GATE Technical: MAIN PAPER - EE questions with solutions

Q1. The equivalent circuits of a diode, during forward biased and reverse biased conditions, are shown in the figure. If such a diode is used in clipper circuit of figure given above, the output voltage (vₒ) of the circuit will be

1. A waveform clipped at +5 V and -5 V
2. A waveform clipped at +5.7 V and -10 V
3. A waveform clipped at +5.7 V and -10 V
4. A waveform clipped at 10 V and -5.7 V

Answer: A waveform clipped at +5.7 V and -10 V

The correct option indicates that the diode's forward voltage drop and reverse characteristics effectively limit the output voltage to the specified clipping levels, which are determined by the diode's behavior in the circuit configuration.

Q2. An input device is interfaced with Intel 8085A microprocessor as memory mapped I/O. The address of the device is 2500H. In order to input data from the device to accumulator, the sequence of instructions will be

1. LXI H, 2500H MOV A, M
2. LXI H, 2500H MOV M, A
3. LHLD 2500H MOV A, M
4. LHLD 2500H MOV M, A

Answer: LXI H, 2500H MOV A, M

The correct option is right because it correctly loads the address of the input device into the HL register pair using LXI, and then retrieves the data from that memory location into the accumulator with the MOV A, M instruction.

Q3. Distributed winding and short chording employed in AC machines will result in

1. increase in emf and reduction in harmonics.
2. reduction in emf and increase in harmonics.
3. increase in both emf and harmonics.
4. reduction in both emf and harmonics.

Answer: reduction in both emf and harmonics.

Distributed winding and short chording help to smooth out the magnetic field in AC machines, which leads to a reduction in the generated electromotive force (emf) and minimizes the presence of harmonics, resulting in a more efficient operation.

Q4. In a stepper motor, the detent torque means

1. minimum of the static torque with the phase winding excited.
2. maximum of the static torque with the phase winding excited.
3. minimum of the static torque with the phase winding unexcited.
4. maximum of the static torque with the phase winding unexcited.

Answer: maximum of the static torque with the phase winding unexcited.

Detent torque refers to the maximum static torque that a stepper motor can hold when the windings are not energized, allowing the motor to maintain its position against external forces without any electrical input.

Q5. A two machine power system is shown below. Transmission line XY has positive sequence impedance of Z1 Ω and zero sequence impedance of Z0 Ω. An ‘a’ phase to ground fault with zero fault impedance occurs at the centre of the transmission line. Bus voltage at X and line current from X to F for the phase ‘a’, are given by Va Volts and Ia Amperes, respectively. Then, the impedance measured by the ground distance relay located at the terminal X of line XY will be given by

1. Z1/2 Ω
2. Z0/2 Ω
3. (Z0+Z1)/2 Ω
4. Va/Ia Ω

Answer: Va/Ia Ω

The impedance measured by the ground distance relay is calculated using the voltage at the relay location divided by the current flowing through the fault, which is represented by Va/Ia. This relationship holds true for any fault condition, making this option the correct choice.

Q6. The impulse response of a causal linear time-invariant system is given as h(t). Now consider the following two statements: Statement (I): Principle of superposition holds Statement (II): h(t) = 0 for t < 0 Which one of the following statements is correct?

1. Statement (I) is correct and Statement (II) is wrong
2. Statement (II) is correct and Statement (I) is wrong
3. Both Statement (I) and Statement (II) are wrong
4. Both Statement (I) and Statement (II) are correct

Answer: Both Statement (I) and Statement (II) are correct

The principle of superposition applies to linear systems, meaning that the response to a combination of inputs is the sum of the responses to each input individually, which is true for causal linear time-invariant systems. Additionally, the definition of a causal system requires that the impulse response h(t) is zero for all times before t=0, confirming that both statements are indeed correct.

Q7. It is desired to measure parameters of 230 V / 115 V, 2 kVA, single-phase transformer. The following wattmeters are available in a laboratory: W1: 250 V, 10 A, Low Power Factor W2: 250 V, 5 A, Low Power Factor W3: 150 V, 10 A, High Power Factor W4: 150 V, 5 A, High Power Factor The wattmeters used in open circuit test and short circuit test of the transformer will respectively be

1. W1 and W2
2. W2 and W4
3. W1 and W4
4. W2 and W3

Answer: W2 and W4

W2 is suitable for the open circuit test due to its lower current rating, which is ideal for measuring low power at high voltage, while W4 is appropriate for the short circuit test as it can handle higher current with a high power factor, ensuring accurate measurement under those conditions.

Q8. A coil of 300 turns is wound on a non-magnetic core having a mean circumference of 300 mm and a cross-sectional area of 300 mm². The inductance of the coil corresponding to a magnetizing current of 3A will be (Given that μ0 = 4π × 10⁻⁷ H/m)

1. 37.68 μH
2. 113.04 μH
3. 37.68 mH
4. 113.04 mH

Answer: 113.04 μH

The inductance of a coil can be calculated using the formula L = (N² * μ * A) / l, where N is the number of turns, μ is the permeability of the core, A is the cross-sectional area, and l is the mean length of the coil. Given the values and using the appropriate calculations, the inductance is determined to be 113.04 μH.

Q9. A 230 V, 50 Hz, 4-pole, single-phase induction motor is rotating in the clockwise (forward) direction at a speed of 1425 rpm. If the rotor resistance at standstill is 7.8 Ω, then the effective rotor resistance in the backward branch of the equivalent circuit will be

1. (A) 2 Ω
2. (B) 4 Ω
3. (C) 78 Ω
4. (D) 156 Ω

Answer: (A) 2 Ω

Ns=1500 rpm, so slip s=(1500-1425)/1500=0.05 and backward slip 2-s=1.95. The effective rotor resistance in the backward branch is R2/[2(2-s)] = 7.8/(2 x 1.95) = 2 ohm.

Q10. A 400 V, 50 Hz, 30 hp, three-phase induction motor is drawing 50 A current at 0.8 power factor lagging. The stator and rotor copper losses are 1.5 kW and 900 W respectively. The friction and windage losses are 1050 W and the core losses are 1200 W. The air-gap power of the motor will be

1. (A) 23.06 kW
2. (B) 24.11 kW
3. (C) 25.01 kW
4. (D) 26.21 kW

Answer: (C) 25.01 kW

The air-gap power is calculated by subtracting all losses from the input power. The input power can be determined from the current, voltage, and power factor, and after accounting for stator and rotor copper losses, friction and windage losses, and core losses, the remaining power is the air-gap power, which in this case is 25.01 kW.

Q11. Voltage phasors at the two terminals of a transmission line of length 70 km have a magnitude of 1.0 per unit but are 180 degrees out of phase. Assuming that the maximum load current in the line is 1/5th of minimum 3-phase fault current, which one of these transmission line protection schemes will NOT pick up for this condition?

1. (A) Distance protection using mho relays with zone-1 set to 80% of the line impedance
2. (B) Directional overcurrent protection set to pick up at 1.25 times the maximum load current
3. (C) Pilot relaying system with directional comparison scheme
4. (D) Pilot relaying system with segregated phase comparison scheme

Answer: (B) Directional overcurrent protection set to pick up at 1.25 times the maximum load current

The directional overcurrent protection set to pick up at 1.25 times the maximum load current will not activate because the current during the fault condition is significantly higher than the maximum load current, and the protection scheme is designed to respond only to currents above that threshold.