An $80 Ω$ galvanometer deflects full-scale for a potential of 20 mV. A voltmeter deflecting full-scale of 5 V is to be made using this galvanometer. We must connect

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a resistance of 20 $k Ω$ parallel to the galvanometer

a resistance of 19.92 $k Ω$ parallel to the galvanometer

a resistance of 19.92 $k Ω$ in series with the galvanometer

a resistance of 20 $k Ω$ in series with the galvanometer

answer is B.

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Detailed Solution

The current through the galvanometer producing full-scale deflection is

$I = \frac{V}{R} = \frac{20 \times 10^{- 3}}{80} = 2.5 \times 10^{- 4} A$

To convert the galvanometer into a voltmeter, a high resistance connected in series with the galvanometer. Therefore,

$5 V = (2.5 \times 10^{- 4}) (R + 80) o r R = 19.92 k Ω$

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