Two resistors of $200 \Omega$ and $400 \Omega$ are connected in series with a battery of 100 V. A bulb rated at $200 \mathrm{~V}, 100 \mathrm{~W}$ is connected across the $400 \Omega$ resistance. The potential drop across the bulb is $\_\_\_\_$ V.

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Current Electricity

Two resistors of

200Ω

and

400Ω

are connected in series with a battery of 100 V. A bulb rated at

200 V, 100 W

is connected across the

400Ω

resistance. The potential drop across the bulb is

____

Answer: (B)

The bulb's resistance follows from its rating,

R_{b} = \frac{V ^{2}}{P} = \frac{20 0 ^{2}}{100} = 400 Ω

. Connected across the

400 Ω

resistor it forms a parallel combination of

\frac{400 \times 400}{400 + 400} = 200 Ω

. This combination is in series with the other

200 Ω

resistor across the

100

V battery, so the supply divides equally and the voltage across the parallel section is

V = \frac{200}{200 + 200} \times 100 = 50 V

. The potential drop across the bulb is

50

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option