A) a resistance of \[19.92\,k\Omega \]parallel to the galvanometer
B) a resistance of \[19.92\,k\Omega \]in series with the galvanoyneter
C) a resistance of \[20\,k\Omega \]parallel to the galvanometer
D) a resistance of \[20\,k\Omega \]in series with galvanometer
Correct Answer: B
Solution :
The current through galvanometer producing full scale deflection is \[I=\frac{V}{R}=\frac{20\times {{10}^{-3}}}{80}\] \[=2.5\times {{10}^{-4}}A\] To convert galvanometer into. a voltmeter, a high resistance is connected in series with the galvanbmeter \[5=(2.5\times {{10}^{-4}})(R+80)\] \[R=19.92\,k\Omega \]
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