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(B) To find the current through the external resistor $R$,we first find the equivalent emf $(E_{eq})$ and equivalent internal resistance $(r_{eq})$ of

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(B) To find the current through the external resistor $R$,we first find the equivalent emf $(E_{eq})$ and equivalent internal resistance $(r_{eq})$ of the two cells connected in parallel.The formula for equivalent emf of cells in parallel is:$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}}$Given $E_1 = 12 \, V$,$r_1 = 3 \, \Omega$ and $E_2 = 6 \, V$,$r_2 = 6 \, \Omega$. Note that the cells are connected such that their polarities oppose each other in the loop,so we use:$E_{eq} = \frac{\frac{12}{3} - \frac{6}{6}}{\frac{1}{3} + \frac{1}{6}} = \frac{4 - 1}{\frac{2+1}{6}} = \frac{3}{\frac{3}{6}} = \frac{3}{0.5} = 6 \, V$The equivalent internal resistance is:$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2} \implies r_{eq} = 2 \, \Omega$Now,the circuit simplifies to a single cell of $6 \, V$ and $2 \, \Omega$ in series with an external resistor $R = 4 \, \Omega$.The current $i$ is given by:$i = \frac{E_{eq}}{r_{eq} + R} = \frac{6}{2 + 4} = \frac{6}{6} = 1 \, A$

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