Question

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

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

Answer 1

Answer: (C)

Internal resistance of voltmeter = $R$.
Effective resistance across $B$ and $C,R^{\prime }$
$\frac{1}{R^{\prime }}=\frac{1}{R}+\frac{1}{50}=\frac{50+R}{50R}$
or $R^{\prime }=\left(\frac{50R}{50+R}\right)$
According to Ohm's law,
$V^{\prime }=I{R}^{\prime }$
or $\frac{100}{3}=I.\left(\frac{50R}{50+R}\right)$
or $\frac{100}{3}\left(\frac{50+R}{50R}\right)=I...(i)$
Now, total resistance of circuit
$R^{\prime \prime }=50+\frac{50R}{50+R}$
or $R"=\frac{\left(2500+100R\right)}{50+R}$
now, $V"=IR"$
$\Rightarrow 100=\frac{100}{3}\left(\frac{50+R}{50R}\right)\frac{2500+100R}{\left(50+R\right)}$
or $150R=2500+100R$
or $50R=2500$
or $R=50k\Omega$