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Series Combination of Cells

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Question

The maximum current that can be measured by a galvanometer of resistance $40 Ω$ is $10 m A$ . 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= $\frac{V}{I_{g}} - R_{g}$
$= \frac{50 V}{0.01 A} - 40 Ω$ $= 4960 Ω$

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