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Three identical resistors are connected in series with an ideal battery of emf $ε = 62 V$ . A voltmeter having finite resistance is used to measure the potential drop across each resistor. It reads $V_{A B} = V_{B C} = V_{C D} = 20 V$ . If the same voltmeter is used to measure potential drops across other pairs of points, then its reading will be

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V_{A C} = 41 V

V_{A C} = 40 V

V_{A D} = 60 V

V_{A D} = 62 V

answer is B, D.

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Detailed Solution

\frac{20}{R_{v}} + \frac{20}{R} = \frac{62 - 20}{2 R} \Rightarrow \frac{20}{R_{v}} = \frac{1}{R}

For AC: $\frac{V}{2 R} + \frac{V}{R_{v}} = \frac{62 - V}{R} \Rightarrow \frac{V}{2} + V \times \frac{1}{20} = 62 - V$

$\Rightarrow 31 V = 62 \times 20 \Rightarrow V_{A C} = 40$ volt

$V_{A D} = 62 v o l t$ (fill emf)

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