Electricity — Class 10 NCERT: 18 MCQs with answers

The rate of flow of electric charge

Electric current is the rate of flow of electric charge, given by $I = Q/t$. The other options describe charge, potential difference and resistance respectively.

15 C

Using $Q = I \times t = 3 \times 5 = 15$ C. The other values come from dividing or adding instead of multiplying.

Ammeter, to measure the current through a component

An ammeter is connected in series so that the same current flows through it as through the component being measured. A voltmeter is connected in parallel.

$1\ \text{V} = 1\ \text{J/C}$

Potential difference is work done per unit charge, so $1\ \text{V} = 1\ \text{J/C}$. The option $1\ \text{C/s}$ is the ampere.

Current

In series, there is only one path, so the current is the same through each resistor; the voltage divides across them.

True

Ohm's law ($V = IR$) holds for metallic conductors only when temperature remains constant.

The battery maintains a potential difference that drives the current

The battery maintains a potential difference; that difference drives the current. Current is not used up — energy is converted into heat and light.

$6\ \Omega$

In series, $R_s = R_1 + R_2 = 2 + 4 = 6\ \Omega$. The value $1.33\ \Omega$ would be the parallel combination.

$1.33\ \Omega$

$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$, so $R_p = \frac{4}{3} \approx 1.33\ \Omega$. The parallel result is always less than the smallest resistor.

2 A

By Ohm's law, $I = V/R = 12/6 = 2$ A. The value 72 A comes from multiplying instead of dividing.

Shape of the wire's cross-section being circular or square

Resistance depends on length, area of cross-section, material and temperature. The shape of the cross-section does not matter as long as the area is the same.

So that each appliance gets the full supply voltage and can be switched independently

In parallel, each appliance receives the full supply voltage and can be operated independently. In series, one failure would break the whole circuit.

Heating effect of current

A fuse works on the heating effect of current: its thin wire melts when the current exceeds a safe value, breaking the circuit.

It becomes four times the original

Since $R = \rho l / A$, doubling $l$ doubles $R$ and halving $A$ doubles it again, giving $4R$. Both changes act in the same direction.

500 W

Using $P = I^2R = 5^2 \times 20 = 25 \times 20 = 500$ W. The value 100 W comes from using $IR$ instead of $I^2R$.

The same current enters and leaves a resistor; energy is converted

Current is not used up in a circuit. The same current enters and leaves a resistor; what gets converted is electrical energy into heat or light.

2 A

To find the total current, first calculate the equivalent resistance \( R_p \) of the parallel resistors using \( rac{1}{R_p} = rac{1}{R_1} + rac{1}{R_2} \). Thus, \( rac{1}{R_p} = rac{1}{3} + rac{1}{6} = rac{2}{6} + rac{1}{6} = rac{3}{6} \), which gives \( R_p = 2 \, \Omega \). Now, use Ohm's law \( I = rac{…

15 V

First, calculate the total resistance \( R_s = R_1 + R_2 + R_3 = 10 + 5 + 15 = 30 \, \Omega \). The total current \( I \) through the circuit is \( I = rac{V}{R_s} = rac{30}{30} = 1 \, A \). The voltage drop across the \( 10 \, \Omega \) resistor is given by \( V = IR = 1 imes 10 = 10 \, V \). Therefore, the voltage…