NCERT Solutions for Class 10 Science Chapter 12 Electricity: Example Questions
Here, we have solved Examples 12.1 to 12.13 of Class 10 Science Chapter 12 Electricity in step-by-step manner. In CBSE and SEBA board exams, many numerical problems are asked directly from these solved examples or follow their exact patterns. However, the textbook often skips intermediate calculation steps; hence, students cannot understand the logic clearly.
In this article, every example is solved in a systematic way. Whether you are practising Ohm’s law, series and parallel resistor combinations, Joule’s law of heating, or electric power bills, these simplified solutions will help you master numerical problem-solving with confidence.
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NCERT Solutions for Class 10 Science Chapter 12 Electricity
Example 12.1 (Page No: 200)
A current of $0.5\text{ A}$ is drawn by a filament of an electric bulb for $10\text{ minutes}$. Find the amount of electric charge that flows through the circuit.
Answer: Given:
Current, $I = 0.5\text{ A}$
Time, $t = 10\text{ min} = 10 \times 60 = 600\text{ s}$
We know that, $I = \frac{Q}{t}$
Therefore, electric charge: $Q = I \times t$
$Q = 0.5 \times 600$
$Q = 300\text{ C}$
Hence, the amount of electric charge that flows through the circuit is $300\text{ C}$.
Example 12.2 (Page No: 202)
How much work is done in moving a charge of $2\text{ C}$ across two points having a potential difference $12\text{ V}$?
Answer: Given:
Charge, $Q = 2\text{ C}$
Potential difference, $V = 12\text{ V}$
We know that, $V = \frac{W}{Q}$
Therefore, the work done is: $W = V \times Q$
$W = 12 \times 2$
$W = 24\text{ J}$
Hence, the amount of work done in moving the charge is $24\text{ J}$.
Example 12.3 (Page No: 208)
(a) How much current will an electric bulb draw from a $220\text{ V}$ source, if the resistance of the bulb filament is $1200\ \Omega$? (b) How much current will an electric heater coil draw from a $220\text{ V}$ source, if the resistance of the heater coil is $100\ \Omega$?
Answer: (a) Given:
Potential difference, $V = 220\text{ V}$
Resistance of the bulb filament, $R = 1200\ \Omega$
According to Ohm’s law, $I = \frac{V}{R}$
$I = \frac{220}{1200}$
$I = 0.18\text{ A}$
Therefore, the current drawn by the electric bulb is $0.18\text{ A}$.
(b) Given:
Potential difference, $V = 220\text{ V}$
Resistance of the heater coil, $R = 100\ \Omega$
According to Ohm’s law, $I = \frac{V}{R}$
$I = \frac{220}{100}$
$I = 2.2\text{ A}$
Therefore, the current drawn by the electric heater coil is $2.2\text{ A}$.
Example 12.4 (Page No: 208)
The potential difference between the terminals of an electric heater is $60\text{ V}$ when it draws a current of $4\text{ A}$ from the source. What current will the heater draw if the potential difference is increased to $120\text{ V}$?
Answer: Given:
Initial potential difference, $V_1 = 60\text{ V}$
Initial current, $I_1 = 4\text{ A}$
New potential difference, $V_2 = 120\text{ V}$
According to Ohm’s law, $R = \frac{V_1}{I_1}$
$R = \frac{60}{4} = 15\ \Omega$
Now, when the potential difference is increased to $120\text{ V}$, the resistance remains constant ($15\ \Omega$). The new current drawn by the heater will be:
$I_2 = \frac{V_2}{R}$
$I_2 = \frac{120}{15}$
$I_2 = 8\text{ A}$
Hence, the heater will draw a current of $8\text{ A}$.
Example 12.5 (Page No: 208)
Resistance of a metal wire of length $1\text{ m}$ is $26\ \Omega$ at $20^\circ\text{C}$. If the diameter of the wire is $0.3\text{ mm}$, what will be the resistivity of the metal at that temperature? Using Table 12.2, predict the material of the wire.
Answer: Given:
Length of the wire, $\ell = 1\text{ m}$
Resistance, $R = 26\ \Omega$
Diameter, $d = 0.3\text{ mm} = 0.3 \times 10^{-3}\text{ m} = 3 \times 10^{-4}\text{ m}$
Radius, $r = \frac{d}{2} = \frac{3 \times 10^{-4}}{2} = 1.5 \times 10^{-4}\text{ m}$
We know that, $R = \rho \frac{\ell}{A}$
Therefore, the resistivity is given by:
$\rho = \frac{R A}{\ell} = \frac{R (\pi r^2)}{\ell}$
$\rho = \frac{26 \times 3.1416 \times (1.5 \times 10^{-4})^2}{1}$
$\rho = \frac{26 \times 3.1416 \times 2.25 \times 10^{-8}}{1}$
$\rho = 1.84 \times 10^{-6}\ \Omega\,\text{m}$
So, the resistivity of the metal at $20^\circ\text{C}$ is $1.84 \times 10^{-6}\ \Omega\,\text{m}$
From Table 12.2, we see that this is the resistivity of manganese.
