CBSE Class 10 Science 2019 · Paper 31/5/1 · Question 19(b)

Current drawn by the $40$ W; $220$ V bulb: $I_1=\frac{P}{V}=\frac{40}{220}=\frac{2}{11}$ A $\approx 0.18$ A. Its resistance: $R_1=\frac{V^2}{P}=\frac{220\times 220}{40}=1210\Omega$. With the $25$ W; $220$ V bulb: $I_2=\frac{25}{220}=\frac{5}{44}$ A $\approx 0.11$ A and $R_2=\frac{220\times 220}{25}=1936\Omega$. Yes, both change. At the same voltage, $I=\frac{P}{V}$ is proportional to the power and $R=\frac{V^2}{P}$ is inversely proportional to it, so the lower-power bulb draws less current and has a higher resistance. The current decreases by $\frac{3}{44}$ A $\approx 0.07$ A (from $0.18$ A to $0.11$ A), and the resistance increases by $726\Omega$ (from $1210\Omega$ to $1936\Omega$).