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

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:

The angular acceleration is:

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

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.

Both instruments are needed at the moment of paralleling: the voltmeter confirms that the incoming machine's voltage magnitude equals the grid voltage, and the synchroscope confirms equality of frequency and coincidence of phase.

Correct answer: A and C (Voltmeter and Synchroscope)

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

Step 2: Calculate phase voltage For star connection:

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:

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

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

The voltage difference:

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

Alternatively, using the approximation for small angles:

Answer: 66.52 A

SSolution

The figure shows a star-connected primary (A, B and C taken to a common neutral) feeding a delta-connected secondary whose corners are brought out as a, c and b. The turns ratio is \(N_1 : N_2 = 2 : 1\).

Step 1: Ampere-turn balance

The primary is star connected, so \(I_A\) is both its line current and its winding current. On the limb carrying the secondary winding \(a\),

and \(I_a\) is in phase with \(I_A\).

Step 2: Delta corner to line current

At corner a the winding current \(I_a\) enters and \(I_c\) leaves, so KCL gives

With phase sequence \(abc\), \(I_a = I\angle 0^\circ\) and \(I_c = I\angle 120^\circ\), hence

Step 3: Combine

and \(I_{ad}\) lags \(I_a\), and therefore \(I_A\), by \(30^\circ\).

Correct answer: A

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

Initially, with negligible resistance:

Step 2: Back EMF relation

Therefore:

Step 3: Torque relationship

For DC series motor:

Load torque:

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

Taking ratio:

Step 4: Back EMF at new condition

Step 5: Calculate external resistance

At new condition:

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

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:

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

At starting, \(s = 1\):

Step 2: Condition for maximum torque

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

At starting, \(s = 1\):

But reactance \(X \propto f\):

Step 3: Find frequency for maximum starting torque

For maximum starting torque:

Step 4: Find voltage for constant V/f control

With constant V/f control:

At \(f = 12.5\) Hz:

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)

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):

Step 2: Calculate initial reactive power

Or alternatively:

Step 3: Calculate final reactive power

Step 4: Calculate capacitor bank kVAR

Required capacitive reactive power:

Step 5: Calculate line current of capacitor bank

For 3-phase capacitor bank:

Alternative calculation:

Using the formula:

Answer: 9.1 A (rounded to one decimal place)