Question

A 100 V voltmeter having an internal resistance of 20 kΩ when connected in series with a large resistance R across a 110 V line reads 5 V. Find out the magnitude of R-
(A) $210 k{\Omega}$
(B) $315 k{\Omega}$
(C) $420 k{\Omega}$
(D) $440 k{\Omega}$

Answer 1

Hint A voltmeter is an instrument used for measuring electric potential difference between two points in an electric circuit. The main principle of voltmeter is that it must be connected in parallel in which we want to measure the voltage. Parallel connection is used because a voltmeter is constructed in such a way that it has a very high value of resistance.

Complete step by step answer
Current through the galvanometer is
${i_g} = \dfrac{V}{r}$
${i_g} = 2.5 \times {10^{ - 4}}$ A
Now, the law of galvanometer we can say,
$V = {i_g}(G + R)$ .............[G stands for galvanometer]
Now, put the values that we know in the equation,
$\Rightarrow 110 = 2.5 \times {10^{ - 4}}(20 \times {10^3} + R)$
$\Rightarrow R = 440000 - 20000 = 420K\Omega$
$∴ R = 420 KΩ.$
Option (C) is correct.

Note
A galvanometer is an electromechanical instrument used for detecting and indicating an electric current. A galvanometer works as an actuator, by producing a rotary deflection (of a "pointer"), in response to electric current flowing through a coil in a constant magnetic field.