GATE 2026 Power Systems Questions and Solutions

A completely transposed single-circuit three-phase overhead line has its three conductors placed in one horizontal plane. The spacings are \(D_{ab}=4\) m, \(D_{bc}=4\) m and \(D_{ca}=8\) m. Each conductor has a geometric mean radius of 1.2 cm. The inductance per phase per kilometre of the line is

1. 0.804 mH/km
2. 1.208 mH/km
3. 1.398 mH/km
4. 2.416 mH/km

Solution

For a transposed line the mutual spacing is replaced by the geometric mean of the three pairwise distances (see Inductance of a Transmission Line):

The inductance per phase per unit length of a transposed line with conductors of geometric mean radius \(D_s\) is

A three-phase, 50 Hz medium transmission line is represented by its nominal-\(\pi\) equivalent. The total series impedance is \(Z = 20 + j60~\Omega\) per phase and the total shunt admittance is \(Y = j4 \times 10^{-4}~\text{S}\) per phase, the latter being split equally between the two ends. The line delivers 100 MVA at 0.8 power factor lagging at a receiving-end line voltage of 220 kV. The sending-end line voltage, in kV, is _____ (round off to one decimal place).

Solution

For the nominal-\(\pi\) model the generalised constants are \(A = D = 1 + \tfrac{ZY}{2}\) and \(B = Z\) (see Medium Transmission Line). First evaluate \(A\):

Take the receiving-end phase voltage as the reference phasor:

At 0.8 power factor lagging the current lags \(V_R\) by \(36.87^\circ\):

Now apply \(V_S = A V_R + B I_R\):

This is the phase value; the sending-end line voltage is \(\sqrt{3}\) times as large:

A synchronous generator is rated 100 MVA, 13.8 kV and has a subtransient reactance of 0.20 pu on its own rating as base. When this machine is included in a system study that uses a base of 200 MVA and 13.2 kV at the generator terminals, its subtransient reactance in per unit is _____ (round off to three decimal places).

Solution

Per-unit impedance scales directly with the base MVA and inversely with the square of the base voltage (see Per-Unit Quantities):

The reactance grows because the new base MVA is larger, and grows a little further because the new base voltage is smaller than the machine rating.

A two-bus system has bus 1 as the slack bus with \(V_1 = 1.0\angle 0^\circ\) pu. The two buses are joined by a single line of series admittance \(y_{12} = 2 - j6\) pu, and the line charging is neglected. Bus 2 is a load bus drawing \(0.5 + j0.2\) pu. Starting from a flat start \(V_2^{(0)} = 1.0\angle 0^\circ\) pu, the magnitude of \(V_2\) after one Gauss-Seidel iteration is _____ pu (round off to four decimal places).

Solution

With no shunt elements the bus admittance matrix entries for bus 2 are \(Y_{22} = y_{12} = 2 - j6\) and \(Y_{21} = -y_{12} = -2 + j6\). The scheduled injection at a load bus is negative: \(P_2 = -0.5\) pu and \(Q_2 = -0.2\) pu. The Gauss-Seidel update (see Gauss-Seidel Load Flow) is

Evaluate the two terms in the bracket:

Divide by \(Y_{22}\), rationalising with the conjugate \(2 + j6\):

An unloaded 50 MVA, 11 kV synchronous generator has its neutral solidly grounded. Its sequence reactances on the machine rating as base are \(X_1 = 0.25\) pu, \(X_2 = 0.20\) pu and \(X_0 = 0.05\) pu; all resistances are negligible. With the prefault terminal voltage at 1.0 pu, a single line-to-ground fault occurs on phase a at the machine terminals. The magnitude of the fault current is

1. 2.62 kA
2. 5.25 kA
3. 15.75 kA
4. 47.24 kA

Solution

For a single line-to-ground fault the three sequence networks are connected in series, so the three sequence currents are equal (see Single Line-to-Ground Fault):

The faulted phase current is the sum of the three sequence components:

