Q. 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 10 V (refer circuit digram) the resistance $R_{1}$ and $R_{2}$ respectively are
A) 950 $\Omega$ and 9150$\ \Omega$
B) 900 $\Omega$ and 9950$\ \Omega$
C) 900 $\Omega\ and\ 9000\Omega$
D) 950 $\Omega$ and 9950$\ \Omega$
Explanation
(d) Given, $V = 1V$
$I_{g} = 1mA$
$I_{g} = 1 \times 10^{- 3}A$
Resistance of galvanometer
$R_{g} = 50\Omega$
$R_{g} = \frac{V}{I_{g}} - R_{g}$
$or\ \ \ \ R_{s} = \frac{1}{10^{- 3}} - 50$
$or\ \ \ R_{1} = 950\Omega\ $
${R^{'}}_{2} = \frac{10}{10^{- 3}} - 50 = 9950\Omega$
$or\ \ R_{2} = {R'}_{2} - R_{1} = 9950 - 950\ $
$or\ \ \ R_{2} = 9000\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.