StreakPeaked· Practice

ExamsGATETechnical › ELECTRICAL – EE

GATE Technical: ELECTRICAL – EE questions with solutions

28 questions with worked solutions.

Questions

Q1. C0 is the capacitance of a parallel plate capacitor with air as dielectric (as in figure (a)). If, half of the entire gap as shown in figure (b), is filled with a dielectric of permittivity εr, the expression for the modified capacitance is

  1. C0/2 (1 + εr)
  2. (C0 + εr)
  3. C0/2 εr
  4. C0(1 + εr)

Answer: C0/2 (1 + εr)

The correct option is derived from the fact that filling half of the capacitor's gap with a dielectric increases the capacitance by a factor related to the dielectric's permittivity, while the other half remains air. The total capacitance can be calculated as the sum of the capacitance contributions from both sections, leading to the expression C0/2 (1 + εr).

Q2. The undesirable property of an electrical insulating material is

  1. high dielectric strength
  2. high relative permittivity
  3. high thermal conductivity
  4. high insulation resistivity

Answer: high relative permittivity

High relative permittivity indicates that a material can store more electric charge, which can lead to increased leakage currents and reduced effectiveness as an insulator, making it undesirable for electrical insulating applications.

Q3. Q.17 In the formation of Routh-Hurwitz array for a polynomial, all the elements of a row have zero values. This premature termination of the array indicates the presence of

  1. only one root at the origin
  2. imaginary roots
  3. only positive real roots
  4. only negative real roots

Answer: imaginary roots

When a row in the Routh-Hurwitz array consists entirely of zeros, it indicates that the polynomial has roots that are not real, which typically means the presence of imaginary roots. This situation arises because the polynomial's behavior suggests oscillatory characteristics, leading to complex conjugate pairs.

Q4. Q.18 The root locus of a unity feedback system is shown in the figure

  1. C(s)/R(s) = K / ((s+1)(s+2))
  2. C(s)/R(s) = -K / ((s+1)(s+2)+K)
  3. C(s)/R(s) = K / ((s+1)(s+2)-K)
  4. C(s)/R(s) = K / ((s+1)(s+2)+K)

Answer: C(s)/R(s) = K / ((s+1)(s+2)+K)

The correct option represents a system where the feedback is negative and the gain K is added to the denominator, which is consistent with the behavior of the root locus as it indicates stability and the movement of poles in the complex plane as K varies.

Q5. Q.19 Power consumed by a balanced 3-phase, 3-wire load is measured by the two wattmeter method. The first wattmeter reads twice that of the second. Then the load impedance angle in radians is

  1. π/12
  2. π/8
  3. π/6
  4. π/3

Answer: π/6

With W1=2W2, tan(phi)=sqrt(3)(2W2-W2)/(2W2+W2)=sqrt(3)/3=1/sqrt(3), so phi=30 deg = pi/6 radians.

Q6. Q.20 In an oscilloscope screen, linear sweep is applied at the

  1. vertical axis
  2. horizontal axis
  3. origin
  4. both horizontal and vertical axis

Answer: horizontal axis

The linear sweep is applied to the horizontal axis of an oscilloscope to represent time, allowing the waveform to be displayed as it changes over time. This horizontal movement is essential for visualizing the signal's behavior in a time-based format.

Q7. A cascade of three identical modulo-5 counters has an overall modulus of

  1. 5
  2. 25
  3. 125
  4. 625

Answer: 125

The overall modulus of a cascade of counters is calculated by raising the modulus of the individual counter to the power of the number of counters. Since each modulo-5 counter has a modulus of 5 and there are three counters, the overall modulus is 5³, which equals 125.

Q8. In an unbalanced three phase system, phase current Iₐ = 1∠(-90°) pu, negative sequence current I_b2 = 4∠(-150°) pu, zero sequence current I_c0 = 3∠90° pu. The magnitude of phase current I_b in pu is

  1. 1.00
  2. 7.81
  3. 11.53
  4. 13.00

Answer: 11.53

From the given quantities the zero, positive and negative sequence components of phase a are I0=3<90, I1=8<-90, I2=4<90. Then Ib = I0 + a^2 I1 + a I2 evaluates to about -10.39 + j5.0, whose magnitude is 11.53 pu, not 7.81 pu.

