Q. The emf of a generator is 6 V and internal resistance is 0.5 $k\Omega$. The reading of a voltmeter having an internal resistance of 2.5 $k\Omega$ is
A) $10^{- 3}V$
B) $10\ V$
C) $5\ V$
D) $0.5\ V$
Explanation
(c) Total resistant$= 2.5 + 0.5 = 3.0\ k\Omega = 3000\ \Omega$
$\therefore$ Current, $i = \frac{6}{3000}A$;
Reading of voltmeter $= i \times (2.5 \times 1000)$
$= \frac{6}{3000} \times 2500 = 5V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.