A) a resistance of \[19.92\,\,k\Omega \] parallel to the galvanometer
B) assistance of \[19.92\,\,k\Omega \] in series with the galvanometer
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 galvanometer \[5=(2.5\times {{10}^{-4}})(R+80)\] \[R=19.92\,\,k\Omega \]
You need to login to perform this action.
You will be redirected in
3 sec
