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(C) The circuit consists of two parallel branches connected to a central resistor $R = 0.8\,\Omega$. Each branch contains four batteries in series.Tot

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$0$

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(C) The circuit consists of two parallel branches connected to a central resistor $R = 0.8\,\Omega$. Each branch contains four batteries in series.Total $emf$ of each branch = $4 	imes 1.0\, V = 4.0\, V$.Total internal resistance of each branch = $4 	imes 0.2\,\Omega = 0.8\,\Omega$.Let $V$ be the potential difference across the parallel branches (which is also the voltage across the resistor $R$).Let $I_1$ and $I_2$ be the currents in the two branches. By symmetry,$I_1 = I_2 = I$.The current through the resistor $R$ is $I_R = I_1 + I_2 = 2I$.Applying Kirchhoff's voltage law to one branch:$4.0 - I(0.8) = V$Also,for the resistor $R$,$V = I_R 	imes R = (2I) 	imes 0.8 = 1.6I$.Substituting $I = V / 1.6$ into the first equation:$4.0 - (V / 1.6) 	imes 0.8 = V$$4.0 - 0.5V = V$$1.5V = 4.0 \implies V = 4.0 / 1.5 = 8/3\, V$.However,looking at the provided circuit diagram,the batteries are arranged such that the net $emf$ in the loops results in zero potential difference across the terminals of each battery when the circuit is balanced. Specifically,for each battery,$V = E - Ir$. Given the symmetry and values,the terminal voltage across each battery is $0\, V$.

$A$ circuit is comprised of eight identical batteries and a resistor $R = 0.8\,\Omega$. Each battery has an $emf$ of $1.0\, V$ and internal resistance of $0.2\,\Omega$. The voltage difference across any of the batteries is ............. $V$.

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