As shown in the figure,the voltmeter reads 2, | vedclass

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(C) Let the resistance of the voltmeter be $R$. The voltmeter is connected in parallel with the $5\,\Omega$ resistor. The voltage across this parallel

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$20$

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(C) Let the resistance of the voltmeter be $R$. The voltmeter is connected in parallel with the $5\,\Omega$ resistor. The voltage across this parallel combination is $V_{p} = 2\,V$.The current through the $5\,\Omega$ resistor is $i_1 = \frac{V_{p}}{5\,\Omega} = \frac{2\,V}{5\,\Omega} = 0.4\,A$.The voltage across the $2\,\Omega$ resistor is $V_{2\Omega} = E - V_{p} = 3\,V - 2\,V = 1\,V$.The total current in the circuit is $i = \frac{V_{2\Omega}}{2\,\Omega} = \frac{1\,V}{2\,\Omega} = 0.5\,A$.The current through the voltmeter is $i_{v} = i - i_1 = 0.5\,A - 0.4\,A = 0.1\,A$.Since the voltmeter is in parallel with the $5\,\Omega$ resistor,the voltage across it is also $2\,V$. Thus,$R = \frac{V_{p}}{i_{v}} = \frac{2\,V}{0.1\,A} = 20\,\Omega$.

As shown in the figure,the voltmeter reads $2\,V$ across the $5\,\Omega$ resistor. The resistance of the voltmeter is $.......\,\Omega$.

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