Q1. [1] medium exam-ready
The heat produced in a conductor when current I flows through it for time t is H = I²Rt. If the current is doubled and the time is halved, the heat produced will be:
(A) remain the same
(B) doubled
(C) halved
(D) quadrupled
- A remain the same
- B doubled
- C halved
- D quadrupled
Generated by claude-sonnet-4-6 · 2026-06-26 01:08 · grounding rag
Model Answer
(A) remain the same
New heat: $H' = (2I)^2 \cdot R \cdot (t/2) = 4I^2R \cdot t/2 = 2I^2Rt$. Wait — correct option is (B) doubled.
When $I' = 2I$ and $t' = t/2$:
$$H' = (2I)^2 \cdot R \cdot \frac{t}{2} = 4I^2R \cdot \frac{t}{2} = 2I^2Rt = 2H$$
Answer: (B) doubled
Source: Chapter 11 – Electricity, Section 11.7 Heating Effect of Electric Current, Eq. (11.21)
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Explanation
- Use Joule's law: $H = I^2Rt$.
- Doubling $I$ increases $H$ by factor $4$ (since $I$ is squared); halving $t$ reduces it by factor $2$.
- Net effect: $4 \div 2 = 2$, so heat is doubled.
- A common mistake is forgetting that current is squared — don't just double it linearly.