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Question

The maximum current that can be measured by a galvanometer of resistance 40Ω is 10 mA. If it is converted into a voltmeter that can read upto 50 V. the resistance to be connected in series with the galvanometer (in ohms) is:

A
2010
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B
4050
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C
5040
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D
4960
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Solution

The correct option is D 4960
Resistance connected in series with a galvanometer to make it a voltmeter=VIgRg
=50 V0.01 A40 Ω =4960 Ω

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