In the circuit shown, if the current through the resistor $R$ is $\frac{1}{5} \mathrm{~A}$, the value of $R$ is

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Aicharya · Accepted Answer

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$ \begin{aligned} & \text { Now, equivalent emf, }=\frac{\frac{V_1}{R_1}-\frac{V_2}{R_2}}{\frac{1}{R_1}+\frac{1}{R_2}} \\ & \therefore \quad e=\frac{\frac{5}{2}-\frac{2}{1}}{\frac{1}{2}+\frac{1}{1}}=\frac{1}{3} \mathrm{~V} \end{aligned} $ $\therefore$ Internal resistance, $ r=\frac{R_1 \cdot R_2}{R_1+R_2} $ $\begin{array}{rlrl} & r & =\frac{2 \times 1}{2+1}=\frac{2}{3} \Omega \\ \therefore \text { Current, } \quad & I & =\frac{e}{r+R} \text { or } \frac{1}{5}=\frac{\frac{1}{3}}{\frac{2}{3}+R} \\ & & \frac{1}{5} & =\frac{\frac{1}{3}}{\frac{2+3 R}{3}} \\ \Rightarrow & & \frac{1}{5} & =\frac{3}{3(2+3 R)} \\ \Rightarrow & & 9 R & =15 \\ \Rightarrow & & 9 R & =15-6 \\ \text { or } & & & =9 \\R= & & 1 \Omega\end{array}$

In the circuit shown, if the current through the resistor $R$ is $\frac{1}{5} \mathrm{~A}$, the value of $R$…

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