Question

A 100 V voltmeter having an internal resistance of 20 k$\Omega$ when connected in series with a large resistance R across a 110 V line reads 5 V. The magnitude of R is :

• A. 210 k $\Omega$
• B. 315 k$\Omega$
• C. 420 k$\Omega$
• D. 440 k$\Omega$

Answer 1

Answer: (C)

420 k$\Omega$

Current through the galvanometer is,
$i_g=\frac{5}{20\times {10}^3}$
$i_g=2.5\times {10}^{-4}A$
$V=i_g(G+R)$
$110=2.5\times {10}^{-4}(20\times {10}^3+R)$
$440000=20000+R$
$R=420k\Omega$