Q. A $100\ V$ voltmeter of internal resistance $20\ k\Omega$ in series with a high resistance $R$ is connected to a $110\ V$ line. The voltmeter reads $5\ V,$ the value of $R$ is
A) $210\ k\Omega$
B) $315\ k\Omega$
C) $420\ k\Omega$
D) $440\ k\Omega$
Explanation
(c) Here $i = \frac{110}{20 \times 10^{3} + R}$
$\because V = iR \Rightarrow 5 = \left( \frac{110}{20 \times 10^{3} + R} \right) \times 20 \times 10^{3}$
$\Rightarrow 10^{5} + 5R = 22 \times 10^{5} \Rightarrow R = 21 \times \frac{10^{5}}{5} = 420\ K\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.