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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 resistance of the bulb.
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Solution
Given: E = 6V, r = 2Ω, i = 0.5A, t = 10 minute = 10 × 60s = 600s.
Let the resistance of bulb is R. Then i =`"E"/("R + r")`
or `0.5 = 6/("R" + 2)` or 0.5 (E + 2) = 6
or 0.5 R + 1 = 6 or 0.5 R = 6 - 1 = 5
or R = `5/0.5 = 10 Omega`
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