Question

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

• A. $100 \, k \Omega $
• B. $75 \, k \Omega $
• C. $50 \, k \Omega $
• D. $25 \, k \Omega $

Answer 1

Answer: (C)

Internal resistance of voltmeter is $R$.
Therefore effective resistance across $B$ and $C, R'$ is given by
$\frac{1}{R'} = \frac{1}{R} + \frac{1}{50} = \frac{50 +R}{50R } $
Or $ R ' = \frac{50 R}{50+R} $
According to Ohm's law
$ V' = IR' $
or $ \frac{100}{3} = I. \frac{50R}{50 +R} $
or $ \frac{100}{3} \frac{50 +R}{50R} = I $
Now, total resistance of circuit
$ R'' = 50 + \frac{50 R}{50+ R} $
OR $ R'' = \frac{2500 + 100 R}{50 +R}$
Now, $ V'' = I R''$
$ \Rightarrow 100 = \frac{100}{3} \frac{50+R}{50R} \frac{2500 + 100R}{50 +R} $
or $150R = 2500 + 100R $
or $50R = 2500$
or $R = 50 \,k\Omega $