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: Understanding the Concept:
Kirchhoff's junction rule says the total current entering a junction equals the total current leaving it. In the figure, the network has four corners: top-left, top-right, bottom-left and bottom-right.

Step 2: Key Formula or Approach:
Directions from the figure: \(I_6\) flows left to right along the top. \(I_3\) flows along the diagonal from the top-right corner to the bottom-left corner. \(I_2\) and \(I_1\) flow downward from the top-right corner to the bottom-right corner. \(I_4\) flows leftwards along the bottom, and \(I_5\) flows upward on the left side.

Step 3: Step A: the bottom-right corner:
Entering: \(I_1\) and \(I_2\). Leaving: \(I_4\).
\[ I_1 + I_2 = I_4 \Rightarrow -0.4 + I_2 = 1 \Rightarrow I_2 = 1.4 \text{ A} \]

Step 4: Step B: the bottom-left corner:
Entering: \(I_3\) and \(I_4\). Leaving: \(I_5\).
\[ I_3 + I_4 = I_5 \Rightarrow I_3 + 1 = 0.4 \Rightarrow I_3 = -0.6 \text{ A} \]

Step 5: Step C: the top-left corner:
Entering: \(I_5\). Leaving: \(I_6\).
\[ I_6 = I_5 = 0.4 \text{ A} \]
Check at the top-right corner: entering \(I_6 = 0.4\), leaving \(I_1 + I_2 + I_3 = -0.4 + 1.4 - 0.6 = 0.4\). The rule holds.

Final Answer:
\(I_2 = 1.4\) A, \(I_3 = -0.6\) A and \(I_6 = 0.4\) A, option (C). \[ \boxed{1.4\text{ A},\ -0.6\text{ A},\ 0.4\text{ A} \text{ (C)}} \]