Physics Circle home Student Login

Home›Chapters›Current Electricity PYQs›Electrical Power & Heating Effect›Question 5

JEE Main 2026 (2 Apr, Shift 2) · Electrical Power & Heating Effect

Electrical Power & Heating Effect question from JEE Main 2026 (2 Apr, Shift 2)

Q5JEE Main 2026 · 2 APR · SHIFT 2Medium

Electrical Power & Heating Effect

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

Show answer and solution

Answer: (b)

  1. Bulb: Rb=2002100=400 ΩR_b = \dfrac{200^2}{100} = 400\ \Omega.
  2. 400∥400=200 Ω400 \parallel 400 = 200\ \Omega; total 200+200=400 Ω200 + 200 = 400\ \Omega.
  3. I=100/400=0.25 AI = 100/400 = 0.25\text{ A}, so Vbulb=0.25×200=50 VV_{bulb} = 0.25 \times 200 = 50\text{ V}.
  4. So the answer is (b) 50.
Report a mistake

More questions on Electrical Power & Heating Effect