A resistor develops 300 J of thermal energy in 15 s, when a current of 2 A is passed through it. If the current increases to 3 A, the energy developed in 10 s is ____________ J.

Solution

Physics sir · 4 steps

Find R from the first trial with Joule's law, then reuse it for 3 A and 10 s.

1. Joule heating. Heat developed in a resistor is H = I² R t. The resistor is the same in both cases, so R stays fixed.

   H = I² R t

2. Find R. From the first trial, 300 = (2)² × R × 15 = 60R. So R = 5 Ω.
3. New heat. With I = 3 A and t = 10 s: H = (3)² × 5 × 10 = 450 J. As a check, 300 × 9/4 × 10/15 = 450.
4. So. The energy developed in 10 s is 450 J.

The common slip: 675. Scaling only the current and keeping the time at 15 s gives 300 × 9/4 = 675 J. The time also changes, from 15 s to 10 s.

The idea to keep. For the same resistor, heat goes as I² t. Scale by the square of the current ratio and by the time ratio, and you never need R itself.

Answer from the NTA final answer key. Solution by PrepWiser.