Question

The figure below shows a part of an electric circuit. The current marked I_A​ is:

• A. 1 A
• B. 1.3 A
• C. 1.7 A
• D. 3 A

Answer 1

The Correct Option is A

From the circuit figure, we can observe the following currents at junction point I:

Branch	Current	Direction
A	0.2 A	Entering
A	0.2 A	Entering
B	1.2 A	Entering
B	0.5 A	Leaving

Applying Kirchhoff's Current Law (KCL) at junction I:

ΣI_entering = ΣI_leaving

(0.2 A + 0.2 A + 1.2 A) = (0.5 A + I_A)

1.6 A = 0.5 A + I_A

I_A = 1.6 A - 0.5 A = 1.1 A

The current marked I_A is approximately 1.1 A.

Answer 2

Applying Kirchhoff's Current Law (KCL) at node A: The sum of currents entering a node is equal to the sum of currents leaving the node.

In this case:

Current entering A: \( 0.2 \, \text{A} + 0.2 \, \text{A} = 0.4 \, \text{A} \)

Current leaving A: \( I_A \)

Therefore: \( 0.4 \, \text{A} = I_A + I_{AB} \)

At node B:

Current entering B: \( I_{AB} + 1.2 \, \text{A} \)

Current leaving B: \( 0.5 \, \text{A} \)

Therefore: \(I_{AB} + 1.2 \, \text{A} = 0.5 \, \text{A} \) which means \( I_{AB} = 0.5 -1.2 = -0.7 A \)

Substituting the value of \(I_{AB}\) into the equaition \( 0.4 \, \text{A} = I_A + I_{AB} \)

Therefore: \( 0.4 = I_A - 0.7 \)

\( I_A = 0.4 + 0.7 = 1.1 A \)