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(D) The circuit consists of two cells with EMFs $E_1 = 10\,V$ and $E_2 = 4\,V$ connected in opposition.Since $E_1 > E_2$,the net $EMF$ in the circuit

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$1A$ from $a$ to $b$ through $e$

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(D) The circuit consists of two cells with EMFs $E_1 = 10\,V$ and $E_2 = 4\,V$ connected in opposition.Since $E_1 > E_2$,the net $EMF$ in the circuit is $E_{net} = E_1 - E_2 = 10\,V - 4\,V = 6\,V$.The total resistance of the circuit is $R_{total} = 1\,\Omega + 2\,\Omega + 3\,\Omega = 6\,\Omega$.Using Ohm's law,the magnitude of the current is $i = \frac{E_{net}}{R_{total}} = \frac{6\,V}{6\,\Omega} = 1\,A$.Since $E_1$ is the dominant source,the current flows in the direction of $E_1$,which is clockwise. Thus,the current flows from $a$ to $b$ through $e$.

Order	$V \ (V)$	$I \ (A)$
$1$	$1.0$	$0.08$
$2$	$0.5$	$0.18$
$3$	$0.8$	$0.12$

The magnitude and direction of the current in the circuit shown will be

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