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A 100 ohm, 1 mA galvanometer becomes a 10 V voltmeter. Series resistance?
A10000 ohm
B9000 ohm
C9900 ohm
D9990 ohm
Answer & Solution
Correct answer: C. 9900 ohm
Total resistance needed is 10 V / 1 mA = 10000 ohm, so the series part is 10000 minus 100 = 9900 ohm.
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