Example 12.6 (Page No: 209)
A wire of given material having length $\ell$ 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 $\ell/2$ and area of cross-section $2A$?
Answer: For the first wire:
Length $= \ell$,
Area of cross-section $= A$,
Resistance $= R_1 = 4\ \Omega$.
We know that, $R_1 = \rho \frac{\ell}{A} = 4\ \Omega$ — (Equation 1)
For the second wire of the same material:
Length, $\ell’ = \frac{\ell}{2}$
Area of cross-section, $A’ = 2A$
Its resistance $R_2$ is given by:
$R_2 = \rho \frac{\ell’}{A’}$
$R_2 = \rho \frac{\ell/2}{2A} = \frac{1}{4} \left(\rho \frac{\ell}{A}\right)$ [from Equation 1]
$R_2 = \frac{1}{4} \times R_1$
$R_2 = \frac{1}{4} \times 4$
$R_2 = 1\ \Omega$
Hence, the resistance of the new wire is $1\ \Omega$.
Example 12.7 (Page No: 212)
An electric lamp, whose resistance is $20\ \Omega$, and a conductor of $4\ \Omega$ resistance are connected to a $6\text{ V}$ battery (Fig. 12.9). Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.
Answer: Given:
Resistance of the electric lamp, $R_1 = 20\ \Omega$
Resistance of the conductor, $R_2 = 4\ \Omega$
Potential difference of the battery, $V = 6\text{ V}$
(a) The total resistance of the circuit is: $R_s = R_1 + R_2$ $R_s = 20 + 4 = 24\ \Omega$
(b) According to Ohm’s law, the current through the circuit is: $I = \frac{V}{R_s}$
$I = \frac{6}{24} = 0.25\text{ A}$
(c) Potential difference across the electric lamp: $V_1 = I \times R_1$
$V_1 = 0.25 \times 20 = 5\text{ V}$
Potential difference across the conductor: $V_2 = I \times R_2$
$V_2 = 0.25 \times 4 = 1\text{ V}$
Hence, the potential difference across the electric lamp is $5\text{ V}$ and across the conductor is $1\text{ V}$.
Example 12.8 (Page No: 214)
In the circuit diagram given in Fig. 12.10, suppose the resistors $R_1$, $R_2$ and $R_3$ have the values $5\ \Omega$, $10\ \Omega$, $30\ \Omega$, respectively, which have been connected to a battery of $12\text{ V}$. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.
Answer: Given:
$R_1 = 5\ \Omega$, $R_2 = 10\ \Omega$, $R_3 = 30\ \Omega$
Potential difference, $V = 12\text{ V}$
(a)
Current through $R_1$: $I_1 = \frac{V}{R_1} = \frac{12}{5} = 2.4\text{ A}$
Current through $R_2$: $I_2 = \frac{V}{R_2} = \frac{12}{10} = 1.2\text{ A}$
Current through $R_3$: $I_3 = \frac{V}{R_3} = \frac{12}{30} = 0.4\text{ A}$
(b) The total current in the circuit,
$I = I_1 + I_2 + I_3$
$I = 2.4 + 1.2 + 0.4$
$I = 4\text{ A}$
Therefore, the total current in the circuit is $4\text{ A}$.
(c) The equivalent resistance $R_p$ of the parallel combination is given by:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
$\frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30}$
$\frac{1}{R_p} = \frac{6 + 3 + 1}{30} = \frac{10}{30} = \frac{1}{3}$
$\therefore R_p = 3\ \Omega$
Hence, the total circuit resistance is $3\ \Omega$.
Example 12.9 (Page No: 215)
In Fig. 12.12, $R_1 = 10\ \Omega$, $R_2 = 40\ \Omega$, $R_3 = 30\ \Omega$, $R_4 = 20\ \Omega$, $R_5 = 60\ \Omega$, and a $12\text{ V}$ battery is connected to the arrangement. Calculate (a) the total resistance in the circuit, and (b) the total current flowing in the circuit.
