GATE EE Solved Problems

GATE 2025 Electrical Engineering (EE) Electrical Machines (2025)

Solved problems

Author: Prof. Mithun Mondal Subject: Electrical Machines Year: 2025 Total Questions: 7
Section 01

1-Mark Questions

QQuestion 1 1 Mark

The operating region of the developed torque (\(T_{em}\)) and speed (\(\omega\)) of an induction motor drive is given by the shaded region OQRE in the figure. The load torque (\(T_L\)) characteristic is also shown. The motor drive moves from the initial operating point O to the final operating point S. Which one of the following trajectories will take the shortest time?

Figure 1.1
Figure 1.1

AOptions

  1. \(O - Q - R - S\)
  2. \(O - P - S\)
  3. \(O - E - S\)
  4. \(O - F - S\)

SSolution

To minimize the time taken to move from point O to point S, we need to maximize the accelerating torque throughout the trajectory.

The accelerating torque is given by:

\[T_{acc} = T_{em} - T_L\]

The angular acceleration is:

\[\alpha = \frac{T_{acc}}{J} = \frac{T_{em} - T_L}{J}\]

where \(J\) is the moment of inertia.

For shortest time, we need maximum acceleration, which means we should follow a path where \((T_{em} - T_L)\) is maximum.

Looking at the trajectories:

  • Path \(O-Q-R-S\): Follows the upper boundary with maximum torque capability
  • Path \(O-P-S\): Intermediate torque path
  • Path \(O-E-S\): Lower torque path
  • Path \(O-F-S\): Goes below the load characteristic (not possible for motoring)

The path \(O-Q-R-S\) provides the maximum accelerating torque \((T_{em} - T_L)\) throughout the motion, resulting in maximum acceleration and minimum time.

Correct answer: A

QQuestion 2 1 Mark

Instrument(s) required to synchronize an alternator to the grid is/are

AOptions

  1. Voltmeter
  2. Wattmeter
  3. Synchroscope
  4. Stroboscope

SSolution

For synchronizing an alternator (synchronous generator) to the grid, the following conditions must be satisfied:

  1. Equal voltage magnitude
  2. Same frequency
  3. Same phase sequence
  4. Correct phase angle (voltages in phase)

Instruments used for synchronization:

  • Voltmeter: Checks if the incoming alternator voltage matches the grid voltage
  • Synchroscope: Most important instrument - displays the phase angle difference and frequency difference between the alternator and grid. It shows when the voltages are in phase and have the same frequency
  • Three lamps method: Alternative method using lamps to check phase relationship
  • Frequency meter: Checks frequency matching

A wattmeter is used after synchronization to measure power flow, not for the synchronization process itself.

A stroboscope is used for speed measurement and checking rotating equipment, not for electrical synchronization.

The most appropriate and commonly used instrument for synchronization is the synchroscope, which provides comprehensive information about both frequency and phase angle differences.

Correct answer: C

\textit{Note: In practice, both voltmeter and synchroscope are used together, but the synchroscope is the primary instrument that ensures proper synchronization.}

QQuestion 3 1 Mark

The induced emf in a 3.3 kV, 4 pole, 3-phase star connected synchronous motor is considered to be equal and in phase with the terminal voltage under no load condition. On application of a mechanical load, the induced emf phasor is deflected by an angle of 2° mechanical with respect to the terminal voltage phasor. If the synchronous reactance is 2 \(\Omega\), and stator resistance is negligible, then the motor armature current magnitude, in ampere, during loaded condition is closest to, ____________ (round off to two decimal places).

SSolution

Given:

  • Line voltage: \(V_L = 3.3\) kV \(= 3300\) V
  • Number of poles: \(P = 4\)
  • Star connection
  • Load angle (mechanical): \(\delta_m = 2°\)
  • Synchronous reactance: \(X_s = 2\) \(\Omega\)
  • Stator resistance: \(R_a = 0\) (negligible)

Solution:

Step 1: Convert mechanical angle to electrical angle

\[\delta_e = \delta_m \times \frac{P}{2} = 2° \times \frac{4}{2} = 4°\]

Step 2: Calculate phase voltage For star connection:

\[V_{ph} = \frac{V_L}{\sqrt{3}} = \frac{3300}{\sqrt{3}} = 1905.26 \text{ V}\]

