GATE 2026 Electrical Machines Questions and Solutions

1 mark, numerical answer. A 10 kVA, 2000 V / 200 V, 50 Hz single-phase transformer is tested as follows. The open-circuit test, performed on the LV side with the HV side open, reads 200 V, 1.2 A and 90 W. The short-circuit test, performed on the HV side with the LV side shorted, reads 60 V, 5 A and 120 W. The equivalent leakage reactance of the transformer referred to the HV side, in ohms rounded off to two decimal places, is _____.

Solution

The short-circuit test excites the series branch of the equivalent circuit at rated current while the flux, and therefore the shunt branch, is negligible. The wattmeter reading is then entirely the copper loss. Note that the applied 60 V drives \(5\,\mathrm{A}\), which is exactly the rated HV current \(10000/2000 = 5\,\mathrm{A}\), so the test is at full load.

The open-circuit data are not needed here; they fix the shunt branch (core loss 90 W) and are deliberately included as a distractor, exactly as GATE does. See Open-circuit and short-circuit testing of transformers.

2 marks, numerical answer. A 25 kVA, 2400 V / 240 V, 50 Hz single-phase transformer has an equivalent resistance of \(1.5\,\Omega\) and an equivalent leakage reactance of \(4.0\,\Omega\), both referred to the HV side. The transformer supplies its rated kVA at a power factor of 0.8 lagging. Taking the terminal voltage as reference and computing the induced emf as an exact phasor sum, the percentage voltage regulation, rounded off to two decimal places, is _____.

Solution

Work entirely on the HV side. The rated HV current is

With \(V = 2400\angle 0^\circ\) taken as reference and a lagging power factor of 0.8, so that \(\cos\phi = 0.8\) and \(\sin\phi = 0.6\),

The internal drop across \(Z_{eq} = 1.5 + j4.0\;\Omega\) is

The familiar approximate expression keeps only the in-phase part of the drop,

and differs from the exact value only in the second decimal, because the neglected quadrature term contributes \(23.9583^{2}/(2\times 2437.5) = 0.118\,\mathrm{V}\). Further reading: Voltage regulation of a transformer.

2 marks, multiple choice. Two single-phase transformers of identical voltage ratio and identical impedance angle are operated in parallel on a common busbar. Transformer A is rated 100 kVA and has a per-unit equivalent impedance of 0.04 pu on its own rating. Transformer B is rated 200 kVA and has a per-unit equivalent impedance of 0.06 pu on its own rating. The bank supplies a total load of 250 kVA. The load shared by transformer A is

1. 83.33 kVA
2. 107.14 kVA
3. 125.00 kVA
4. 142.86 kVA

Solution

Per-unit impedances quoted on different bases cannot be compared directly, so both are first referred to a common base. Choosing 100 kVA,

Both transformers see the same terminal voltage and the same internal drop, so the currents, and therefore the apparent powers, divide in inverse proportion to the impedances. Because the impedance angles are equal, the two shares add arithmetically.

The two shares sum to the 250 kVA load, as they must. Note that transformer A carries 107.14 kVA against a 100 kVA rating and is therefore overloaded by 7.1 %, while B runs at only 71 % of its rating. This is the standard consequence of paralleling units whose per-unit impedances are unequal. See Parallel operation of transformers.

1 mark, numerical answer. A 220 V DC shunt motor has an armature circuit resistance of \(0.25\,\Omega\). When the armature draws 40 A the motor runs at 1000 rpm. The field current is held constant and the magnetic circuit is assumed linear, so the flux per pole does not change. Brush contact drop and armature reaction are neglected. If the mechanical load is reduced until the armature current falls to 20 A, the new speed in rpm, rounded off to two decimal places, is _____.

Solution

The back emf of a DC machine is \(E_b = K\phi N\). With \(\phi\) held constant, speed is directly proportional to back emf, and the back emf follows from the armature circuit equation \(E_b = V - I_a R_a\).

Unloading a shunt motor therefore raises its speed only slightly, which is what gives the shunt machine its near-constant-speed characteristic. Background: DC motor characteristics.

2 marks, numerical answer. A 240 V DC shunt motor drives a load whose torque is independent of speed. The armature circuit resistance is \(0.4\,\Omega\) and the field current is held fixed, so the flux per pole is constant. At rated conditions the armature draws 30 A and the motor runs at 1200 rpm. The speed is to be reduced to 900 rpm by inserting a resistance \(R\) in series with the armature, the supply voltage and field current being unchanged. The value of \(R\), in ohms rounded off to two decimal places, is _____.

Solution

The developed torque is \(T = K\phi I_a\). The load torque is the same at both speeds and the flux is unchanged, so the armature current is unchanged:

At the original operating point,

Since \(E_b \propto N\) at constant flux, the back emf required at 900 rpm is

The armature circuit equation with the added resistance gives

The price of this method is dissipation: the added resistor burns \(I_a^2 R = 30^2 \times 1.9 = 1710\,\mathrm{W}\), which is why armature-resistance control is used only for short-duration or low-power duty. See Speed control of DC motors.

