Question

In the following network,
\(I_1 = -0.4\,\text{A}\) , \(I_4 = 1\,\text{A}\) , \(I_5 = 0.4\,\text{A}\)
The values of \(I_2\), \(I_3\) and \(I_6\) are respectively

• A. \(1.4\,\text{A}\) , \(0.4\,\text{A}\) , \(-0.6\,\text{A}\)
• B. \(0.4\,\text{A}\) , \(-0.6\,\text{A}\) , \(1.4\,\text{A}\)
• C. \(1.4\,\text{A}\) , \(-0.6\,\text{A}\) , \(0.4\,\text{A}\)
• D. \(-0.6\,\text{A}\) , \(1.4\,\text{A}\) , \(0.4\,\text{A}\)

Hint:

Apply Kirchhoff's current law at each corner of the network, using the arrow directions in the figure.

Answer 1

The Correct Option is C

Step 1: Count what leaves and enters the right-hand part:
The two lower corners share three wires. Using the junction law on the group made of the bottom-right corner alone: $I_4 = I_1 + I_2$, so $I_2 = 1 - (-0.4) = 1.4$ A.

Step 2: Left side:
The top-left corner has only two wires, $I_5$ going in and $I_6$ going out, so $I_6 = I_5 = 0.4$ A.

Step 3: Remaining current:
The top-right corner has $I_6$ entering and $I_1, I_2, I_3$ leaving: $0.4 = -0.4 + 1.4 + I_3$, so $I_3 = -0.6$ A. The negative sign means the actual current flows opposite to the arrow shown.

Step 4: Match:
$(1.4, -0.6, 0.4)$ is option (C).

Final Answer:
Option (C). \[ \boxed{1.4\text{ A},\ -0.6\text{ A},\ 0.4\text{ A} \text{ (C)}} \]