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