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. \[{{i}_{g}}=\frac{V}{G+R}\]or \[R=\frac{V}{{{i}_{g}}}=-G\] Given, \[G=50\,\Omega ,\] \[{{i}_{g}}=25\times 4\times {{10}^{-4}}={{10}^{-2}}A,V=25V\] \[\therefore \]\[R=\frac{25}{{{10}^{-2}}}-50=2500-50=2450\Omega \]