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 $

Answer 1

The Correct Option is 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 $