A resistor dissipates 192 J of energy in 1 s when a current of 4A is passed through it. Now, when the current is doubled, the amount of thermal energy dissipated in 5 s in______J.
Write Your Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
E = i²Rt
192 = 16 (R) (1)
R = 12 Ω
E₁ = (8)² (12) (5)
= 3840 J
Hello 👋 Welcome to Competishun – India’s most trusted platform for JEE & NEET preparation. Need help with JEE / NEET courses, fees, batches, test series or free study material? Chat with us now 👇