Q9. The following four vector fields are given in Cartesian co-ordinate system. The vector field which does not satisfy the property of magnetic flux density is

  1. y² aₓ + z² a_y + x² a_z
  2. z² aₓ + x² a_y + y² a_z
  3. x² aₓ + y² a_y + z² a_z
  4. y² z² aₓ + x² z² a_y + x² y² a_z

Answer: x² aₓ + y² a_y + z² a_z

The correct option does not satisfy the condition of magnetic flux density because it does not have the appropriate divergence properties, specifically that the divergence of a magnetic field must equal zero. This means that the vector field must be solenoidal, which is not the case for the given option.

Q10. Let X(z) = 1/(1 - z⁻³) be the Z-transform of a causal signal x[n]. Then, the values of x[2] and x[3] are

  1. 0 and 0
  2. 0 and 1
  3. 1 and 0
  4. 1 and 1

Answer: 0 and 1

Expanding X(z)=1/(1-z^-3)=1+z^-3+z^-6+... gives x[n]=1 at n=0,3,6,... and 0 elsewhere. Thus x[2]=0 and x[3]=1, which is option 1; the stored '0 and 0' is wrong.

Q11. Let f(t) be a continuous time signal and let F(ω) be its Fourier Transform defined by F(ω) = ∫ from -∞ to ∞ f(t)e^-jωt dt. Define g(t) by g(t) = ∫ from -∞ to ∞ F(u)e^-jut du. What is the relationship between f(t) and g(t)?

  1. g(t) would always be proportional to f(t).
  2. g(t) would be proportional to f(t) if f(t) is an even function.
  3. g(t) would be proportional to f(t) only if f(t) is a sinusoidal function.
  4. g(t) would never be proportional to f(t).

Answer: g(t) would be proportional to f(t) if f(t) is an even function.

Using e^{-jut} in both transforms, g(t)=2pi f(-t). This is proportional to f(t) precisely when f is even (f(-t)=f(t)). So g is proportional to f only for even f.

Q12. The core loss of a single phase, 230/115 V, 50 Hz power transformer is measured from 230 V side by feeding the primary (230 V side) from a variable voltage variable frequency source while keeping the secondary open circuited. The core loss is measured to be 1050 W for 230 V, 50 Hz input. The core loss is again measured to be 500 W for 138 V, 30 Hz input. The hysteresis and eddy current losses of the transformer for 230 V, 50 Hz input are respectively,

  1. 508 W and 542 W.
  2. 468 W and 582 W.
  3. 498 W and 552 W.
  4. 488 W and 562 W.

Answer: 508 W and 542 W.

The core loss in a transformer consists of hysteresis and eddy current losses, which can be calculated using the Steinmetz equation. By analyzing the losses at different voltages and frequencies, the specific contributions of hysteresis and eddy current losses can be determined, leading to the correct values of 508 W for hysteresis and 542 W for eddy currents.

Q13. A 3 phase, 50 Hz, six pole induction motor has a rotor resistance of 0.1 Ω and reactance of 0.92 Ω. Neglect the voltage drop in stator and assume that the rotor resistance is constant. Given that the full load slip is 3%, the ratio of maximum torque to full load torque is

  1. (A) 1.567
  2. (B) 1.712
  3. (C) 1.948
  4. (D) 2.134

Answer: (C) 1.948

Slip at max torque s_m=R2/X2=0.1/0.92=0.1087. Tmax/Tfl=(s_fl^2+s_m^2)/(2*s_fl*s_m)=(0.0009+0.01181)/(2*0.03*0.1087)=0.01271/0.006522=1.95. So the ratio is about 1.948.

Q14. A two bus power system shown in the figure supplies load of 1.0+j0.5 p.u. Bus 1 V1 ∠0° Bus 2 1∠δ2. The values of V1 in p.u. and δ2 respectively are

  1. 0.95 and 6.00°
  2. 1.05 and -5.44°
  3. 1.1 and -6.00°
  4. 1.1 and -27.12°

Answer: 1.05 and -5.44°

The correct option is right because it accurately reflects the voltage magnitude and angle at Bus 2 required to balance the power flow in the system while supplying the specified load, ensuring that the power equations are satisfied under the given conditions.

