In the circuit shown in figure, the ammeter shows 5 A current, voltmeter shows 250 V and the internal resistance of the voltmeter is 2500Ω, then the value of R is :

51Ω

Current flowing through the voltmeter is $I_V = \frac{250 V}{2500 \Omega}=\frac{1}{10} A$

Current flowering through the resistor R is

$I_{R}=5 A-\frac{1}{10} A=\frac{49}{10} A$

Potential difference across resistance R is $V_{R}$ = 250 V [∵ R is connected in parallel with voltmeter]

∴ $R=\frac{V_{R}}{I_{R}}=\frac{250 V}{\frac{49}{10} A}=51 \Omega$