The figure show the single line diagram of a 4 - bus power network. Branches $b_2,b_3$ and $b_4$ have impedance $4z,z,2z$ 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 Ipu, where $I\ne 0$. If the network to operate with minimum loss, the branch that should be opened is

The correct option is b₃.

Step-by-Step Analysis:

1. Network Parameters and Notation
Let us label the buses in the diagram as follows:

• Bus 1: The source bus at the top, connected to the generator.
• Bus 2: The load bus on the left, drawing a load current of I.
• Bus 3: The load bus on the right, drawing a load current of I.
• Bus 4: The load bus at the bottom, drawing a load current of I.

The branch impedances are given as:
$z_{b_1}=4z$
$z_{b_2}=z$
$z_{b_3}=2z$
$z_{b_4}=4z$
where $z=r+jx$. The resistance of each branch is directly proportional to the real part of its impedance:
$R_{b_1}=4r$
$R_{b_2}=r$
$R_{b_3}=2r$
$R_{b_4}=4r$

The total power loss in the network is given by:
$P_{\text{loss}}=I_{b_1}^2{R}_{b_1}+I_{b_2}^2{R}_{b_2}+I_{b_3}^2{R}_{b_3}+I_{b_4}^2{R}_{b_4}$

2. Calculating Losses when Different Branches are Opened

Case 1: Branch b₁ is opened ($I_{b_1}=0$)
All power to the load buses must flow through branch b₂.

• Current in branch b₃: carries the load of Bus 2 ⇒ $I_{b_3}=I$
• Current in branch b₄: carries the loads of Bus 4 and Bus 2 ⇒ $I_{b_4}=2I$
• Current in branch b₂: carries the loads of Bus 3, Bus 4, and Bus 2 ⇒ $I_{b_2}=3I$

Case 2: Branch b₂ is opened ($I_{b_2}=0$)
All power to the load buses must flow through branch b₁.

• Current in branch b₄: carries the load of Bus 3 ⇒
• Current in branch b₃: carries the loads of Bus 4 and Bus 3 ⇒
• Current in branch b₁: carries the loads of Bus 2, Bus 4, and Bus 3 ⇒

Case 3: Branch b₃ is opened ($I_{b_3}=0$)
The network splits into two independent radial paths:

• Path 1: Bus 1 → b₁ → Bus 2. Current in branch b₁ carries the load of Bus 2 ⇒ $I_{b_1}=I$.
• Path 2: Bus 1 → b₂ → Bus 3 → b₄ → Bus 4.
  • Current in branch b₄ carries the load of Bus 4 ⇒ $I_{b_4}=I$.
  • Current in branch b₂ carries the loads of Bus 3 and Bus 4 ⇒ .

Case 4: Branch b₄ is opened ($I_{b_4}=0$)
The network splits into two independent radial paths:

• Path 1: Bus 1 → b₂ → Bus 3. Current in branch b₂ carries the load of Bus 3 ⇒ $I_{b_2}=I$.
• Path 2: Bus 1 → b₁ → Bus 2 → b₃ → Bus 4.
  • Current in branch b₃ carries the load of Bus 4 ⇒ $I_{b_3}=I$.
  • Current in branch b₁ carries the loads of Bus 2 and Bus 4 ⇒ .

Ploss = (2I)2 (4r) + (I)2 (r) + (I)2 (2r) + (0)2 (4r) = I2r [ 16 + 1 + 2 + 0 ] = 19 I2r

Conclusion:
Comparing the four scenarios, the minimum power loss is $12I^{2}r$, which occurs when branch b₃ is opened.

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