$

Answer

(d) 5 A

Explanation

$P=V\times I$

Or $1000w=220vxI$

$I={\displaystyle \frac{1000w}{220v}}=4.54A$

$=5A$

Answer

(b) same current flowing through them when connected in series

Explanation

In series combination current does not get divided into branches because resistor receives a common current.

Answer

(b) kilowatt hour

Explanation

Volt-ampere (VA) is the unit used for the apparent power in an electrical circuit. A watt second (also watt-second, symbol W s or W. s) is a derived unit of energy equivalent to the joule. The joule-second is the unit used for Planck’s constant.

Answer

Answer

Current $P=I^{2}R$

$18W=I^2\times 2\mathrm{\Omega }$

$I^2=18W/2\mathrm{\Omega }$

$=9$

$I=3A$

This is the maximum current that can flow through the three resistors.

Answer

Resistance of ammeter should be zero because ammeter should not affect the flow of current.

Answer

Total resistance for parallel combination of 40 resistors can be calculated as follows:

${\displaystyle \frac{1}{R}}=\underset{―}{1}+{\displaystyle \frac{1}{4}}={\displaystyle \frac{1}{2}}$

$R=2\mathrm{\Omega }$

Thus, resistance of parallel combination is equal to resistance of resistors in series. So, potential difference across 20 resistance will be same as potential difference across the other two resistors which are connected in parallel.

Answer

Fuse wire has great resistance than the main wiring. When there is significant increase in the electric current. Fuse wire melts to break the circuit. This prevents damage of electrical appliance.

Answer

Property of the conductor which resists the flow of electric current is called resistivity. Resistance for a particular material is unique. Resistance is directly proportional to length of conductor and inversely proportional to current flow.

When length is doubled resistance becomes double and current flow reduces to half. This is the reason for the decrease in ammeter reading.

Answer

Commercial unit of electrical energy is kilowatt/hr

$\begin{array}{rl}1kw/hr& =1kWh\\ & =1000W\times 60\times 60s\\ & =3.6\times {10}^{6}J\end{array}$

Answer

1. Let $R$ be the resistance of the electric lamp. In series total resistance $=5+R$

$I=v/r$

$1=10/5+R$

$R=5ohm$

2. $V$ across Lamp + conductor $=10V$

V acoess Lamp $=I\times R=1\ast 5=5$ Volt

Answer

Parallel arrangement used in domestic wiring because it provides the same potential difference across each electrical appliance.

Answer

i) Potential difference does not get divided in parallel circuit. Hence glowing of other bulbs will not get affected when bulb one is fused.

ii) Ammeter A shows a reading of 3A. This means each of the ${\mathrm{A}}_1,{\mathrm{A}}_2$ and ${\mathrm{A}}_3$ show IA reading.

iii) $R={\displaystyle \frac{V}{I}}={\displaystyle \frac{4.5V}{3A}}=1.5\mathrm{\Omega }$

Now $P=I^{2}R$

$\begin{array}{rl}& =(3A{)}^2\times 1.5\mathrm{\Omega }\\ & =13.5W\end{array}$

Answer

(a) Resistance of the bulbs in series will be three times the resistance of single bulb. Hence, the current in the series combination will be one-third compared to current in each bulb in parallel combination. The parallel combination bulbs will glow more brightly.

(b) The bulbs in series combination will stop glowing as the circuit is broken and current is zero. However the bulbs in parallel combination shall continue to glow with the same brightness.

Answer

Ohm’s law states that at constant temperature potential difference (voltage) across an ideal conductor is proportional to the current through it.

$\underset{―}{V}=R$

I

Verification of Ohm’s law

Set up a circuit as shown in Fig. consisting of a nichrome wire XY of length, say $0.5m$, an ammeter, a voltmeter and four cells of $1.5V$ 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.

Answer

Resistivity is an inherent property of a conductor which resists the flow of electric current. Resistivity of each material is unique. SI unit of resistance is $\mathrm{\Omega }m$.