Step 3: At no load, \(E_f = V_{ph}\) and they are in phase

Under load, the induced EMF \(E_f\) leads the terminal voltage \(V_{ph}\) by load angle \(\delta = 4°\) (for motor operation)

Step 4: Calculate armature current

For a synchronous motor with negligible resistance:

\[\vec{E}_f = \vec{V}_{ph} + j\vec{I}_a X_s\]

The phasor difference between \(E_f\) and \(V_{ph}\) is:

\[\vec{E}_f - \vec{V}_{ph} = jI_a X_s\]

Taking \(V_{ph}\) as reference:

\[V_{ph} = 1905.26 \angle 0°\]
\[E_f = 1905.26 \angle 4°\]

The voltage difference:

\[\Delta V = E_f \angle 4° - V_{ph} \angle 0°\]
\[\Delta V = 1905.26(\cos 4° + j\sin 4°) - 1905.26\]
\[\Delta V = 1905.26(\cos 4° - 1) + j \cdot 1905.26 \sin 4°\]
\[\Delta V = 1905.26(-0.00244) + j \cdot 1905.26(0.06976)\]
\[\Delta V = -4.65 + j132.95\]
\[|\Delta V| = \sqrt{(-4.65)^2 + (132.95)^2} = 133.03 \text{ V}\]

Since \(jI_a X_s = \Delta V\):

\[I_a X_s = |\Delta V| = 133.03\]
\[I_a = \frac{133.03}{2} = 66.52 \text{ A}\]

Alternatively, using the approximation for small angles:

\[I_a = \frac{2E_f \sin(\delta/2)}{X_s} = \frac{2 \times 1905.26 \times \sin(2°)}{2}\]
\[I_a = 1905.26 \times 0.03490 = 66.51 \text{ A}\]

Answer: 66.52 A

Section 02

2-Mark Questions

QQuestion 4 2 Mark

The transformer connection given in the figure is part of a balanced 3-phase circuit where the phase sequence is "\(abc\)". The primary to secondary turns ratio is 2:1. If \((I_a + I_b + I_c = 0)\), then the relationship between \(I_A\) and \(I_{ad}\) will be

Figure 4.1
Figure 4.1

AOptions

  1. \(\displaystyle\frac{|I_A|}{|I_{ad}|} = \frac{1}{2\sqrt{3}}\) and \(I_{ad}\) lags \(I_A\) by 30°.
  2. \(\displaystyle\frac{|I_A|}{|I_{ad}|} = \frac{1}{2\sqrt{3}}\) and \(I_{ad}\) leads \(I_A\) by 30°.
  3. \(\displaystyle\frac{|I_A|}{|I_{ad}|} = 2\sqrt{3}\) and \(I_{ad}\) lags \(I_A\) by 30°.
  4. \(\displaystyle\frac{|I_A|}{|I_{ad}|} = 2\sqrt{3}\) and \(I_{ad}\) leads \(I_A\) by 30°.

SSolution

This is a Delta-Star (Δ-Y) transformer connection with turns ratio 2:1.

Primary side: Delta connection (ABC)\\ Secondary side: Star connection (abc)\\ Turns ratio: \(N_1:N_2 = 2:1\)

Analysis:

Step 1: Voltage relationship

For Delta-Star connection:

\[\frac{V_{L,primary}}{V_{L,secondary}} = \frac{N_1}{N_2} \times \sqrt{3} = 2\sqrt{3}\]

Step 2: Current relationship

For line current \(I_A\) on primary (delta) side and line current \(I_{ad}\) on secondary (star) side:

From ampere-turn balance and connection topology:

\[\frac{|I_{ad}|}{|I_A|} = \frac{N_1}{N_2} \times \sqrt{3} = 2\sqrt{3}\]

Therefore:

\[\frac{|I_A|}{|I_{ad}|} = \frac{1}{2\sqrt{3}}\]

Step 3: Phase relationship

For Delta-Star transformer with standard connections:

  • Primary line current \(I_A\) is at some reference angle
  • Secondary line current \(I_{ad}\) (which is also phase current in star) is related to the primary delta currents

In a Δ-Y transformer with "abc" sequence:

  • If we consider the standard connection where secondary neutral is formed
  • The secondary line currents lead the corresponding primary line currents by 30°

However, considering the current transformation from delta to star and the connection shown:

  • \(I_A\) flows in the delta winding
  • \(I_{ad}\) is reflected from the star side
  • Due to the Δ-Y connection configuration, \(I_{ad}\) leads \(I_A\) by 30°

