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

Answer: (D)

$950\,\Omega$ and $9000\,\Omega$

Answer 2

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, ${{R}_{1}}=950\,\Omega ,{{R}_{2}}=9000\,\Omega$