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Question

In the circuit shown in figure reading of voltmeter is $${V}_{1}$$ when only $${S}_{1}$$ is closed, reading of voltmeter is $${V}_{2}$$ when only $${S}_{2}$$ is closed and reading of voltmeter is $${V}_{3}$$ when both $${S}_{1}$$ and $${S}_{2}$$ are closed. Then
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A
V2>V1>V3
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B
V1>V3>V2
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C
V3>V1>V2
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D
V1>V2>V3
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Solution

The correct option is A $${V}_{2}> {V}_{1}> {V}_{3}$$
In series: Potential difference $$\propto R$$
When only $$S_1$$ is closed $$V_1=\dfrac 34E=0.75\ E$$
When only $$S_2$$ is closed $$V_2=\dfrac 67E=0.86\ E$$
and when both $$S_1$$ and $$S_2$$ are closed combined resistance of $$ 6\ R$$ and $$3\ R$$ is $$2\ R$$
$$\therefore V_3 =\left( \dfrac 23 \right)E=0.67\ $$
$$E\Rightarrow V_2 > V_1 > V_3$$

Physics

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