In the adjoining circuit diagram,the readings | vedclass

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(C) Let $I$ be the total current,$I_g$ be the current through the voltmeter,and $R_v$ be the resistance of the voltmeter.From the circuit diagram,the

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$300$

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(C) Let $I$ be the total current,$I_g$ be the current through the voltmeter,and $R_v$ be the resistance of the voltmeter.From the circuit diagram,the total current $I = 2 \, A$ splits into the current through the resistor $R$ and the current through the voltmeter $I_g$.The voltage across the parallel combination is $V = 120 \, V$.The current through the resistor $R = 75 \, \Omega$ is given by $I_R = \frac{V}{R} = \frac{120}{75} = 1.6 \, A$.Since $I = I_R + I_g$,we have $I_g = I - I_R = 2 - 1.6 = 0.4 \, A$.The resistance of the voltmeter is $R_v = \frac{V}{I_g} = \frac{120}{0.4} = 300 \, \Omega$.

In the adjoining circuit diagram,the readings of the ammeter and voltmeter are $2 \, A$ and $120 \, V$,respectively. If the value of $R$ is $75 \, \Omega$,then the voltmeter resistance will be (in $\Omega$):

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