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

Question

In the circuit diagram shown below, the magnitude and direction of the flow of current, respectively, would be <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-8ol9h6lh0fgqqfxa.jpg" /> (A) (7/3)A from a to b via e (B) (7/3)A from b to a via e (C) 1A from b to a via e (D) 1A from a to b via e

Tardigrade Admin · Accepted Answer

I=(10 - 4/1 + 2 + 3)=1 A (Clockwise so a to b via e )

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

$\frac{7}{3} A$ from $a$ to $b$ via $e$

$\frac{7}{3} A$ from $b$ to $a$ via $e$

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

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

$I = \frac{10 - 4}{1 + 2 + 3} = 1 A$ (Clockwise so $a$ to $b$ via $e$ )

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

37​A from a to b via e

37​A from b to a via e

1A from b to a via e

1A from a to b via e

I=1+2+310−4​=1A (Clockwise so a to b via e )

$\frac{7}{3} A$ from $a$ to $b$ via $e$

$\frac{7}{3} A$ from $b$ to $a$ via $e$

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

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

$I = \frac{10 - 4}{1 + 2 + 3} = 1 A$ (Clockwise so $a$ to $b$ via $e$ )