Question

The voltmeter shown in figure reads 18 V across 50 ohm resistor. The resistance of voltmeter is nearly

• A. 140 $\Omega$
• B. 128.5 $\Omega$
• C. 103 $\Omega$
• D. 162 $\Omega$

Answer 1

Answer: (B)

Voltmeter reads 18 V
$\Rightarrow$ Potential difference across 24 $\Omega$ resistor
= 30 - 18 = 12 V
$\Rightarrow$ Current through battery = $\frac{12}{24}=0.5A$
$\Rightarrow$ Let voltmeter resistance = x
Now current through 50 ohm is $\frac{18}{50}$
and current through voltmeter = $\frac{18}{x}$
$\frac{18}{50}+\frac{18}{x}=0.5$
$\Rightarrow x=128.5\Omega$