Q. A voltmeter has a range $0 - V$ with a series resistance $R.$ With a series resistance $2R,$ the range is $0 - V^{'}.$ The correct relation between $V$and $V'$ is
A) $V^{'} = 2V$
B) $V^{'} > 2V$
C) $V^{'} \gg 2V$
D) $V^{'} < 2V$
Explanation
(d) For conversion of galvanometer (of resistance) into voltmeter, a resistance $R$ is connected in series
$\therefore i_{g} = \frac{V_{1}}{R + G}$ and $i_{g} = \frac{V_{2}}{2R + G}$
$\Rightarrow \frac{V_{1}}{R + G} = \frac{V_{2}}{2R + G} \Rightarrow \frac{V_{2}}{V_{1}} = \frac{2R + G}{R + G} = \frac{2(R + G) - G}{(R + G)}$
$= 2 - \frac{G}{(R + G)} \Rightarrow V_{2} = 2V_{1} - \frac{V_{1}G}{(R + G)} \Rightarrow V_{2} < 2V_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.