Question

The resistance of a galvanometer is $ 50\,\Omega $ . and it shows full scale deflection for a current of $1 \,mA$. To convert it into a voltmeter to measure $1 \,V$ and as well as $10\, V$ (Refer circuit diagram) the resistances $ {{R}_{1}} $ and $ {{R}_{2}} $ respectively are

• A. $ 950\,\Omega $ and $ 9150\,\Omega $
• B. $ 900\,\Omega $ and $ 9950\,\Omega $
• C. $ 900\,\Omega $ and $ 9900\,\Omega $
• D. $ 950\,\Omega $ and $ 9000\,\Omega $
• E. $ 950\,\Omega $ and $ 9950\,\Omega $

Answer 1

Answer: (D)

Given: $ V=1 $ volt $ {{I}_{g}}=1\,mA $
$ =1\times {{10}^{-3}}A $
Resistance of a galvanometer $ G=50\,\Omega $
$ \therefore $ $ {{R}_{T}}=\frac{V}{{{I}_{g}}}-G $
or $ {{R}_{T}}=\frac{1}{{{10}^{-3}}}-50 $
or $ {{R}_{1}}=950\,\Omega $
AISO $ {{R}_{2}}=\frac{10}{{{10}^{-3}}}-50 $
$ =9950\,\Omega $ Additional resistance for attaining
$ 10V=9950-950 $
$ =9000\,\Omega $
Hence, $ {{R}_{1}}=950\,\Omega ,{{R}_{2}}=9000\,\Omega $