GATE 2024 EE Q.19: solution and answer

GATE 2024 · EEQ.19 · 1 mark · MCQ

The figure shows the single line diagram of a 4-bus power network. Branches b1, b2, b3, and b4 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 I pu, where I ≠ 0. If the network is to operate with minimum loss, the branch that should be opened is

GATE 2024 EE question 19: A generator (source) feeds a top bus. — Figure from the GATE 2024 EE paper. It is part of the question.

(A)
b1
(B)
b2
(C)
b3
(D)
b4

Answer and solution

ANSWER

Open branch b3, which is option (C). It gives the lowest radial loss, 12 I²r, against 27, 48 and 19 for opening b1, b2 and b4.

Official answer: CSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

Label the load buses $L$ (left), $R$ (right) and $B$ (bottom), each drawing current $I$ . The ring is Source– $b_{1}$ – $L$ – $b_{3}$ – $B$ – $b_{4}$ – $R$ – $b_{2}$ –Source.
Opening one branch leaves a radial network. Each remaining branch then carries the total current of the loads it feeds, so the branch currents do not depend on $x$ .
Loss in a branch is $∣ I_{k} ∣^{2} R_{k}$ . Every branch resistance is its multiplier times $r$ , so the total loss is $r I^{2} \sum k (I_{k} / I)^{2}$ , with $k = 4, 1, 2, 4$ for $b_{1}, b_{2}, b_{3}, b_{4}$ . Only the sum $S = \sum k (I_{k} / I)^{2}$ needs comparing.
Open $b_{1}$ : $b_{2}$ carries $3 I$ , $b_{4}$ carries $2 I$ , $b_{3}$ carries $I$ . $S = 1 \cdot 9 + 4 \cdot 4 + 2 \cdot 1 = 27$ .
Open $b_{2}$ : $b_{1}$ carries $3 I$ , $b_{3}$ carries $2 I$ , $b_{4}$ carries $I$ . $S = 4 \cdot 9 + 2 \cdot 4 + 4 \cdot 1 = 48$ .
Open $b_{3}$ : $b_{1}$ carries $I$ (feeds $L$ only). $b_{2}$ carries $2 I$ (feeds $R$ and $B$ ). $b_{4}$ carries $I$ (feeds $B$ ). $S = 4 \cdot 1 + 1 \cdot 4 + 4 \cdot 1 = 12$ .
Open $b_{4}$ : $b_{1}$ carries $2 I$ (feeds $L$ and $B$ ), $b_{3}$ carries $I$ , $b_{2}$ carries $I$ (feeds $R$ ). $S = 4 \cdot 4 + 2 \cdot 1 + 1 \cdot 1 = 19$ .
The smallest value is $S = 12$ , so the loss is least when $b_{3}$ is opened. This holds for any $r > 0$ and $x > 0$ . Answer: (C) $b_{3}$ .

KEY IDEA

A ring network is opened at one point to make it radial. The loss is then the sum of I²R over the branches, with each branch carrying the load current beyond it. The best branch to open is the one that gives the smallest sum of resistance times current squared. Reactance does not matter here.

WATCH OUT FOR

Opening the highest-impedance branch, b1 or b4, because it looks like the weakest link. The loss depends on the current each branch carries after opening. Opening b3, the middle branch between two loads, splits the load best and gives the smallest loss.

From the GATE 2024 EE paper. Every question of this paper · Download the full paper · All 6 Transmission & distribution questions, with notes

GATE 2024 EE Q.19: solution and answer

Answer and solution

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