JEE Main 2026 — Apr 2, Shift 2 — Physics solution

JEE Main2026Apr 2, Shift 2Physics

Two resistors of $200 Ω$ and $400 Ω$ are connected in series with a battery of $100$ V. A bulb rated at $200$ V, $100$ W is connected across the $400 Ω$ resistance. The potential drop across the bulb is_______V.

A
$25$
B
$50$
C
$66.6$
D
$100$

Solution

Bulb resistance: $R_{b} = \frac{V ^{2}}{P} = \frac{20 0 ^{2}}{100} = 400 Ω$ .

The bulb in parallel with the $400 Ω$ resistor gives $\frac{400 \cdot 400}{800} = 200 Ω$ .

Now the circuit has $200 Ω$ (series) and $200 Ω$ (parallel combination) across $100$ V. By symmetry of equal resistances, the voltage divides equally: $V_{bulb} = 50$ V.

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