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

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Aicharya · Accepted Answer

Internal resistance of voltmeter is $R$. Therefore effective resistance across $B$ and $C, R^{\prime}$ is given by $\begin{aligned} \frac{1}{R^{\prime}} & =\frac{1}{R}+\frac{1}{50} \\ & =\frac{50+R}{50 R} \end{aligned}$ or $\quad R^{\prime}=\left(\frac{50 R}{50+R}\right)$ According to Ohm's law $\begin{aligned} V^{\prime} & =I R^{\prime} \\ \text { or } \quad \frac{100}{3} & =I \cdot\left(\frac{50 R}{50+R}\right) \end{aligned}$

Now, total resistance of circuit $\begin{aligned} R^{\prime \prime} & =50+\frac{50 R}{50+R} \\ \text { or } \quad R^{\prime \prime} & =\frac{(2500+100 R)}{(50+R)} \end{aligned}$ $\begin{array}{ll}\text { Now, } & V^{\prime \prime}=I R^{\prime \prime} \\ \Rightarrow & 100=\frac{100}{3}\left(\frac{50+R}{50 R}\right) \frac{2500+100 R}{(50+R)}\end{array}$ $\begin{array}{rlrl}\text { or } & 150 R =2500+100 R \\ \text { or } &50 R =2500 \\ \text { or } & R=50 \mathrm{k} \Omega\end{array}$

In the adjacent shown circuit, a voltmeter of internal resistance $R$, when connected across $B$ and $C$…

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