2 marks, multiple choice. A three-phase, 4-pole, 50 Hz induction motor has a rotor resistance of \(0.12\,\Omega\) per phase and a standstill rotor leakage reactance of \(0.60\,\Omega\) per phase, both referred to the stator. The stator resistance and leakage reactance and the magnetising branch are all neglected, so the applied phase voltage appears entirely across the rotor branch. The ratio of the starting torque to the maximum torque is

1. 0.385
2. 0.400
3. 0.500
4. 0.769

Solution

With the stator impedance neglected, the torque at slip \(s\) is proportional to the rotor power transferred across the air gap:

Differentiating with respect to \(s\) and equating to zero gives the slip at which the torque peaks:

Forming the ratio of the torque at any slip \(s\) to the maximum torque and simplifying with \(R_2 = s_{mT}X_2\),

At starting the rotor is stationary, so \(s = 1\):

For completeness, the synchronous speed is \(N_s = 120f/P = 120\times 50/4 = 1500\,\mathrm{rpm}\), so the maximum torque occurs at \(N = 1500(1-0.2) = 1200\,\mathrm{rpm}\). More on this in Torque-slip characteristics of an induction motor.

1 mark, multiple choice. A three-phase induction motor operates at a slip of 4 %. The power crossing the air gap is 12 kW and the combined friction and windage loss is 300 W. Stray load loss is neglected. The mechanical power available at the shaft is

1. 11.52 kW
2. 11.70 kW
3. 11.22 kW
4. 12.00 kW

Solution

Of the power that crosses the air gap, a fraction equal to the slip is dissipated as rotor copper loss and the remainder is converted to mechanical form:

The friction and windage loss is subtracted from the developed mechanical power to give the shaft output:

Option (A) is the trap for a candidate who stops at the developed power and forgets the rotational loss. See Power stages of a three-phase induction motor.

2 marks, numerical answer. A 5 MVA, 6.6 kV (line), three-phase star-connected synchronous generator has a synchronous reactance of \(4\,\Omega\) per phase. The armature resistance is negligible and magnetic saturation is ignored. The machine delivers rated current at a power factor of 0.8 lagging. Using the synchronous impedance method, the percentage voltage regulation, rounded off to two decimal places, is _____.

Solution

All quantities are taken per phase. For a star connection,

Taking \(V_{ph}\) as reference and the current lagging by \(\phi = \cos^{-1}0.8 = 36.87^\circ\), the generator equation with \(R_a = 0\) is \(E = V_{ph} + jIX_s\). Multiplying the current phasor by \(j\) advances it by \(90^\circ\), so the drop lies at \(-36.87^\circ + 90^\circ = 53.13^\circ\):

An equivalent single-line form is \(|E| = \sqrt{(V\cos\phi)^2 + (V\sin\phi + IX_s)^2}\), which gives \(\sqrt{3048.41^2 + (2286.31+1749.55)^2} = 5057.76\,\mathrm{V}\), confirming the result. Further reading: Voltage regulation of an alternator.

2 marks, numerical answer. A 500 kVA, 11 kV / 433 V, 50 Hz three-phase transformer has its HV winding connected in delta and its LV winding connected in star. Magnetising current, losses and internal impedance drops are neglected, so the rated line values may be used directly. At rated load, the current flowing in each HV winding, in amperes rounded off to two decimal places, is _____.

Solution

In a delta connection each winding is connected directly between two lines, so the winding voltage equals the line voltage, while the winding current is the line current divided by \(\sqrt{3}\). The rated HV line current is

The same figure follows directly from a per-limb power balance, which is the quicker route: each of the three limbs handles one third of the rating at the full 11 kV winding voltage,

For reference, the LV star winding voltage is \(433/\sqrt{3} = 250\,\mathrm{V}\), so the turns ratio per limb is \(11000/250 = 44\), which is not the same as the 11000/433 line-voltage ratio of 25.4. Confusing the two is the usual error. See Three-phase transformer winding connections.

2 marks, multiple choice. A three-phase squirrel-cage induction motor draws a starting current equal to 6 times its full-load current when switched direct-on-line at rated voltage. The full-load slip is 4 %. The motor is instead started through a star-delta starter, the machine being designed to run with its stator in delta. Neglecting the change in rotor resistance with frequency, the ratio of the starting torque to the full-load torque is

1. 0.16
2. 0.24
3. 0.32
4. 0.48

Solution

Torque is proportional to the square of the rotor current and, at a given slip, to the square of the applied voltage. Comparing the starting condition (slip 1) with the full-load condition (slip \(s_{fl}\)) gives the standard result

Direct-on-line, with \(I_{sc}/I_{fl} = 6\) and \(s_{fl} = 0.04\),

A star-delta starter applies the line voltage to a star-connected stator at start, so each winding receives \(V_L/\sqrt{3}\) instead of \(V_L\). The winding current, and hence the line current, falls to one third of its delta value, and the torque, being proportional to the square of the winding voltage, also falls to one third:

The starter therefore reduces both the starting current and the starting torque by the same factor of three, which is exactly why star-delta starting is restricted to loads that start light. Background: Starting methods for induction motors.