In the circuit shown, the current through the 5 Ω resistor is

Question

In the circuit shown, the current through the 5 Ω resistor is (A) (8/3) A (B) (9/13) A (C) (4/13) A (D) (1/3) A

Tardigrade Admin · Accepted Answer

Applying Kirchhoff's second law for closed loop

A EFB A

, we get

- (l_{1} + l_{2}) \times 5 - l_{1} \times 2 + 2 = 0

or

7 l_{1} + 5 l_{2} = 2

Again, applying Kirchhoff's second law for a closed loop

D EFC D

, we get

- (l_{1} + l_{2}) \times 5 - l_{2} \times 2 + 2 = 0

or

5 l_{1} + 7 l_{2} = 2

On solving, we get

I_{1} = \frac{1}{6} A

and

I_{2} = \frac{1}{6} A

I = I_{1} + I_{2} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} A

Q. In the circuit shown, the current through the $5 Ω$ resistor is