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 $G$ be the resistance of the voltmeter.From the circuit diagram,the cu

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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 $G$ be the resistance of the voltmeter.From the circuit diagram,the current through the resistor $R$ is $(I - I_g)$.The voltage across the resistor $R$ is equal to the voltage across the voltmeter,which is $V = 120 \ V$.Using Ohm's law for the resistor $R$:$V = (I - I_g) R$$120 = (2 - I_g) 	imes 75$$1.6 = 2 - I_g$$I_g = 2 - 1.6 = 0.4 \ A$Now,for the voltmeter,the voltage $V$ is given by $V = I_g 	imes G$:$120 = 0.4 	imes G$$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 $\Omega$.

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