Convert to amperes using the base current at the 11 kV level:

The line currents of a three-phase, four-wire system are measured as \(I_a = 6\angle 0^\circ\) A, \(I_b = 4\angle -90^\circ\) A and \(I_c = 8\angle 150^\circ\) A. The zero-sequence component of the phase-a current is

1. 0 A
2. \(0.309\angle 180^\circ\) A
3. \(0.928\angle 180^\circ\) A
4. \(3.000\angle 0^\circ\) A

Solution

The zero-sequence current is the average, not the sum, of the three phase currents (see Symmetrical Components):

Resolve each phasor into rectangular form:

The neutral current, for reference, is \(I_n = 3I_{a0} = 0.928\angle 180^\circ\) A, which is choice C. That is the trap: option C omits the factor of one third.

A lossless synchronous generator with inertia constant \(H = 5\) MJ/MVA is connected to an infinite bus through a transmission link. In the prefault condition the power-angle characteristic is \(P_e = 2.0\sin\delta\) pu and the constant mechanical input is \(P_m = 1.0\) pu. A three-phase fault at the generator terminals reduces the electrical power output to zero, and the fault is cleared by restoring the original prefault network. For a system frequency of 50 Hz, the critical clearing time is _____ s (round off to three decimal places).

Solution

The initial rotor angle follows from \(P_m = P_{max}\sin\delta_0\):

Because the postfault characteristic is identical to the prefault one, the maximum swing angle is

Applying the equal-area criterion with zero power transfer during the fault, the accelerating area from \(\delta_0\) to \(\delta_{cc}\) must equal the decelerating area from \(\delta_{cc}\) to \(\delta_{max}\) (see Equal Area Criterion):

During the fault the electrical power is zero, so the swing equation integrates in closed form and the clearing time corresponding to \(\delta_{cc}\) is

Two thermal units supply a total load of 250 MW. Their fuel cost characteristics, in Rs./h with the outputs in MW, are

The generation limits are \(50 \le P_1 \le 200\) MW and \(30 \le P_2 \le 100\) MW. Transmission losses are neglected. For the economic dispatch, the output of unit 1 is _____ MW.

Solution

The incremental cost of each unit is the derivative of its cost characteristic:

Ignoring the limits for the moment, equal incremental cost with \(P_1 + P_2 = 250\) gives (see Economic Load Dispatch)

This trial value of \(P_2\) exceeds its 100 MW upper limit, so it is not feasible. Unit 2 is therefore held at its limit and the balance is taken by unit 1:

The system incremental cost is now set by the free unit:

The solution is consistent because the fixed unit sits at its maximum and its incremental cost there, \(0.08(100) + 12 = 20\) Rs./MWh, is below \(\lambda\); loading it further would still be cheap, but the limit forbids it. Both outputs lie within the stated ranges.

A feeder is protected by an IDMT overcurrent relay of the normal inverse type fed from a current transformer of ratio 400/5 A. The relay plug setting is 125 percent and the time multiplier setting is 0.3. The relay operating time is given by

For a symmetrical fault current of 4000 A in the primary, the relay operating time is _____ s (round off to three decimal places).

Solution

Refer the fault current to the relay coil through the CT ratio:

A plug setting of 125 percent means the relay picks up at 1.25 times the CT secondary rated current:

The plug setting multiplier is the ratio of the two:

Evaluate the characteristic. Since \(8^{0.02} = e^{0.02\ln 8} = e^{0.041589} = 1.042466\),

A lossless 400 kV three-phase transmission line has a series inductance of 1.0 mH/km and a shunt capacitance of 11.1 nF/km per phase. The surge impedance loading of the line is

1. 267 MW
2. 533 MW
3. 800 MW
4. 1066 MW

Solution

For a lossless line the surge impedance depends only on the ratio of the distributed parameters, not on the line length (see Surge Impedance Loading):

The surge impedance loading is the three-phase power delivered when the line is terminated in \(Z_c\), with \(V_L\) the rated line-to-line voltage: