Question

The resistance of a galvanometer is $50\Omega$ . and it shows full scale deflection for a current of $1mA$. To convert it into a voltmeter to measure $1V$ and as well as $10V$ (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 $I_g=1mA$
$=1\times {10}^{-3}A$
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, $R_1=950\Omega ,R_2=9000\Omega$