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(A) The voltmeter measures the potential difference across the parallel combination of the resistors.$1$. When only $S_1$ is closed,the circuit consis

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$V_2 > V_1 > V_3$

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(A) The voltmeter measures the potential difference across the parallel combination of the resistors.$1$. When only $S_1$ is closed,the circuit consists of a resistor $R$ in series with $3R$. The total resistance is $R_{eq} = R + 3R = 4R$. The voltage across $3R$ is $V_1 = E 	imes \frac{3R}{4R} = \frac{3}{4} E = 0.75 E$.$2$. When only $S_2$ is closed,the circuit consists of a resistor $R$ in series with $6R$. The total resistance is $R_{eq} = R + 6R = 7R$. The voltage across $6R$ is $V_2 = E 	imes \frac{6R}{7R} = \frac{6}{7} E \approx 0.857 E$.$3$. When both $S_1$ and $S_2$ are closed,the resistors $3R$ and $6R$ are in parallel. Their equivalent resistance is $R_p = \frac{3R 	imes 6R}{3R + 6R} = \frac{18R^2}{9R} = 2R$. The total resistance is $R_{eq} = R + 2R = 3R$. The voltage across the parallel combination is $V_3 = E 	imes \frac{2R}{3R} = \frac{2}{3} E \approx 0.667 E$.Comparing the values: $V_2 \approx 0.857 E$,$V_1 = 0.75 E$,and $V_3 \approx 0.667 E$. Therefore,$V_2 > V_1 > V_3$.

In the circuit shown in the figure,the reading of the voltmeter is $V_1$ when only $S_1$ is closed,and the reading of the voltmeter is $V_2$ when only $S_2$ is closed. The reading of the voltmeter is $V_3$ when both $S_1$ and $S_2$ are closed. Then:

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