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(D) To find the current in the circuit,we apply Kirchhoff's Voltage Law $(KVL)$ to the loop $abcda$.Let the current in the circuit be $I$ flowing in t

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$1.0$ amp from $a$ to $b$ via $e$

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(D) To find the current in the circuit,we apply Kirchhoff's Voltage Law $(KVL)$ to the loop $abcda$.Let the current in the circuit be $I$ flowing in the clockwise direction.Starting from point $a$ and moving clockwise:Moving through the $1 \ \Omega$ resistor,the potential drop is $-I(1)$.Moving through the $10 \ V$ battery from negative to positive,the potential gain is $+10 \ V$.Moving through the $4 \ V$ battery from positive to negative,the potential drop is $-4 \ V$.Moving through the $2 \ \Omega$ resistor,the potential drop is $-2I$.Moving through the $3 \ \Omega$ resistor,the potential drop is $-3I$.Applying $KVL$: $-I + 10 - 4 - 2I - 3I = 0$.Simplifying the equation: $6 - 6I = 0$.$6I = 6$,which gives $I = 1 \ A$.Since the current $I$ is positive,the assumed clockwise direction is correct.In the upper branch containing point $e$,the current flows from $a$ to $b$ through $e$.

In the circuit diagram shown below,the magnitude and direction of the flow of current respectively would be

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