When a resistance of 60,000 ohm is connected in
series with a voltmeter, the combination now can
measure five times more voltage than the original
voltmeter. The initial internal resistance of the
voltmeter is
Answers
Answered by
0
Explanation:
240,000
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Answered by
1
Answer: Let us consider resistance=G
V1/V2 = R1/R2
1/5 = G/ G+60000
G+60000 = 5G
4G = 60000
G = 15000ohm
Explanation:
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