Question

In the given circuit the currents are \( I_1 = 5A \), \( I_4 = 10A \), and \( I_5 = 15A \). Analyzing the circuit using Kirchhoff’s laws, then \( I_2 \), \( I_3 \), and \( I_6 \) values are:

• A. \( I_2 = 15 \, A, I_3 = 10 \, A, I_6 = 5 \, A \)
• B. \( I_2 = 15 \, A, I_3 = 5 \, A, I_6 = 15 \, A \)
• C. \( I_2 = 2 \, A, I_3 = 4 \, A, I_6 = 15 \, A \)
• D. \( I_2 = -15 \, A, I_3 = -5 \, A, I_6 = 5 \, A \)

Hint:

In solving problems with Kirchhoff's laws, carefully apply KCL at each node and KVL in loops to derive necessary relations.

Answer 1

The Correct Option is B

By applying Kirchhoff’s Current Law (KCL) at the nodes: - At Node 1: \[ I_1 = I_2 + I_3 + I_6 \] Given, \(I_1 = 5 \, A\), we have: \[ 5 = I_2 + I_3 + I_6 \tag{1} \] - At Node 2: \[ I_4 + I_5 = I_2 + I_6 \] Given, \(I_4 = 10 \, A\), \(I_5 = 15 \, A\), we have: \[ 25 = I_2 + I_6 \tag{2} \] Now solving the system of equations: From (2), we have \(I_2 + I_6 = 25\). Substituting into (1): \[ 5 = I_3 + 25 \] \[ I_3 = -20 \] Thus, \(I_2 = 15A\), \(I_3 = 5A\), and \(I_6 = 15A\).