Question

The figure shows the single-line diagram of a 4-bus power network. Branches \( b_1, b_2, b_3 \), and \( b_4 \) have impedances \( 4z, z, 2z \), and \( 4z \) per-unit (pu), respectively, where \( z = r + jx \) with \( r>0 \) and \( x>0 \). The current drawn from each load bus (marked as arrows) is equal to \( 1 \, \text{pu} \), where \( I \neq 0 \). If the network is to operate with minimum loss, the branch that should be opened is:
\includegraphics[width=0.5\linewidth]{19.png}

• A. \( b_1 \)
• B. \( b_2 \)
• C. \( b_3 \)
• D. \( b_4 \)

Hint:

For power networks, remove the branch that results in the minimum total power loss.

Answer 1

The Correct Option is C

Step 1: Understanding the power loss calculation. The power loss (\( P_L \)) in a branch is proportional to: \[ P_L = I^2 R, \] where \( R \) is the resistance of the branch, and \( I \) is the current through it. Step 2: Analyzing branch removal. - Removing \( b_1 \): Total power loss = \( 27r \). - Removing \( b_2 \): Total power loss = \( 48r \). - Removing \( b_3 \): Total power loss = \( 12r \). - Removing \( b_4 \): Total power loss = \( 19r \). Step 3: Selecting the branch to minimize losses.
Removing \( b_3 \) results in the minimum power loss (\( 12r \)).
Hence, the correct option is (C) \( b_3 \).