Question

Find magnitude and direction of current in 1 Ω resistor in the given circuit.

Answer 1

We use Kirchhoff’s Voltage Law (KVL) to solve for the currents in the multi-loop circuit.

The sum of potentials in a loop is zero:

∑V = 0

Junction rule:

I_total = I₁ + I₂

Let the current in the left loop be x (counter-clockwise) and the current in the right loop be y (clockwise).

Applying KVL to the left loop (starting from the 6 V battery):

6 − 3x − 1(x + y) = 0

4x + y = 6    ...(i)

Applying KVL to the right loop (starting from the 9 V battery):

9 − (2 + 3)y − 1(x + y) = 0

9 − 5y − x − y = 0

x + 6y = 9    ...(ii)

From equation (i), y = 6 − 4x. Substituting into equation (ii):

x + 6(6 − 4x) = 9

⇒ x + 36 − 24x = 9

⇒ −23x = −27

⇒ x = `27/23 A`

≈ 1.17 A

Now find y:

y = `6 - 4(27/23)`

= `(138 - 108)/23`

= `30/23 A`

≈ 1.30 A

The current in the 1 Ω resistor is the sum of x + y:

I_PQ = `27/23 + 30/23`

= `57/23 A`

≈ 2.48 A

Since both x and y contribute in the same direction from P to Q, the direction is downward.

∴ The current is 2.48 A flowing from P to Q.