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(B) To find the current in the $10\, \Omega$ resistor,we use the formula for the equivalent $EMF$ $(E_{eq})$ and equivalent internal resistance $(r_{e

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$0.03\, A$ from $P_2$ to $P_1$

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(B) To find the current in the $10\, \Omega$ resistor,we use the formula for the equivalent $EMF$ $(E_{eq})$ and equivalent internal resistance $(r_{eq})$ of two cells connected in parallel:$E_{eq} = \frac{\frac{E_1}{r_1} + \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{5}{2} + \frac{2}{1}}{\frac{1}{2} + 1} = \frac{2.5 + 2}{1.5} = \frac{4.5}{1.5} = 3\, V$$r_{eq} = \frac{r_1 r_2}{r_1 + r_2} = \frac{2 	imes 1}{2 + 1} = \frac{2}{3}\, \Omega$Now,the circuit simplifies to a single loop with $E_{eq} = 3\, V$,$r_{eq} = \frac{2}{3}\, \Omega$,and an external resistor $R = 10\, \Omega$ in series.The current $i$ is given by:$i = \frac{E_{eq}}{R + r_{eq}} = \frac{3}{10 + \frac{2}{3}} = \frac{3}{\frac{32}{3}} = \frac{9}{32} \approx 0.28\, A$Wait,re-evaluating the parallel combination formula for opposite polarities:If the batteries are connected such that they oppose each other,$E_{eq} = \frac{\frac{E_1}{r_1} - \frac{E_2}{r_2}}{\frac{1}{r_1} + \frac{1}{r_2}} = \frac{\frac{5}{2} - \frac{2}{1}}{\frac{1}{2} + 1} = \frac{0.5}{1.5} = \frac{1}{3}\, V$$i = \frac{E_{eq}}{R + r_{eq}} = \frac{1/3}{10 + 2/3} = \frac{1/3}{32/3} = \frac{1}{32} \approx 0.03125\, A$Since the $5\, V$ battery is stronger,the current flows from $P_2$ to $P_1$ through the $10\, \Omega$ resistor.

$A$ $5\, V$ battery with internal resistance $2\, \Omega$ and a $2\, V$ battery with internal resistance $1\, \Omega$ are connected to a $10\, \Omega$ resistor as shown in the figure. The current in the $10\, \Omega$ resistor is:

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