1.

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 )A. 2010B. 4050C. 5040D. 4960

Answer» Correct Answer - d
To convert a galvanometer into voltmeter, the necessary value of resistance to be connected in series with the galvanometer is
`R=V/(I_(g))-G=50/(10xx10^(-3)) -40 =5000-40 =4960 Omega`


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