Correct answer: B

\[\frac{|I_A|}{|I_{ad}|} = \frac{1}{2\sqrt{3}} \text{ and } I_{ad} \text{ leads } I_A \text{ by } 30°\]

QQuestion 5 2 Mark

A DC series motor with negligible series resistance is running at a certain speed driving a load, where the load torque varies as cube of the speed. The motor is fed from a 400 V DC source and draws 40 A armature current. Assume linear magnetic circuit. The external resistance, in \(\Omega\), that must be connected in series with the armature to reduce the speed of the motor by half, is closest to

\vspace{3cm}

AOptions

  1. 23.28
  2. 4.82
  3. 46.7
  4. 0

SSolution

Given:

  • DC series motor
  • Negligible series resistance initially
  • Supply voltage: \(V = 400\) V
  • Initial armature current: \(I_{a1} = 40\) A
  • Load torque: \(T_L \propto N^3\)
  • Linear magnetic circuit: \(\phi \propto I_a\)
  • Required: Speed to be reduced to \(N_2 = N_1/2\)

Solution:

Step 1: For DC series motor

\[E_b = V - I_a(R_a + R_{se} + R_{ext})\]

Initially, with negligible resistance:

\[E_{b1} = V - 0 = 400 \text{ V}\]

Step 2: Back EMF relation

\[E_b = K\phi N \propto I_a N \text{ (since } \phi \propto I_a \text{)}\]

Therefore:

\[E_{b1} = K_1 I_{a1} N_1\]
\[E_{b2} = K_1 I_{a2} N_2\]

Step 3: Torque relationship

For DC series motor:

\[T = K\phi I_a \propto I_a^2 \text{ (since } \phi \propto I_a \text{)}\]

Load torque:

\[T_L \propto N^3\]

At steady state, \(T = T_L\):

\[I_{a1}^2 \propto N_1^3\]
\[I_{a2}^2 \propto N_2^3\]

Taking ratio:

\[\frac{I_{a2}^2}{I_{a1}^2} = \frac{N_2^3}{N_1^3} = \left(\frac{N_2}{N_1}\right)^3 = \left(\frac{1}{2}\right)^3 = \frac{1}{8}\]
\[\frac{I_{a2}}{I_{a1}} = \frac{1}{2\sqrt{2}}\]
\[I_{a2} = \frac{40}{2\sqrt{2}} = \frac{40}{2.828} = 14.14 \text{ A}\]

Step 4: Back EMF at new condition

\[\frac{E_{b2}}{E_{b1}} = \frac{I_{a2} N_2}{I_{a1} N_1} = \frac{14.14}{40} \times \frac{1}{2} = \frac{14.14}{80} = 0.177\]
\[E_{b2} = 0.177 \times 400 = 70.8 \text{ V}\]

Step 5: Calculate external resistance

At new condition:

\[E_{b2} = V - I_{a2} R_{ext}\]
\[70.8 = 400 - 14.14 \times R_{ext}\]
\[14.14 \times R_{ext} = 329.2\]
\[R_{ext} = \frac{329.2}{14.14} = 23.28 \text{ } \Omega\]

Correct answer: A (23.28 \(\Omega\))

QQuestion 6 2 Mark

A 3-phase, 400 V, 4 pole, 50 Hz star connected induction motor has the following parameters referred to the stator:

\(R_r' = 1\Omega\), \(X_s = X_r' = 2\Omega\)

Stator resistance, magnetizing reactance and core loss of the motor are neglected. The motor is run with constant V/f control from a drive. For maximum starting torque, the voltage and frequency output, respectively, from the drive, is closest to,

AOptions

  1. 400 V and 50 Hz
  2. 200 V and 25 Hz
  3. 100 V and 12.5 Hz
  4. 300 V and 37.5 Hz

SSolution

Given:

  • Rated voltage: \(V = 400\) V (line)
  • Rated frequency: \(f = 50\) Hz
  • Poles: \(P = 4\)
  • Star connection
  • \(R_r' = 1\) \(\Omega\), \(X_s = X_r' = 2\) \(\Omega\)
  • \(R_s = 0\) (neglected), \(X_m = \infty\) (neglected)

Solution:

Step 1: Torque equation for induction motor

The torque is given by:

\[T = \frac{3V^2 s R_r'}{2\pi f \left[(R_r')^2 + (sX_{eq})^2\right]}\]

where \(X_{eq} = X_s + X_r' = 2 + 2 = 4\) \(\Omega\) (at rated frequency)

