Gate Mock Test: Electrical Engineering(EE)- 2
65 Questions MCQ Test GATE Electrical Engineering (EE) 2023 Mock Test Series | Gate Mock Test: Electrical Engineering(EE)- 2
A number of Indian goods face a _____ competition from Chinese goods in terms of prices and looks.
Hence, option C is the correct answer.
In the following question, out of the five alternatives, select the word similar in meaning to the given word.
Appease
Agitate = make (someone) troubled or nervous.
Pacify = to cause someone who is angry or upset to be calm and satisfied.
Interrupt = stop the continuous progress of (an activity or process).
So, option B is the correct answer.
Select the most appropriate option to substitute the underlined segment in the given sentence. If there is no need to substitute it, select No improvement.
It is not wise relying to anybody too much.
Therefore, we can say option C is the correct answer.
Select the most appropriate option to substitute the underlined segment in the given sentence. If no substitution is required, select No improvement.
His miserable condition made us wept.
because they cause something else to happen. The grammatical structure for make is given below:
MAKE + PERSON + VERB (base form).
The structure given in option A follows the above rule.
So, it is the correct answer.
Select the most appropriate word to fill in the blank.
Many items made of ivory were _____ from a dealer in antiques by the custom authorities at the Delhi airport.
‘Confiscate’ means ‘take or seize (someone's property) with authority’. This word fits best in the sentence.
Therefore, option A is the correct answer.
Select the most appropriate option to substitute the underlined segment in the given sentence. If there is no need to substitute it, select No improvement
The man who seen the accident to occur telephoned the police.
Therefore, option B is the correct response.
Select the most appropriate option to substitute the underlined segment in the given sentence. If no substitution is required, select No improvement.
If you listen to English news, it improve your English.
Hence, option B is the correct answer.
Select the most appropriate word to fill in the blank.
Scientists at Cambridge University are ______ how plants can give us sustainable energy.
The sentence implies that the scientists at Cambridge University are carrying out the research to find how plants can give us sustainable energy.
‘Investigating’ meaning ‘carry out research or study into (a subject or problem, typically one in a scientific or academic field)’ perfectly fits in the blank.
Hence, option B is the correct answer.
In the following question, out of the five alternatives, select the word similar in meaning to the given word.
Acrimonious
Acrimonious (adj.): (typically of speech or discussion) angry and bitter.
Tawdry (adj.): showy but cheap and of poor quality.
Gaudy (adj.): extravagantly bright or showy, typically so as to be tasteless.
Vitriolic (adj.): violent hate and anger expressed through severe criticism.
Ameliorate (v.):
make (something bad or unsatisfactory) better.
Courteous (adj.): polite, respectful, or considerate in manner.
So, the correct answer is option B.
Direction: In the following question, a part of the sentence is bold. Below the sentence alternatives to the bold part are given at (A), (B) and (C) which may improve the sentence. Choose the correct alternative. In case ‘No improvement’ is needed, your answer is (D).
The professor has agreed to take remediable classes for the weaker students.
'Remedial' should be used to make the sentence grammatically correct. It means giving or intended as a remedy or cure while remediable means capable of being cured; treatable. The classes don't need to be cured.
Thus, option B is the correct answer.
The open loop transfer function of a unity negative feedback system is given by 
Find the value of 'K' such that the phase margin is 60▫
The starting current of a star-connected, 3-phase induction motor at rated voltage is 6 times the full load current and full load slip of 5%. Auto-transformer is used to limit the starting current from mains to 4 times the full load current. Determine the ratio of starting torque to the full load torque.
As we known
A DC voltage of 50 V is applied to a coil having R = 10 Ω, L = 10 H. The time taken by the current to reach 75% of its final value is –
The charging current will be
i = 1 (1 - e - t/T)
i = 5 [ l - e-7]
Time required to reach 75% of its final value,
0.75 x 5 = 5 [1 - e-]
e-t = 0.25
t = 1.386 sec
For the operational amplifier circuit shown in the figure below, what is the maximum possible value of R1, if the voltage gain required is between – 10 and – 25? (The upper limit on RF is 1 MΩ)
So lesser the gain higher will be R1
So fo r AV = - 10
→ R1 will be maximum
In the circuit of figure, Vs = 2 cos3t volts the VL(t) will be-
Where,
V0 = [2 cos 3f] volts
Voltage across inductor
The power absorbed by 0.5 Ω resistance in the given circuit is:
Since, bridge is in the balanced condition.
