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(C) The circuit consists of two cells of $10\,V$ and $5\,V$ connected in opposition. The total electromotive force $(EMF)$ of the circuit is $E_{net}

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$0.5\,A$ from $A$ to $B$ through $E$

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(C) The circuit consists of two cells of $10\,V$ and $5\,V$ connected in opposition. The total electromotive force $(EMF)$ of the circuit is $E_{net} = 10\,V - 5\,V = 5\,V$.The total resistance of the circuit is $R_{total} = 2\,\Omega + 1\,\Omega + 7\,\Omega = 10\,\Omega$.Using Ohm's law,the current $i$ in the circuit is given by $i = \frac{E_{net}}{R_{total}} = \frac{5\,V}{10\,\Omega} = 0.5\,A$.Since the $10\,V$ battery is stronger than the $5\,V$ battery,the current flows in the direction determined by the $10\,V$ battery,which is from $A$ to $B$ through $E$.

The magnitude and direction of the current in the following circuit is:

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