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A galvanometer of resistance 20Ω requires a current of 6mA for full scale deflection. How do you modify it to measure a potential difference of 3 volts ?

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480Ω in parallel

520Ω In series

520Ω in parallel

480Ω in series

answer is B.

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Detailed Solution

$V = i_{g} (R + G); 3 = 6 \times 10^{- 3} \times (20 + R); R = 480 Ω$

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