Q. A voltmeter with a resistance $50 \times 10^{3}\Omega$ is used to measure voltage in a circuit. To increase its range to 3 times, the additional resistance to be put in series is
A) $9\ \times 10^{6}\ \Omega$
B) $10^{5}\ \Omega$
C) $1.5\ \times 10^{5}\ \Omega$
D) $9\ \times 10^{5}\ \Omega$
Explanation
(b) $V = i_{g}R\ \ \ \ \ \ \ \ and\ \ V^{'} = i_{g}R^{'}\ \ \ \ \ or\ \frac{R^{'}}{R} = \frac{V^{'}}{V}$
Or $R^{'} = \frac{V^{'}}{V}R = \frac{3V}{V} \times 50 \times 10^{3} = 1.5\ \times 10^{5}\Omega$
$\therefore\ \ \ \ \ $Additional resistance
$= 1.5 \times 10^{5} - 0.5\ \times 10^{5} = 10^{5}\ \Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.