question_answer

    A moving coil galvanometer has a resistance of\[10\text{ }\Omega \]. and full scale deflection of 0.01 A. It can be converted into voltmeter of 10 V full scale by connecting into resistance of

A)  \[\text{9}\text{.90 }\Omega \] in series

B)  \[\text{10 }\Omega \] in series

C)  \[\text{990 }\Omega \] in series

D)  \[\text{0}\text{.10 }\Omega \] in series

Correct Answer: C

Solution :

                Let G be resistance of galvanometer and\[{{i}_{g}}\]the current through it. Let Vis maximum potential difference, then from Ohms law                 \[{{i}_{g}}=\frac{V}{G+R}\] \[\Rightarrow \]               \[R=\frac{V}{{{i}_{g}}}-G\] Given,     \[G=10\text{ }\Omega ,\text{ }{{i}_{g}}=0.01\text{ }A\] \[V=10\text{ }volt\] \[\therefore \]  \[R=\frac{10}{0.01}-10=990\,\Omega \] Thus, on connecting a resistance R of\[990\,\Omega \]in series   with   the   galvanometer,   the galvanometer will become a voltmeter of range zero to 10 V. Note: For the voltmeter, a high resistance is series with the galvanometer and so the resistance of a voltmeter is very high compared to that of galvanometer.