First slide

KVL

Question

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

Moderate

A

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

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

I A from b to a through e
D

I A from a to b through e

Solution

Since E₁(10V) > E₂(4V)
So current in the circuit will be clockwise.
Applying Kirchhoff's voltage law,
$\begin{array}{l} - 1 \times i + 10 - 4 - 2 \times i - 3 i = 0 \Rightarrow i = 1 A (a to b via e) \\ ∴ Current = \frac{V}{R} = \frac{10 - 4}{6} = 1 .0 ampere \end{array}$

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