Two resistors of 100 Omega and 200 Omega ar | vedclass

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(B) Let the resistance of the voltmeter be $R_V$. The voltmeter is connected in parallel with the $100 \ \Omega$ resistor. The equivalent resistance o

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$200$

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(B) Let the resistance of the voltmeter be $R_V$. The voltmeter is connected in parallel with the $100 \ \Omega$ resistor. The equivalent resistance of this parallel combination is $R_p = \frac{100 \cdot R_V}{100 + R_V}$.The circuit now consists of $R_p$ in series with the $200 \ \Omega$ resistor,connected to a $4 \ V$ battery.According to the voltage divider rule,the voltage across the parallel combination $(R_p)$ is given by $V_p = V \cdot \frac{R_p}{R_p + 200}$.Given $V_p = 1 \ V$ and $V = 4 \ V$,we have $1 = 4 \cdot \frac{R_p}{R_p + 200}$.$R_p + 200 = 4 R_p \implies 3 R_p = 200 \implies R_p = \frac{200}{3} \ \Omega$.Equating the two expressions for $R_p$: $\frac{100 R_V}{100 + R_V} = \frac{200}{3}$.$300 R_V = 200(100 + R_V) \implies 300 R_V = 20000 + 200 R_V$.$100 R_V = 20000 \implies R_V = 200 \ \Omega$.

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