In the adjacent shown circuit, a voltmeter of internal resistance $R,$ when connected across $B$ and $C$ reads $\frac{100}{3}V.$ Neglecting the internal resistance of the battery, the value of $R$ is [image:3_image89.emf]

Question

Dr. Abubacker · Accepted Answer

(c) Internal resistance of voltmeter is $R$
Therefore effective resistance across $B$ and $C,$
[image:3_image176.emf]
$R''$ is given by
$\frac{1}{R^{''}} = \frac{1}{R} + \frac{1}{50} = \frac{50 + R}{50R}$
$\Rightarrow R^{''} = \left( \frac{50R}{50 + R} \right)$
According to ohm’s law
$V^{''} = IR''$
$\Rightarrow \frac{100}{3} = I.\left( \frac{50R}{50 + R} \right)$
$\Rightarrow \frac{100}{3}\left( \frac{50 + R}{50R} \right) = I$ …(ii)
Now, total resistance of circuit
$R^{''''} = 50 + \frac{50R}{50 + R}$
$\Rightarrow R" = \frac{(2500 + 100R)}{(50 + R)}$
Now, $V" = IR"$
$\Rightarrow 100 = \frac{100}{3}\left( \frac{50 + R}{50R} \right)\frac{2500 + 100R}{(50 + R)}$
$\Rightarrow 150R = 2500 + 100R$
$\Rightarrow R = 50\ k\Omega$

Q. In the adjacent shown circuit, a voltmeter of internal resistance $R,$ when connected across $B$ and $C$ reads $\frac{100}{3}V.$ Neglecting the internal resistance of the battery, the value of $R$ is

Explanation

Reviewed by Dr. Abubacker

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