GATE 2025 EE Q.33: solution and answer

GATE 2025 · EEQ.33 · 1 mark · NAT

The bus impedance matrix of a 3-bus system (in pu) is Z_bus = [[j0.059, j0.061, j0.038],[j0.061, j0.093, j0.066],[j0.038, j0.066, j0.110]]. A symmetrical fault (through a fault impedance of j0.007 pu) occurs at bus 2. Neglecting pre-fault loading conditions, the voltage at bus 1, during the fault is ________ pu (round off to three decimal places).

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Answer and solution

ANSWER

The voltage at bus 1 during the fault is 0.390 pu.

Official answer: 0.380 to 0.400Source: Official answer key, GATE 2025 EE (IIT Roorkee)

No pre-fault loading, so every bus starts at $V_{i} (0) = 1∠ 0^{\circ}$ pu. The fault is at bus 2 ( $k = 2$ ) through $Z_{f} = j 0.007$ pu.
Fault current from the bus impedance matrix: $I_{f} = \frac{V _{2} ( 0 )}{Z _{22} + Z _{f}}$ .
$I_{f} = \frac{1}{j 0.093 + j 0.007} = \frac{1}{j 0.100} = - j 10 pu$
Fault current injected at bus 2 changes every bus voltage by $Δ V_{i} = - Z_{i 2} I_{f}$ , so during the fault $V_{i} (F) = V_{i} (0) - Z_{i 2} I_{f}$ .
For bus 1 use $Z_{12} = j 0.061$ :
$V_{1} (F) = 1 - (j 0.061) (- j 10) = 1 - 0.610$
$V_{1} (F) = 0.390 pu$
Check: $V_{2} (F) = 1 - (j 0.093) (- j 10) = 1 - 0.930 = 0.070$ pu, which equals $Z_{f} I_{f}$ in magnitude ( $0.007 \times 10$ ), as it should.

KEY IDEA

For a fault at bus k through impedance Zf, the fault current is I_f = V_k(0)/(Z_kk + Zf), and every bus voltage falls by Z_ik times I_f, so V_i(F) = 1 - Z_ik/(Z_kk + Zf).

WATCH OUT FOR

Using Z_11 (j0.059) instead of the mutual term Z_12, or leaving out the fault impedance (dividing by 0.093 instead of 0.100), which gives 0.344 instead of 0.390.

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GATE 2025 EE Q.33: solution and answer

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