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

Answer 1

Answer: (C)

Internal resistance of voltmeter = $R$.
Effective resistance across $B$ and $C,R'$
$\frac{1}{R'}=\frac{1}{R}+\frac{1}{50}=\frac{50+R}{50R}$
or $R' =\left(\frac{50R}{50+R}\right)$
According to Ohm's law,
$V ' =IR '$
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
$R ''=50+\frac{50R}{50+R}$
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 $150 R=2500+100 R$
or $50 R=2500$
or $R=50\, k \,\Omega$