question_answer

A galvanometer of resistance \[10\text{ }\Omega \] gives full-scale deflection when I mA current passes through it. The resistance required to convert it into a voltmeter reading upto 2.5 V is:

A) \[24.9\,\Omega \]                       

B) \[249\,\Omega \]

C) \[2490\,\Omega \]                      

D) \[24900\,\Omega \]

Correct Answer: C

Solution :

(c) \[2490\,\Omega \] Here,    \[{{l}_{g}}=1mA=1\times {{10}^{-3}}A,\,{{R}_{G}}=10\,\Omega \]             \[V=2.5\,V\] From the figure,         \[V={{l}_{g}}({{R}_{G}}+R)\] or \[R=\frac{V}{{{l}_{g}}}{{R}_{G}}\] Substituting the given values, we get, \[R=\frac{2.5\,V}{1\times {{10}^{-3}}A}-10\,\Omega =2500\,\Omega -10\,\Omega =2490\,\Omega \]