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  4. The maximum current that can be measured by a galvanometer of resistance 40 Ω 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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Q. 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)

KCETKCET 2004

Solution:

To convert a galvanometer into voltmeter, the necessary value of resistance to be connected in series with the galvanometer is
$ R = \frac{V}{I_g} - G $
$ = \frac{50}{10 \times 10^{-3}} - 40 $
$ = 5000 - 40 $
$ = 4960\Omega$