A , B and C are voltmeters of resistance R , 1 .5 R and 3 R respectively as shown in the figure. When some…

$\text{A}$ , $\text{B}$ and $\text{C}$ are voltmeters of resistance $\text{R}$ , $\text{1}\text{.5 R}$ and $\text{3 R}$ respectively as shown in the figure. When some potential difference is applied between $\text{X}$ and $\text{Y}$ , the voltmeter readings are ${\mathrm{V}}_{\mathrm{A}},\mathrm{ }{\mathrm{V}}_{\mathrm{B}}$ and ${\mathrm{V}}_{\mathrm{C}}$ respectively. Then:

1. ${\mathrm{V}}_{\mathrm{A}}={\mathrm{V}}_{\mathrm{B}}={\mathrm{V}}_{\mathrm{C}}$
2. ${\mathrm{V}}_{\mathrm{A}}\ne {\mathrm{V}}_{\mathrm{B}}={\mathrm{V}}_{\mathrm{C}}$
3. ${\mathrm{V}}_{\mathrm{A}}={\mathrm{V}}_{\mathrm{B}}\ne {\mathrm{V}}_{\mathrm{C}}$
4. ${\mathrm{V}}_{\mathrm{A}}\ne {\mathrm{V}}_{\mathrm{B}}\ne {\mathrm{V}}_{\mathrm{C}}$

Solution

Voltage in parallel is constant
$\begin{aligned} & i_{P}=i_{B}+i_{C}, \quad V_{B}=V_{c} \\ & \frac{V_{p}}{R_{i}}=\frac{V_{B}}{R_{B}}+\frac{V_{C}}{R_{C}} \\ & \frac{1}{R_{p}}=\frac{1}{R_{B}}+\frac{1}{R_{C}} \Rightarrow \frac{1}{R^{\prime}}=\frac{2}{3 R}+\frac{1}{3 R} \\ \Rightarrow R^{\prime} & =R \\ & V_{X P}=V_{A}=i R \\ & V_{P Q}=V_{B}=V_{C}=i R \\ \Rightarrow V_{A} & =V_{B}=V_{C} \end{aligned}$

Asked in: NEET 2015 (Phase 1)

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