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In the adjacent shown circuit, a voltmeter of internal resistance $R$, when connected across B and $\mathrm{C}$ reads $\frac{100}{3} \mathrm{~V}$. Neglecting the internal resistance of the battery, the value of $R$ is
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Verified Answer
The correct answer is:
$50 \mathrm{k} \Omega$
Internal resistance of voltmeter $=R$. Effective resistance across $B$ and $C$
$$
\frac{1}{R^{\prime}}=\frac{1}{R}+\frac{1}{50}=\frac{50+R}{50 R}
$$
or $R^{\prime}=\left(\frac{50 R}{50+R}\right)$
According to Ohm's law, $V^{\prime}=I R^{\prime}$
$$
\begin{array}{l}
\text { or } \frac{100}{3}=I \cdot\left(\frac{50 R}{50+R}\right) \\
\text { or } \frac{100}{3}\left(\frac{50+R}{50 R}\right)=I...(i)
\end{array}
$$
Now, total resistance of circuit
$$
\begin{array}{l}
R^{n}=50+\frac{50 R}{50+R} \\
\text { or } R^{\prime \prime}=\frac{(2500+100 R)}{(50+R)}
\end{array}
$$
Now, $V^{\prime \prime}=I R^{\prime \prime}$
$$
\Rightarrow 100=\frac{100}{3}\left(\frac{50+R}{50 R}\right) \frac{2500+100 R}{(50+R)}
$$
$$
\Rightarrow 150 R=2500+100 R \text { or } R=50 \mathrm{k} \Omega
$$
$$
\frac{1}{R^{\prime}}=\frac{1}{R}+\frac{1}{50}=\frac{50+R}{50 R}
$$
or $R^{\prime}=\left(\frac{50 R}{50+R}\right)$
According to Ohm's law, $V^{\prime}=I R^{\prime}$
$$
\begin{array}{l}
\text { or } \frac{100}{3}=I \cdot\left(\frac{50 R}{50+R}\right) \\
\text { or } \frac{100}{3}\left(\frac{50+R}{50 R}\right)=I...(i)
\end{array}
$$
Now, total resistance of circuit
$$
\begin{array}{l}
R^{n}=50+\frac{50 R}{50+R} \\
\text { or } R^{\prime \prime}=\frac{(2500+100 R)}{(50+R)}
\end{array}
$$
Now, $V^{\prime \prime}=I R^{\prime \prime}$
$$
\Rightarrow 100=\frac{100}{3}\left(\frac{50+R}{50 R}\right) \frac{2500+100 R}{(50+R)}
$$
$$
\Rightarrow 150 R=2500+100 R \text { or } R=50 \mathrm{k} \Omega
$$
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