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(D) Let the two batteries be $V_1 = 5 \, V$ with internal resistance $r_1 = 2 \, \Omega$ and $V_2 = 2 \, V$ with internal resistance $r_2 = 1 \, \Omeg

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(D) Let the two batteries be $V_1 = 5 \, V$ with internal resistance $r_1 = 2 \, \Omega$ and $V_2 = 2 \, V$ with internal resistance $r_2 = 1 \, \Omega$.Using the formula for equivalent $EMF$ $(E_{eq})$ and equivalent internal resistance $(r_{eq})$ for two parallel branches:$E_{eq} = \frac{\frac{V_1}{r_1} + \frac{V_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{5}{2} + \frac{2}{1}}{\frac{1}{2} + 1} = \frac{4.5}{1.5} = 3 \, V$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 	imes 1}{2 + 1} = \frac{2}{3} \, \Omega$The current $I$ through the resistor $R$ is given by $I = \frac{E_{eq}}{r_{eq} + R}$.Given $I = \frac{1}{5} \, A$, we have:$\frac{1}{5} = \frac{3}{\frac{2}{3} + R}$$\frac{2}{3} + R = 15$$R = 15 - \frac{2}{3} = \frac{45 - 2}{3} = \frac{43}{3} \, \Omega \approx 14.33 \, \Omega$.Wait, re-evaluating the circuit polarity: If the batteries are opposing, $E_{eq} = \frac{\frac{V_1}{r_1} - \frac{V_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{2.5 - 2}{1.5} = \frac{0.5}{1.5} = \frac{1}{3} \, V$.Then, $\frac{1}{5} = \frac{1/3}{2/3 + R} \implies \frac{2}{3} + R = \frac{1/3}{1/5} = \frac{5}{3}$.$R = \frac{5}{3} - \frac{2}{3} = \frac{3}{3} = 1 \, \Omega$.

In the circuit shown, if the current through the resistor $R$ is $\frac{1}{5} \, A$, the value of $R$ is (in $\Omega$)

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