So, power absorbed by 0.5Ω resistance is
P = 12R = ((3.43)2 × 0.5) = 5.88W
In a 132 kV system, the series inductance up to the point of circuit breaker location is 50 mH. The shunt capacitance at the circuit breaker terminal is 0.05 μF. The critical value of resistance in ohms required to be connected across the circuit breaker contacts which will give no transient oscillation is _____________.
Critical resistance to have transient free oscillations is
R = 500 Ω
A signal x(t) is denoted by a function which is periodic and continuous in its time domain representation. Then, which of the following would be true for the fourier transform of the signal
(i) Discrete and Periodic
(ii) Discrete and Aperiodic
(iii) Real{x(t)} transforms in to Real{X(jω)}
(iv) Real{x(t)} transforms in to Imag{X(jω)}
Also, Even part of a signal gets transformed into real part of Fourier representation and Odd part of signal gets transformed into Imaginary part of Fourier representation.
Find the differential equation of the system described by the transfer function given as:
⇒ Y (s) [2s2 + 5s] = X(s) • (s + 3)
⇒ 2s2 • y(s) + 5 • y(s) = s • X(s) + 3. X(s)
⇒ 
Which of the following is correct for 1110100 ÷ 1010?
1010) 1110100(1011
(1) 1) 1) 1) 1) 1) 1010
10000
1010
110
Therefore, Quotient= 1011 and remainder= 110
What is the ratio of starting torque(Tst) to full load torque(Tfl) when an induction motor is started using auto transformer and it has a full load slip(Sfl) of 0.02. Its starting current(Ist) is 3 times its full load current(Ifl)?
Fault current of a line to ground fault and LLL fault for an unloaded generator is same. If X1 =0.3 pu, X2 = 0.2 pu and Xo = 0.04 pu. Then calculate value of inductive reactance (Xn) required for natural grounding (in pu)
for LG fault:-
IfLG = 3 ER1 / (X1 eq + X 2eq + X oeq)
For LLL fault:-
ILL = E R1 / X 1eq
ILL = 1/0.3
Both faults currents are same
IFLG = LFLL
0 .54 + 3 X n = 0 .9
3 Xn = 0.9 - 0.54
Grounding reactance (Xn) = 0.12 pu
Determine the value of V and I in the given network.
⇒ 2V + 4(V - 6) + V = 0
⇒ 2V + 4V - 24 + V = 0
The instantaneous polarity of primary winding of an ideal transformer is shown in figure below. The direction of flux and polarity of instantaneous voltage induced in secondary is
Given below is a 3-phase full-bridge rectifier. When the diode D3 is conducting, which of the following diodes cannot conduct at the same time?
D3 is conducting, D1 and D5 cannot conduct. Similarly, only one diode can conduct in the bottom half at the same time. The diode that is conducting has its cathode connected to the lowest phase voltage at that moment. Hence, when D3 is conducting, D6 cannot conduct.
A simple slide wire potentiometer is used for measurement of current in a circuit. The voltage drop across a standard resistor of 0.5 Ω is balanced at 95 cm. find the magnitude of the current (in Ampere) if the standard cell emf of 1.95 volts is balanced at 60 cm.
For the same working current, if 60 cm corresponds to 1.95 volt.
Then 95cm of the slide wire corresponds to =1.95 / 60x10-2 x 95x10-2 = 3.421 volt
So, cross the resistance 0.5 Ω the voltage drop is 3.421 volt.
Then the value of the current is (I) = 3.421 / 0.5 = 6.84 A
Determine V0 ( in Volts) in the given network.
Hence,
V0 = 4 × I2 = 4 × 6.25× = 25V
In the figure shown below, ‘N’ is two port network. If I = 2 A, then the values of V1 and V2 will be
Assume the Y - parameter matrix for ' N′ as 
I1 = 2V1 + V2........(i)
I2 = 2V1 + 2V2 ........ (ii)
Applying KCL at input node,
Applying KCL at output node,
Putting the value of I1 and l2 in equation (iii) and (iv),
Find the value of y{z), n > 0 if y(n) + 
(n - 1 ) - 
(n ~ 2) = 0 and y(-1) = y { - 2) = 1?
y{n - 1) - 
y(n - 2) = 0Y (-1)= y (-2 ) = 1
Apply Z-transform to the above equation, we get
Let the state transition matrix of a system be given as 
Then which of the following is equal to 
Hence, replace ′t′ in state transition matrix by '-t'.
