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In the circuit shown in figure, the ammeter shows $\mathrm{5 A}$ current, voltmeter shows $\mathrm{250 V}$ and the internal resistance of the voltmeter is $\mathrm{2500\Omega}$ , then the value of $\mathrm{R}$ is:

Option: 1
$150\Omega$

Option: 2
$0.51\Omega$

Option: 3
$510\Omega$

Option: 4
$51\Omega$

Answers (1)

best_answer

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

Current flowering through the resistor $\mathrm{R}$ is

$\mathrm{I}_{\mathrm{R}}=5 \mathrm{~A}-\frac{1}{10} \mathrm{~A}=\frac{49}{10} \mathrm{~A}$

Potential difference across resistance $\mathrm{R}$ is $\mathrm{V}_{\mathrm{R}}=250 \mathrm{~V}$ [ $\because \mathrm{R}$ is connected in parallel with voltmeter ]

$\therefore \quad \mathrm{R}=\frac{\mathrm{V}_{\mathrm{R}}}{\mathrm{I}_{\mathrm{R}}}=\frac{250 \mathrm{~V}}{\frac{49}{10} \mathrm{~A}}=51 \Omega$

Hence option 4 is correct.

Posted by

Divya Prakash Singh

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