A battery of 15 V and negligible internal resistance is connected across a 50 Ω resistor. The amount of energy dissipated as heat in the resistor in one minute is :

The energy dissipated as heat in a resistor is given by:
Q = P × t
where power
P is:
P = V²/R
Given:
V =15V,R=50Ω,
t = 1 minute = 60 seconds.
P = 15²/50 = 225/50 = 4.5 W
Q = 4.5 × 60 = 270J
Answer: (b) 270 J.
Class 12th Science Physics NCERT MCQ Questions
NCERT Books MCQ Questions Session 2024-2025.