The full-load torque angle of a synchronous motor at rated voltage and frequency is 30° elect. The stator resistance is negligible. ______ degree will be the torque angle if the load torque and terminal voltage remaining constant, the excitation and frequency are raised by 10%
Assuming Ra to be negligible, 
sin δ = 1.1 sin 30∘ = 0.55
8 = 33.36∘
If 2,3 and 5 are the three Eigen values of a matrix A3 x 3 Then, I Adj A I=
|Adj A| = |A|n−1 = |A|2 = (30)2 = 900
Below figure shows the frequency Response of a lead compensator in which the lead compensator coefficient is _______
The answer is in between 0.49 and 0.51f
on comparing, 20 log α = −6
log α = −0.3
α = 0.501
In a system of 132 kV, the line to ground capacitance is 0.01 μF and the inductance is 5 henries. The voltage appearing across the pole of a C .B, if a magnetising current of 5A (instantaneous value) is interrupted and also the value of resistance to be connected across contacts of C .B . to eliminate the restriking voltage is
V = 111.8 KV
In order to eliminate the transient critically the value of resistance across the breaker contacts required
i.e
= 0. 5 × 109 x√5
R = 11.18 KO
A 10 μF condenser is connected in series with a coil having inductance of 25 mH. If a 100 V source operating at resonance frequency causes a circuit current of 10 mA . What is the Q-factor of the coil?
Given
The transfer characteristics of a Schmitt trigger is shown below:
Hysteresis voltage;
VH = [VUTP – VLTP]
VH = [2 – (–3)] = 5 V
A clockwise circulating noise eliminating loop corresponds to an inverted Schmitt trigger.
An A .C Bridge as shown in the figure is used to measure unknown inductance L1, in companion with capacitance. What would be the quality factor if the frequency is 1500 Hz & the various parameter are as below –
R2 = 360 Ω, R3 = 720 Ω, R4 = 1200 Ω & C4 = 1.2 Μf
The answer is in between 13 and 14
At Balance condition:
On separating real & imaginary part we get,
&L1 = R2R3C4 = 360 × 720 × (1.2 × 10−6) = 0.311H
Two identical unloaded generators are connected in parallel as shown below. Both the generators are having positive, negative and zero sequence impedance of j0.8 p.u., j 0.2 p.u. and j 0.3 p.u. respectively. If pre-fault voltage is 1 p.u. for a line to ground fault at the terminals of the generator, the fault current magnitude (in p.u.) is ___________. (Assume neutral impedance, Zn = j0.02 p.u.)
The positive sequence reactance,
Similarly, the equivalent negative sequence reactance is
The zero-sequence reactance diagram is
For a single line to ground fault, fault current is
In the given figure, a V volt Battery is connected across the capacitor, galvanometer shows the deflection of found. The switch is closed at t = 0. If distance between the plate of capacitor is increased after which the switch is removed from the circuit. What will be effect on Vc capacitor voltage & electric field
.
When the switch is closed for a long line then capacitor will charges up to Vc = V volt.
Now as the switch is removed as shown in figure.
The charge Q across the plate of capacitor are trapped.
After removing the switch distance between the plate of capacitor to increased.
therefore C w ill decrease.
as the charge are trapped i.e; constant.
So 
where VC is decreased therefore Vc will increases.
So, E remain constant.
One transformer is rated at 400 KVA & has per unit impedance of (0.02 + j0.08) p.u. and other is rated at 800 KVA & has per unit impedance of (0.20 + j0.08) p.u. Both transformers are connected in parallel to the load. Find the maximum load (in KVA) that can be imposed to transformer’s such that no transformer is overloaded.
The answer is in between 1197 and 1200
Let choose 800 KVA as base KVA.
NOTE: Transformer having the least p.u. impedance saturates first.
So, the max. load is shared when transformer T2 Share its rated load.
Consider both the op-amps are ideal, find out the value of (V01 – V02) in volts___________V.
