The maximum current that can be measured by a galvanometer of resistance $40 \,\Omega$ is $10\, mA$. It is converted into a voltmeter that can read upto $50\, V$. The resistance to be connected in series with the galvanometer is ... (in $ohm$)
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(b) ${i_g} = \frac{{50}}{{10 \times {{10}^{ - 3}}}} - 40 = 4960\,\Omega $
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