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.You need to login to perform this action.
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