As op-amps are ideal. Its terminal will not draw no current. Thus,
At a particular instant, the R-phase voltage of a balanced 3-Φ system is +60 V, and Y-phase voltage is –120 V. The voltage of B-phase at that instant is ____ (in V).
The answer is in between 58 and 62
For balanced 3 — ф system, V R (t) = V m sinwt
VY ( t ) = V m sin (wt — 120°)
VB(t) = Vmsin (wt + 120°)
Given: V m sinwt = 60 . . . (i) V m sin (wt — 120°) = —120
Vm [sinwtcos 120° — cos wt sin 120°] = —120
cut = 30° Equation
(i) 2+ Equation
(ii) 2 Vm2 sin 2wt + V m 2 cos 2wt = (60)2 + (60√3)2
Vm2[sin 2wt + cos 2wt] = (60)2 + (60-√/3)2
Vm = 120V
VB = V m sin (wt + 120°)
VB = 120 sin (30° + 120°)
VB = 120 sin 150° = 120 x ½ = 60 volts
The value o f function x(t) at t = 
For a 6-bit ADC, having full scale deflection of 10volts, the quantization error in millivolts would be________
The answer is in between 79 and 80
Quantization error
= 0.07936volt
= 79.36 mV
The area bounded by the curve y = x2 and y = 1 – x2 is
The answer is in between 0.9 and 1
Given that R: y = x2 &t/ = 1 — x2
The point of intersection of the curves y = x 2 & y = 1 — x 2 are given by
X2 = 1 - X2
are the point of intersection of a curves.
The required are is given by
If the Size of data bus is 16 bits and the size of address bus is 20 bits. Then the processor will be of:
The size of processor depends only on size of data bus. Here the size of data bus is 16bits. Hence the size of processor will also be 16bits.
For nMOS and its Transfer curve shown in fig. below, Its region of operation is:
From Transfer curve.
VT = 1 V
From the circuit,
VGS = VG – VS = 3 – 1 = 2 V
VDS = VD – VS = 5 – 1 = 4 V
VGS = VT = 1 V
Since, VDS > VGS – VT
The MOSFET operates in saturation region.
Consider the system described by following state space equations
For the state x3, which one of the options is correct.
Since A is a diagonal matrix, Also all the eigen values are distinct.
x3 is controllable because the 3rd element in the row of B is non zero.
x3 is unobservable because the 3rd element in the column of C is zero.
Thus, option D is the correct answer.
Evaluate 
where C is circle \z\ = 3
Poles z = 1&2 both are inside the |z| = 3
f(z) = cos πz2 is an analytic function
= 2πif(1) + 2πif(2)
= 2πi[− cos π + cos 4π]
= 4πi
Consider the following synchronous counter made up of JK, D and T Flip-Flops.
here
Consider characteristic equation of D− Flip-Flop.
(ii) Consider characteristic equation of T - Flip-Flop Consider charactrisic
Using equations (i), (ii) and (iii)
The number of used states = 5
∴ Modulus value = 5
Let the transfer function of a stable discrete time system be H(z). The pole zero diagram of H (z) is shown in the figure below.
Assume that system is causal and stable, then which of the following statements can be true.
Since the system is stable and causal thus all the poles must lie inside the unit circle and for the system to be all-pass system the condition is
⇒ the distance of poles from origin
Considering a clamper circuit, where capacitance C, load R, the cut-in voltage of diode are unknown.
Which is the correct statement?
The shape and peak to peak value of signal remain unchanged in a clamper.
A clamper only affects the DC voltage level of the wave, which can be both moved up and down, not simply up.
Find the differential gain 
IE1 + IE2 = 2mA
In the differential amplifier, the transistors are biased at same operating points.
Ic1 + Ic2 = 1mA
rc = 25Ω
The formula for the differential gain Ad = -gmRc = 
In a 3 phase, 3 wire system the conductors are arranged in a horizontal plane with spacing D13 = 4m, D12=2m=D23. Conductors are transposed having diameter of 2.5cm. The ratio of mutual GMD to self GMD =?
