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

Answer 1

Answer: (C)

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