Question

Sol. Key concept: Rated filament resistance and voltage division in series.

Using the usual constant-resistance model for the hot filaments, R = V²/P. Therefore the 60 W bulb has a larger resistance than the 100 W bulb, although both are rated at 220 V.

$$R_{1} = \frac{220^{2}}{100} = 484\ \Omega,\ \ \ \ \ R_{2} = \frac{220^{2}}{60}$$

$$R_{1}:R_{2} = 3:5,\ \ \ \ \ V_{1} = \frac{3}{8}(440) = 165\ V$$

$$V_{2} = \frac{5}{8}(440) = 275\ V$$

B₂ receives more than its rated voltage and is the bulb expected to fuse. When it breaks, the series circuit opens, so B₁ does not subsequently receive the full supply voltage. Hence (b).