Mutual GMD = 
Self GMD = 0.7788 × 
The ratio of mutual GMD to self GMD = 2.52/9.73x10−3 = 258.84
A 1200V, 6pole, 50Hz, 3-phase star connected induction motor has stator to rotor turns ratio of 3. At standstill, its rotor impedance is (0.4+j8)Ω/phase. What is the gross output of motor at the speed of 960 rpm in kW? [Write the answer upto three decimal Points]
Standstill ratar emf, E 2 = 
Synchronous speed of mator Ns = 
Slip 
Er = sE 2 = 0.04 x 230.94 = 9.4238V
Rator impedance at 0.04 slip Zr = 
18.04ATotalCu loss
= 31 2R = 3 x 18.042 x 0.4 = 3 9 0 .6 W
arassautputpowerafmatar 1 — s
Grassoutput = 
gross output = 
A two generator system supplying a load of 40MW connected at bus 2. The fuel costs of generators are:
C1(PG1) = 8500 Rs/MWh
C2(PG2) = 11000 Rs/MWh
The loss in the line is Ploss(p.u.)=0.5PG1(p.u.)2 where the loss coefficient is specified in p.u. on a 100 MVA base. The most economic power generation schedule (in MW) is:
From coordination equation,
(dFn/dPGn){1/ (1-∂PL/∂PGn)} =λ
Given,
PL=0.5PG12
∂PL/∂PG1=PG1 and ∂PL/∂PG2=0
(dF1/dPG1)(1/ (1-PG1) = (dF2/dPG2) (1/ (1-0)
Or, 8500(1/ (1-PG1) = 11000
Or, PG1=0.227 p.u.
As the base value is 100MVA.
PG1=100x0.227 = 22.7MVA
PL= 0.5x(0.227)2 = 0.0258 p.u. = 0.0258x100 MVA = 2.58MVA
PD=PG1+ PG2- PL
Or, 40=22.7+ PG2-2.58
Or, PG2=19.9 MVA
A unity feedback control system is characterized by the open loop transfer function
The Nyquist path and the corresponding Nyquist plot of G(s) are shown in figures below. If 0< /><1 the="" number="" of="" poles="" of="" the="" closed="" loop="" transfer="" function="" that="" lie="" in="" the="" right="" half="" of="" the="">
Where, N = number of encirclements of the critical point.
P = number of open loop poles lying in the RHS of s-plane.
Z = number of closed loop poles lying in the RHS of s-plane.
Here 0
Now finding value of P
Use Routh array to find the number of roots of open loop system lying in the RHS of s-plane.
S3+as2+b
Since a>0, b>0
Therefore, number of sign changes = 2
Hence, P = 2
So, N = P-Z
0 = 2-Z
Z = 2
Hence, two closed loop poles are lying in the RHS of s-plane.
Find the Thevenin equivalent voltage across x-y terminals.
Energy of signal whose fourier transform is shown in figure equals to
According to Parseval's Energy theorem
A boost converter is shown in the figure.
If Vs = 20V, V0 = 60V, R = 60Ω, L = 30 μH and D (duty cycle) = 0.Then the converter will operate at a switching frequency (in kHz) approx. _________.
V0 >V0
So, the converter is operating in discontinuous mode.
Inductor current waveform will be
When CH is ON
When CH is OFF
From (i) and (ii)
From power balance
For boost converter
Is = IL = 3A
Average inductor current from the waveform
Ip = 8A
From (i)
For the power system shown below
the zero sequence reactance in pu are indicated. What is the zero sequence driving point reactance of the bus 3 in pu?
So, zero sequence driving point reactance at bus 3;
Xo = (0.22 × 0.3)/(0.22 + 0.3)
Xo = 0.1269 ≈ 0.13pu
If z(x,y) = x2 – y2 , x(t) = t – t-2 and y(t) = t2 + t-3
Then dx/dt at t=1 is equla to …………..
Using chain rule
The given rectifier circuit has a source voltage of 150 sin ωt, and a load resistor of 40 Ω. Determine the average load current, if the thyristors are fired at ωt = 30°
Given, Vm = 150 V,Rl = 40Ω,α = 300
The output voltage is computed as,
Substituting values,
The average load current is given as,
A power system is comprised of three elements 0 – 1, 1 – 2, and 2 – 0 of equal pu impedances of 0.25. Determine the bus impedance matrix of the network
where y10 = y12 = y20 = 
The bus admittance matrix can be written as
The bus impedance matrix pan be computed as 
(y bus) det (Ybus) = (8 x 8) - ( - 4 x - 4) = 48
adjoint (ybus) = 
Hence Z bus = 
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