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:                                                                                                                   [AIPMT (S) 2004]

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 :

[d] 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,   \[G=50\Omega \]

\[{{i}_{g}}=25\times 4\times {{10}^{-4}}={{10}^{-2}}A,V=25V\]

\[\therefore \]      \[R=\frac{25}{{{10}^{-2}}}-50=2500-50=2450\,\Omega \]