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Capacitor in an AC Circuit

Three resisto...

Question

Three resistors are connected in series with ideal cell of emf 62 V.Measurements are carried out by a single voltmeter.The results of measurements are $V_{A B} = V_{B C} = V_{C D} = 20 V$ . Then

V_{A C} = 40 V

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V_{A B}

will be less than 40 v

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Resistance of voltemeter is

20 R_{1}

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R_{1} = R_{2} = R_{3}

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Solution

The correct options are
A $V_{A C} = 40 V$
C Resistance of voltemeter is $20 R_{1}$
D $R_{1} = R_{2} = R_{3}$
From symmetry $R_{1} = R_{2} = R_{3} = R$
let resistance of voltmeter be $R_{V}$
then

20= $i R_{V} = (I - i) R$ -----(1)
and $20 = 62 - 2 I R$ -----(2)
By (1) & (2)
$R_{V} = 20 R$

than $\Rightarrow$
$R_{e q} = R + \frac{2 R \times 20 R}{22 R}$
$R_{e q} = \frac{31}{11} R$
$I^{'} = \frac{62}{31} \times 11 = \frac{22}{R}$
$V_{A C} = I^{'} \times \frac{20}{11} R = \frac{20}{11} \times = 40 V$

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