NCERT Class 10 Science Chapter 11 Electricity

Q1

A piece of wire of resistance $R$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is $R’$, then the ratio $R/R’$ is:

(a) 1/25 (b) 1/5 (c) 5 (d) 25

Answer: (d) 25

Explanation: When a wire of resistance $R$ is cut into 5 equal parts, the resistance of each part becomes $R/5$.

↔ Swipe Calculation

$$\frac{1}{R’} = \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5} + \frac{1}{R/5}$$

$$\frac{1}{R’} = \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} + \frac{5}{R} = \frac{25}{R}$$

$$R’ = \frac{R}{25} \Rightarrow \frac{R}{R’} = 25$$

Q2

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

(a) $I^2R$ (b) $IR^2$ (c) $VI$ (d) $V^2/R$

Answer: (b) $IR^2$

Reason: Power is given by $P = VI$. From Ohm’s law ($V=IR$), we can derive:

• $P = (IR)I = I^2R$
• $P = V(V/R) = V^2/R$
Therefore, $IR^2$ is the incorrect formula.

Q3

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 (c) 50 W (d) 25 W

Answer: (d) 25 W

First, we find the constant resistance ($R$) of the bulb.

↔ Swipe Calculation

$$R = \frac{V^2}{P} = \frac{220^2}{100} = \frac{48400}{100} = 484 \, \Omega$$

Now, calculate power at 110 V:

$$P_{new} = \frac{V_{new}^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25 \, \text{W}$$

Q4

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 to parallel combinations would be:

(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1

Answer: (c) 1:4

Let resistance of each wire be $R$.

Series ($R_s$): $R + R = 2R$
Parallel ($R_p$): $1/R_p = 1/R + 1/R \Rightarrow R_p = R/2$

Heat ($H$) = $\frac{V^2}{R} \cdot t$ (since V is constant).

↔ Swipe Calculation

$$\frac{H_s}{H_p} = \frac{V^2/R_s}{V^2/R_p} = \frac{R_p}{R_s} = \frac{R/2}{2R} = \frac{1}{4}$$

Q5

How is a voltmeter connected in the circuit to measure the potential difference between two points?

A voltmeter is always connected in Parallel across the points between which the potential difference is to be measured.

Q6

A copper wire has diameter 0.5 mm and resistivity of $1.6 \times 10^{-8} \, \Omega \text{ m}$. What will be the length of this wire to make its resistance 10 $\Omega$?

Given: $R=10\Omega$, $\rho=1.6\times10^{-8} \Omega\text{m}$, $d=0.5\text{ mm} = 5\times10^{-4}\text{ m}$.

First, find Area ($A$):

$$A = \frac{\pi d^2}{4} = \frac{3.14 \times (5\times10^{-4})^2}{4} = 1.96 \times 10^{-7} \, \text{m}^2$$

Now, find Length ($L$):

↔ Swipe Calculation

$$R = \rho \frac{L}{A} \Rightarrow L = \frac{R \times A}{\rho}$$

$$L = \frac{10 \times 1.96 \times 10^{-7}}{1.6 \times 10^{-8}} = 122.7 \, \text{m}$$

Q8

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.

Given: $V = 12$ V, $I = 2.5$ mA = $2.5 \times 10^{-3}$ A.

↔ Swipe Calculation

$$R = \frac{V}{I} = \frac{12}{2.5 \times 10^{-3}} = \frac{12000}{2.5} = 4800 \, \Omega$$

$$R = 4.8 \, \text{k}\Omega$$

Q9

A battery of 9 V is connected in series with resistors of 0.2 $\Omega$, 0.3 $\Omega$, 0.4 $\Omega$, 0.5 $\Omega$ and 12 $\Omega$, respectively. How much current would flow through the 12 $\Omega$ resistor?

Total Resistance ($R_s$): Since they are in series, add all resistances.

$$R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 \, \Omega$$

Current ($I$):

$$I = \frac{V}{R_s} = \frac{9}{13.4} = 0.67 \, \text{A}$$

In a series circuit, current is the same through all resistors. Therefore, 0.67 A flows through the 12 $\Omega$ resistor.

Q10

How many 176 $\Omega$ resistors (in parallel) are required to carry 5 A on a 220 V line?

Let the number of resistors be $n$.

First, find the required Equivalent Resistance ($R_{eq}$):

$$R_{eq} = \frac{V}{I} = \frac{220}{5} = 44 \, \Omega$$

For $n$ equal resistors in parallel: $R_{eq} = R/n$

$$44 = \frac{176}{n} \Rightarrow n = \frac{176}{44} = 4$$

Answer: 4 resistors.

Q11

Show how you would connect three resistors, each of resistance 6 $\Omega$, so that the combination has a resistance of (i) 9 $\Omega$, (ii) 4 $\Omega$.

(i) To get 9 $\Omega$: Connect two in parallel, and one in series.

$$R_{eq} = 6 + (6 || 6) = 6 + \frac{6 \times 6}{6+6} = 6 + 3 = 9 \, \Omega$$

(ii) To get 4 $\Omega$: Connect two in series, and that combination in parallel with the third.

$$R_{eq} = (6+6) || 6 = 12 || 6 = \frac{12 \times 6}{12+6} = \frac{72}{18} = 4 \, \Omega$$

Q12

Several electric bulbs designed to be used on a 220 V line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

1. Current drawn by one bulb ($I_1$):

$$I_1 = \frac{P}{V} = \frac{10}{220} = \frac{1}{22} \, \text{A}$$

2. Total allowable current ($I_{total}$) = 5 A.

3. Number of bulbs ($n$):

$$n \times I_1 = I_{total} \Rightarrow n \times \frac{1}{22} = 5$$

$$n = 5 \times 22 = 110$$

Answer: 110 bulbs.

Q14

Compare the power used in the 2 $\Omega$ resistor in each of the following circuits:
(i) a 6 V battery in series with 1 $\Omega$ and 2 $\Omega$ resistors, and
(ii) a 4 V battery in parallel with 12 $\Omega$ and 2 $\Omega$ resistors.

Case (i) Series:

$$I = \frac{V}{R_1+R_2} = \frac{6}{1+2} = 2 \, \text{A}$$

$$P = I^2R = (2)^2 \times 2 = 8 \, \text{W}$$

Case (ii) Parallel: Voltage across parallel branches is same (4 V).

$$P = \frac{V^2}{R} = \frac{4^2}{2} = \frac{16}{2} = 8 \, \text{W}$$

Result: Power is equal (8 W) in both cases.

Q16

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

TV Set:

$$E = P \times t = 250 \times 1 = 250 \, \text{Wh}$$

Toaster: (Convert 10 min to hours = 10/60 hr)

$$E = 1200 \times \frac{10}{60} = 1200 \times \frac{1}{6} = 200 \, \text{Wh}$$

Result: The TV set uses more energy.

Q18

Explain the following:

(a) Why is tungsten used almost exclusively for filament of electric lamps?
Tungsten has a very high melting point ($3380^\circ$C) and does not oxidize (burn) easily at high temperatures. This allows it to get very hot and emit light without melting.

(b) Why are alloys used in heating devices like toasters?
Alloys have higher resistivity than pure metals and do not oxidize easily at high temperatures, making them ideal for producing heat.

(d) How does resistance vary with area of cross-section?
Resistance is inversely proportional to the area of cross-section ($R \propto 1/A$). A thicker wire has less resistance.