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(A) Let the potential at the left junction be $V_L$ and at the right junction be $V_R$. Since no current flows through the ammeter,the potential at th

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$5$

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(A) Let the potential at the left junction be $V_L$ and at the right junction be $V_R$. Since no current flows through the ammeter,the potential at the point between $R$ and the battery is the same as the potential between the $40 \, \Omega$ and $20 \, \Omega$ resistors.Let the current through the upper branch be $I_1$ and the lower branch be $I_2$. Given total current is $0.5 \, A$,so $I_1 + I_2 = 0.5 \, A$.In the lower branch,the current $I_2 = 0.1 \, A$ flows through both $40 \, \Omega$ and $20 \, \Omega$ resistors in series.The potential drop across the $40 \, \Omega$ resistor is $V_{40} = I_2 	imes 40 = 0.1 	imes 40 = 4 \, V$.The potential drop across the $20 \, \Omega$ resistor is $V_{20} = I_2 	imes 20 = 0.1 	imes 20 = 2 \, V$.Since no current flows through the ammeter,the potential difference across the upper branch must equal the potential difference across the lower branch.For the upper branch,$I_1 = 0.5 - 0.1 = 0.4 \, A$.The potential difference across the upper branch is $V_R + V_{battery} = I_1 	imes R + 2 \, V$.Equating the potential drops: $I_1 	imes R + 2 = V_{40} = 4 \, V$.$0.4 	imes R + 2 = 4 \Rightarrow 0.4 	imes R = 2 \Rightarrow R = \frac{2}{0.4} = 5 \, \Omega$.

In the circuit shown in the figure,no current flows through the ideal ammeter. If the internal resistance of the cell is negligible,the value of unknown resistance $R$ is .............. $\Omega$.

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