question_answer

A galvanometer of\[50\,\Omega \]resistance has 25 divisions. A current of\[4\times {{10}^{-4}}\]A gives a deflection of one division. To convert this galvanometer into a voltmeter having a range of 25 V, it should be connected with a resistance of

A) \[2500\,\Omega \]as a shunt

B) \[245\,\Omega \]as a shunt

C) \[2550\,\Omega \]in series

D) \[2450\,\Omega \]in series

Correct Answer: D

Solution :

 To convert a galvanometer into .voltmeter, high resistance should be connected in series with it. Let R is the resistance connected in series with the galvanometer. Galvanometer current \[{{i}_{g}}=\frac{V}{G+R}\] Or \[R=\frac{V}{{{i}_{g}}}-G\] Given,      \[G=50\,\Omega ,\] \[{{i}_{g}}=25\times 4\times {{10}^{-4}}={{10}^{-2}}A,V=25V\] \[\therefore \] \[R=\frac{25}{{{10}^{-2}}}-50=2500-50=2450\text{ }\Omega \]