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Question
A bulb is connected to a battery of e.m.f. 6V and internal resistance 2Ω A steady current of 0.5A flows through the bulb. Calculate the heat energy dissipated in the bulb in 10 minutes.
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Solution
Given: E = 6V, r = 2Ω, i = 0.5A, t = 10 minute = 10 × 60s = 600s.
The heat energy dissipated in the bulb = i2Rt
= (0.5)2 × 10 × 600 = 1500 J
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