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)

$50 \, k \Omega$

Answer 2

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 $R = 50 \,k\Omega$