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(B) The circuit consists of two identical cells, each with an electromotive force $(EMF)$ $E = 3 	ext{ V}$ and internal resistance $r = 3 \Omega$, co

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$\frac{2}{5} \,A$

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(B) The circuit consists of two identical cells, each with an electromotive force $(EMF)$ $E = 3 	ext{ V}$ and internal resistance $r = 3 \Omega$, connected in parallel. These are connected to an external resistor $R = 6 \Omega$.For cells in parallel, the equivalent $EMF$ $(E_{eq})$ is given by:$E_{eq} = \frac{E_1/r_1 + E_2/r_2}{1/r_1 + 1/r_2} = \frac{3/3 + 3/3}{1/3 + 1/3} = \frac{1 + 1}{2/3} = \frac{2}{2/3} = 3 	ext{ V}$.The equivalent internal resistance $(r_{eq})$ is given by:$\frac{1}{r_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \implies r_{eq} = 1.5 \Omega$.The total resistance of the circuit is $R_{total} = R + r_{eq} = 6 + 1.5 = 7.5 \Omega$.The current through the $6 \Omega$ resistor is given by Ohm's Law:$I = \frac{E_{eq}}{R_{total}} = \frac{3}{7.5} = \frac{30}{75} = \frac{2}{5} 	ext{ A}$.

In the following circuit, the current through $6 \Omega$ resistor is

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