Electricity — Question 14

(i) The potential difference is 6 V and the resistors 1 Ω and 2 Ω are connected in series:

R_s = 1 Ω + 2 Ω = 3 Ω.

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{6}{3} \\[1em] \Rightarrow \text{I} = 2 \text{ A}$

Power across 2 Ω resistor = I²R = 2 x 2 x 2 = 8 W

Hence, the power consumed by the 2 Ω resistor is 8 W.

(ii) In parallel circuit, voltage across the resistors remains the same. So, voltage across 2 Ω resistor is 4 V:

Power across 2 Ω resistor = $\dfrac{\text{V}^2}{{\text{R}}} = \dfrac{4 \times 4}{2}$ = 8 W

The power consumed by the 2 Ω resistor is 8 W.

Hence, same power is consumed by 2 Ω resistor in both the circuits.