Question

When a resistance of $60,000\Omega$ is connected in series with a voltmeter, the combination now can measure five times more voltage than the original voltmeter. The internal resistance of the voltmeter is :

• A. $12,000\Omega$
• B. $15,000\Omega$
• C. $2,40,000\Omega$
• D. $3,00,000\Omega$

Answer 1

Answer: (B)

$15,000\Omega$

Let Resistance of Voltmeter $=G$
$\frac{V_1}{V_2}=\frac{R_1}{R_2}$
$\frac{1}{5}=\frac{G}{G+60000}$
$G+60000=5G$
$4G=60000$
$G=15000\Omega$