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Three voltmeters A, B and C having resistances $R, 1.5 R$ and $3 R$ respectively are used in a circuit as shown. When a potential difference is applied between $\mathrm{X}$ and $\mathrm{Y}$, the readings of the voltmeters are $V_{1}, V_{2}$, and $V_{3}$ respectively. Then
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The correct answer is:
$\mathrm{V}_{1}=\mathrm{V}_{2}=\mathrm{V}_{3}$
$$
\begin{array}{lc}
\text { Here, } & \mathrm{V}_{2}=\mathrm{V}_{3} \\
\text { ie, } & \mathrm{i}_{2} \times 1.5 \mathrm{R}=3 \mathrm{R} \times \mathrm{i}_{3} \\
\text { and } & \mathrm{i}_{2}+\mathrm{i}_{3}=\mathrm{i} \\
\Rightarrow & \mathrm{i}_{2}=\frac{2 \mathrm{i}}{3} \text { and } \mathrm{i}_{3}=\frac{\mathrm{i}}{3} \\
\text { Now, } & \mathrm{V}_{1}=\mathrm{IR} \\
& \mathrm{V}_{2}=\frac{2 \mathrm{i}}{3} \times 1.5 \mathrm{R}=\mathrm{iR} \\
\text { ie, } & \mathrm{V}_{3}=\frac{\mathrm{i}}{3} \times 3 \mathrm{R}=\mathrm{iR}
\end{array}
$$
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