Experiment to study the factors on which the resistance of conducting wire depends.

Take a nichrome wire, a torch bulb, a $10W$ bulb and an ammeter (0 - $5A$ range), a plug key and some connecting wires.

Set up the circuit by connecting four dry cells of $1.5V$ each in series with the ammeter leaving a gap XY in the circuit, as shown in Fig. 12.4.

Observation:

It is observed that resistance depend on material of conductor

Length of conductor determines resistance

Resistance depends on area of cross section.

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 $10W$ bulb in the gap XY. $v$ 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.

Answer

• Collect three resistors ${\mathrm{R}}_1,{\mathrm{R}}_2$, ${\mathrm{R}}_3$ in seriesto make the circuit.
• Use ammeterto see the potential difference of three resistors.
• Remove ${\mathrm{R}}_1$ and take the reading of potential difference of ${\mathrm{R}}_2$ and ${\mathrm{R}}_3$
• Remove ${\mathrm{R}}_2$ and take the reading of potential difference of ${\mathrm{R}}_1$ and ${\mathrm{R}}_3$

Observation:

Ammeter reading was same in each case which shows same potential difference across three resistors connected in series.

Answer

Take three resistors ${\mathrm{R}}_1,{\mathrm{R}}_2$, and ${\mathrm{R}}_3$ and connect them in parallel to make a circuit; as shown in figure.

Use voltmeter to take reading of potential difference of three resistors in parallel combination.

• Now, remove the resistor ${\mathrm{R}}_1$ and take the reacting of potential difference of remaining resistors combination.
• Then, remove the resistor $R$, and take the reading of potential difference of remaining resistor.

Figure 12.7 Reststors in parallel

Observation:

In each case Voltmeter reading was same which shows that the same potential difference exists across three resistors connected in a parallel arrangement.

Answer

According to Joules heating effect heat produces in a resistor is

a) Directly proportional to square of current for the given resistor.

b) Directly proportional to resistance for a given current,

c) Directly proportional to the time of current flowing through the resistor.

This can be expressed as

$H=I^{2}Rt$

$H$ is heating effect, $I$ is electric current, $R$ is resistance and $t$ is time.

Experiment to demonstrate Joules law of heating

• Take a water heating immersion rod and connect to a socket which is connected to regulator. It Is important to recall that a regulator controls the amount of current flowing through a device.
• Keep the pointer of regulator on minimum and count the time taken by immersion rod to heat a certain amount of water.
• Increase the pointer of regulator to next level. Count the time taken by immersion rod to heat the same amount of water.
• Repeat above

step for higher levels on regulator to count the time.

Observation:

It is seen that with increased amount of electric current, less time is required o heat the same amount of water. This shows Joule’s Law of Heating.

Application:

Electric toaster, oven, electric kettle and electric heater etc. work on the basis of leafing effect of current.

Answer

(a) Effective resistance of the resistors connected in parallel is given by,

$\begin{array}{rl}& {\displaystyle \frac{1}{R}}={\displaystyle \frac{1}{R_1}}+{\displaystyle \frac{1}{R_2}}\\ & {\displaystyle \frac{1}{R}}={\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{8}}\\ & R={\displaystyle \frac{R_1\times R_2}{R_1+R_2}}\\ & R=\left({\displaystyle \frac{8\times 8}{8+8}}\right)=4\mathrm{\Omega }\end{array}$

(b) Total resistance $=4+4=8\mathrm{\Omega }$

$\text{ I }={\displaystyle \frac{\mathrm{V}}{\mathrm{R}}}={\displaystyle \frac{8}{8}}=1\mathrm{A}$

(c) Potential drop $=\mathrm{IR}=1\times 4=4\mathrm{V}$

(d) Power dissipated $=I^2\mathrm{R}=(1{)}^{2}4=4\mathrm{W}$

(e) The same current flows through each element in a series circuit. Therefore, there is no difference.