Q15. The switch SW shown in the circuit is kept at position ‘1’ for a long duration. At t = 0+, the switch is moved to position ‘2’. Assuming |Vo2| > |Vo1|, the voltage v_c(t) across the capacitor is

  1. v_c(t) = −Vo2(1 − e^(−t/2RC)) − Vo1
  2. v_c(t) = Vo2(1 − e^(−t/2RC)) + Vo1
  3. v_c(t) = −(Vo2 + Vo1)(1 − e^(−t/2RC)) − Vo1
  4. v_c(t) = (Vo2 − Vo1)(1 − e^(−t/2RC)) + Vo1

Answer: v_c(t) = (Vo2 − Vo1)(1 − e^(−t/2RC)) + Vo1

The correct option describes the voltage across the capacitor as it transitions from an initial voltage Vo1 to a new voltage Vo2 after the switch is moved. The term (Vo2 - Vo1) represents the change in voltage, and the exponential factor accounts for the charging behavior of the capacitor over time, starting from Vo1.

Q16. A parallel plate capacitor consisting two dielectric materials is shown in the figure. The middle dielectric slab is placed symmetrically with respect to the plates. If the potential difference between one of the plates and the nearest surface of dielectric interface is 2 Volts, then the ratio ε1: ε2 is

  1. 1:4
  2. 2:3
  3. 3:2
  4. 4:1

Answer: 4:1

The ratio of the dielectric constants ε1 and ε2 is determined by the relationship between the electric field strengths in the two dielectrics, which is inversely proportional to their dielectric constants. Given that the potential difference across the dielectric interfaces is equal, the ratio of the distances and the potential differences leads to the conclusion that ε1: ε2 = 4:1.

Q17. Consider an LTI system with transfer function H(s) = 1 / [s (s + 4)] If the input to the system is cos(3t) and the steady state output is A sin(3t + α), then the value of A is

  1. 1/30
  2. 1/15
  3. 3/4
  4. 4/3

Answer: 1/15

The steady-state output amplitude A for an LTI system can be determined using the magnitude of the transfer function evaluated at the frequency of the input. For the given transfer function H(s), substituting s = j3 gives a magnitude of |H(j3)| = 1/15, which corresponds to the output amplitude.

Q18. Consider an LTI system with impulse response h(t) = e⁻⁵t u(t). If the output of the system is y(t) = e⁻³t u(t) - e⁻⁵t u(t) then the input, x(t), is given by

  1. e⁻³t u(t)
  2. 2 e⁻³t u(t)
  3. e⁻⁵t u(t)
  4. 2 e⁻⁵t u(t)

Answer: 2 e⁻³t u(t)

H(s) = 1/(s+5). Y(s) = 1/(s+3) - 1/(s+5) = 2/[(s+3)(s+5)]. So X(s) = Y(s)/H(s) = 2(s+5)/[(s+3)(s+5)] = 2/(s+3), giving x(t) = 2 e^(-3t) u(t).

Q19. Shunt reactors are sometimes used in high voltage transmission systems to

  1. limit the short circuit current through the line.
  2. compensate for the series reactance of the line under heavily loaded condition.
  3. limit over-voltages at the load side under lightly loaded condition.
  4. compensate for the voltage drop in the line under heavily loaded condition.

Answer: limit over-voltages at the load side under lightly loaded condition.

Shunt reactors are employed in high voltage transmission systems to absorb reactive power, which helps to stabilize voltage levels and prevent over-voltages, particularly when the system is lightly loaded.

Q20. The state transition matrix for the system [ẋ1; ẋ2] = [[1,0],[1,1]] [x1; x2] + [1;1] u is

  1. [e^t 0; e^t e^t]
  2. [e^t 0; t² e^t e^t]
  3. [e^t 0; t e^t e^t]
  4. [e^t t e^t; 0 e^t]

Answer: [e^t 0; t e^t e^t]

The correct option represents the state transition matrix derived from the system's dynamics, where the first row reflects the exponential growth of the first state and the second row captures the interaction between the states over time, specifically incorporating the effect of the input through the term 't e^t'.

Q21. Which of the following is an invalid state in an 8-4-2-1 Binary Coded Decimal counter

  1. 1000
  2. 1001
  3. 0011
  4. 1100

Answer: 1100

The binary coded decimal (BCD) system represents decimal digits using four bits, and valid BCD values range from 0000 (0) to 1001 (9). The binary representation 1100 corresponds to the decimal value 12, which is not a valid BCD representation.

Q22. The sinusoidal ac source in the figure has an rms value of 20/√2 V. Considering all possible values of RL, the minimum value of Rs in Ω to avoid burnout of the Zener diode is ________.