Answer: Given:
$R_1 = 10\ \Omega$, $R_2 = 40\ \Omega$, $R_3 = 30\ \Omega$, $R_4 = 20\ \Omega$, $R_5 = 60\ \Omega$
Potential difference, $V = 12\text{ V}$
(a) Let $R’$ be the equivalent resistance of the parallel combination of $R_1$ and $R_2$::
$\frac{1}{R’} = \frac{1}{R_1} + \frac{1}{R_2}$
$\frac{1}{R’} = \frac{1}{10} + \frac{1}{40}$
$\frac{1}{R’}= \frac{4 + 1}{40} = \frac{5}{40} = \frac{1}{8}$
$\therefore R’ = 8\ \Omega$
Let $R”$ be the equivalent resistance of the parallel combination of $R_3$, $R_4$, and $R_5$:
$\frac{1}{R”} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5}$
$\frac{1}{R”} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60}$
$\frac{1}{R”}= \frac{2 + 3 + 1}{60} = \frac{6}{60} = \frac{1}{10}$
$\therefore R” = 10\ \Omega$
Now, since $R’$ and $R”$ are connected in series, the total resistance of the circuit is:
$R_{\text{total}} = R’ + R”$
$R_{\text{total}} = 8 + 10 = 18\ \Omega$
Therefore, the total resistance in the circuit is $18\ \Omega$.
(b) According to Ohm’s law, $I = \frac{V}{R_{\text{total}}}$
$I = \frac{12}{18} = \frac{2}{3} \approx 0.67\text{ A}$
Hence, the total current flowing in the circuit is $0.67\text{ A}$.
Example 12.10 (Page No: 217)
An electric iron consumes energy at a rate of $840\text{ W}$ when heating is at the maximum rate and $360\text{ W}$ when the heating is at the minimum. The voltage is $220\text{ V}$. What are the current and the resistance in each case?
Answer: Given:
- Voltage, $V = 220\text{ V}$
(a) When heating is at the maximum rate:
Power, $P = 840\text{ W}$
We know that power is given by $P = VI$,
Thus, $I = \frac{P}{V}$.
$I = \frac{P}{V} = \frac{840}{220} = 3.82\text{ A}$
Again, according to Ohm’s law, $R = \frac{V}{I}$.
$R = \frac{V}{I} = \frac{220}{3.82} = 57.60\ \Omega$
(b) When heating is at the minimum rate:
Power, $P = 360\text{ W}$
Current, $I = \frac{P}{V} = \frac{360}{220} = 1.64\text{ A}$
Resistance, $R = \frac{V}{I} = \frac{220}{1.64} = 134.15\ \Omega$
Example 12.11 (Page No: 217)
$100\text{ J}$ of heat is produced each second in a $4\ \Omega$ resistance. Find the potential difference across the resistor.
Answer: Given:
Heat produced, $H = 100\text{ J}$
Time, $t = 1\text{ s}$
Resistance, $R = 4\ \Omega$
According to Joule’s law of heating: $H = I^2 R t$
Therefore, $I^2 = \frac{H}{R t}$
$I^2 = \frac{100}{4 \times 1} = 25$
$I = \sqrt{25} = 5\text{ A}$
Now, using Ohm’s law, the potential difference across the resistor is: $V = I \times R$
$V = 5 \times 4$
$V = 20\text{ V}$
Hence, the potential difference across the resistor is $20\text{ V}$.
Example 12.12 (Page No: 219)
An electric bulb is connected to a $220\text{ V}$ generator. The current is $0.50\text{ A}$. What is the power of the bulb?
Answer: Given:
Potential difference, $V = 220\text{ V}$
Current, $I = 0.50\text{ A}$
We know that electric power is given by: $P = V \times I$
$P = 220 \times 0.50$
$P = 110\text{ W}$
Hence, the power of the bulb is $110\text{ W}$.
Example 12.13 (Page No: 219)
An electric refrigerator rated $400\text{ W}$ operates $8\text{ hour/day}$. What is the cost of the energy to operate it for $30\text{ days}$ at $\text{Rs } 3.00\text{ per kWh}$?
Answer: Given:
Power of the refrigerator, $P = 400\text{ W}$
Time, $t = 8\text{ hours/day}$
Total number of days $= 30\text{ days}$
Cost per unit (kWh) $= \text{Rs } 3.00$
Total electrical energy consumed in 30 days is: $E = P \times \text{time}$
$E = 400\text{ W} \times (8\text{ h/day} \times 30\text{ days})$
$E = 400 \times 240\text{ Wh}$
$E = 96000\text{ Wh}$
∴ $E = \frac{96000}{1000} = 96\text{ kWh}$
Now, $\text{Total cost} = \text{Total energy consumed in kWh} \times \text{Rate per kWh}$
$\text{Total cost} = 96 \times 3.00 = \text{Rs } 288.00$
Therefore, the cost of energy to operate the refrigerator for 30 days is $\text{Rs } 288.00$.