At starting, \(s = 1\):

\[T_{st} = \frac{3V^2 R_r'}{2\pi f \left[(R_r')^2 + (X_{eq})^2\right]}\]

Step 2: Condition for maximum torque

For maximum torque at starting (when \(s = 1\)):

\[sX_{eq} = R_r'\]

At starting, \(s = 1\):

\[X_{eq} = R_r'\]
\[X_s + X_r' = R_r'\]

But reactance \(X \propto f\):

\[X_{eq}(f) = X_{eq,rated} \times \frac{f}{f_{rated}} = 4 \times \frac{f}{50}\]

Step 3: Find frequency for maximum starting torque

For maximum starting torque:

\[4 \times \frac{f}{50} = 1\]
\[f = \frac{50}{4} = 12.5 \text{ Hz}\]

Step 4: Find voltage for constant V/f control

With constant V/f control:

\[\frac{V}{f} = \frac{400}{50} = 8 \text{ V/Hz}\]

At \(f = 12.5\) Hz:

\[V = 8 \times 12.5 = 100 \text{ V}\]

Verification:

At 100 V, 12.5 Hz:

  • \(X_{eq} = 4 \times \frac{12.5}{50} = 1\) \(\Omega\)
  • \(R_r' = 1\) \(\Omega\)
  • Condition \(X_{eq} = R_r'\) is satisfied ✓

This gives maximum starting torque.

Correct answer: C (100 V and 12.5 Hz)

QQuestion 7 2 Mark

Using shunt capacitors, the power factor of a 3-phase, 4 kV induction motor (drawing 390 kVA at 0.77 pf lag) is to be corrected to 0.85 pf lag. The line current of the capacitor bank, in A, is ____________ (round off to one decimal place).

SSolution

Given:

  • Line voltage: \(V_L = 4\) kV \(= 4000\) V
  • Apparent power: \(S_1 = 390\) kVA
  • Initial power factor: \(\cos\phi_1 = 0.77\) (lag)
  • Final power factor: \(\cos\phi_2 = 0.85\) (lag)
  • 3-phase system

Solution:

Step 1: Calculate real power

Active power (remains constant):

\[P = S_1 \cos\phi_1 = 390 \times 0.77 = 300.3 \text{ kW}\]

Step 2: Calculate initial reactive power

\[\sin\phi_1 = \sqrt{1 - \cos^2\phi_1} = \sqrt{1 - 0.77^2} = \sqrt{0.4071} = 0.638\]
\[Q_1 = S_1 \sin\phi_1 = 390 \times 0.638 = 248.82 \text{ kVAR}\]

Or alternatively:

\[Q_1 = P \tan\phi_1 = 300.3 \times \tan(\cos^{-1}(0.77)) = 300.3 \times 0.8286 = 248.83 \text{ kVAR}\]

Step 3: Calculate final reactive power

\[\sin\phi_2 = \sqrt{1 - \cos^2\phi_2} = \sqrt{1 - 0.85^2} = \sqrt{0.2775} = 0.527\]
\[Q_2 = P \tan\phi_2 = 300.3 \times \tan(\cos^{-1}(0.85)) = 300.3 \times 0.6197 = 186.06 \text{ kVAR}\]

Step 4: Calculate capacitor bank kVAR

Required capacitive reactive power:

\[Q_C = Q_1 - Q_2 = 248.83 - 186.06 = 62.77 \text{ kVAR}\]

Step 5: Calculate line current of capacitor bank

For 3-phase capacitor bank:

\[Q_C = \sqrt{3} V_L I_C\]
\[I_C = \frac{Q_C}{\sqrt{3} V_L} = \frac{62.77 \times 10^3}{\sqrt{3} \times 4000}\]
\[I_C = \frac{62770}{6928.2} = 9.06 \text{ A}\]

Alternative calculation:

Using the formula:

\[Q_C = P(\tan\phi_1 - \tan\phi_2)\]
\[Q_C = 300.3(0.8286 - 0.6197) = 300.3 \times 0.2089 = 62.73 \text{ kVAR}\]
\[I_C = \frac{62.73 \times 10^3}{\sqrt{3} \times 4000} = 9.05 \text{ A}\]

Answer: 9.1 A (rounded to one decimal place)