  1. 10 Ω
  2. 20 Ω
  3. 30 Ω
  4. 40 Ω

Answer: 20 Ω

The correct option is 20 Ω because it ensures that the Zener diode operates within its safe limits by providing sufficient current regulation, preventing excessive power dissipation that could lead to burnout.

Q23. The magnitude of magnetic flux density (B) at a point having normal distance d meters from an infinitely extended wire carrying current of I A is μ₀ I/(2πd) (in SI units). An infinitely extended wire is laid along the x-axis and is carrying current of 4 A in the +ve x direction. Another infinitely extended wire is laid along the y-axis and is carrying 2 A current in the +ve y direction. μ₀ is permeability of free space. Assume î, ĵ, k̂ to be unit vectors along x, y and z axes respectively. Assuming right handed coordinate system, magnetic field intensity, H at coordinate (2,1,0) will be

  1. 3/(2π) k̂ weber/m²
  2. 4/(3π) î A/m
  3. 3/(2π) k̂ A/m
  4. 0 A/m

Answer: 3/(2π) k̂ A/m

The magnetic field intensity H at a point in space due to multiple currents can be determined using the superposition principle. For the given currents, the contributions from each wire at the specified coordinates combine to yield a resultant magnetic field in the z-direction, specifically resulting in the value of 3/(2π) k A/m.

Q24. A 10 kHz even-symmetric square wave is passed through a bandpass filter with centre frequency at 30 kHz and 3 dB passband of 6 kHz. The filter output is

  1. a highly attenuated square wave at 10 kHz.
  2. nearly zero.
  3. a nearly perfect cosine wave at 30 kHz.
  4. a nearly perfect sine wave at 30 kHz.

Answer: a nearly perfect sine wave at 30 kHz.

The bandpass filter is designed to allow frequencies around its center frequency of 30 kHz to pass through while attenuating frequencies outside its passband. Since the 10 kHz square wave is well below the filter's passband, it is significantly attenuated, and the output primarily consists of the fundamental frequency component at the center frequency, resulting in a nearly perfect sine wave at 30 kHz.

Q25. For a single phase, two winding transformer, the supply frequency and voltage are both increased by 10%. The percentage changes in the hysteresis loss and eddy current loss, respectively, are

  1. 10 and 21
  2. −10 and 21
  3. 21 and 10
  4. −21 and 10

Answer: 10 and 21

Hysteresis loss in a transformer is proportional to the frequency and the maximum flux density, so a 10% increase in frequency results in a 10% increase in hysteresis loss. Eddy current loss, on the other hand, is proportional to the square of the frequency, leading to a 21% increase when frequency is raised by 10%.

Q26. For the transfer function G(s) = 5(s + 4) / [s(s + 0.25)(s² + 4s + 25)] The values of the constant gain term and the highest corner frequency of the Bode plot respectively are

  1. 3.2, 5.0
  2. 16.0, 4.0
  3. 3.2, 4.0
  4. 16.0, 5.0

Answer: 3.2, 5.0

Bode gain constant = (5*4)/(0.25*25)=20/6.25=3.2. Corner frequencies are 0.25, 4 (from s+4), and wn=5 (from s^2+4s+25). The highest corner frequency is 5.0, so the pair is (3.2, 5.0).

Q27. The second order dynamic system dx/dt = PX + Qu y = RX has the matrices P, Q and R as follows: P = [ -1 1 0 -3 ], Q = [ 0 1 ], R = [ 0 1 ] The system has the following controllability and observability properties:

  1. Controllable and observable
  2. Not controllable but observable
  3. Controllable but not observable
  4. Not controllable and not observable

Answer: Controllable but not observable

Controllability matrix [Q PQ] = [[0,1],[1,-3]] has determinant -1, so the system is controllable. Observability matrix [R; RP] = [[0,1],[0,-3]] has determinant 0, so it is not observable. Hence controllable but not observable.

Q28. In an 8085 microprocessor, the following program is executed Address location – Instruction 2000H XRA A 2001H MVI B,04H 2003H MVI A, 03H 2005H RAR 2006H DCR B 2007H JNZ 2005 200AH HLT At the end of program, register A contains

  1. 60H
  2. 30H
  3. 06H
  4. 03H

Answer: 60H

Starting with A=03H (00000011) and CY=0 after XRA A, four RAR (rotate-right-through-carry) operations give 03H->01H->80H->C0H->60H. Final A = 60H, not 30H.

⚔️ Practice GATE Technical free + battle 1v1 →