Electrical Engineering (EE) : Mock Test 9 For GATE
65 Questions MCQ Test Mock Test Series for Electrical Engineering (EE) GATE | Electrical Engineering (EE) : Mock Test 9 For GATE
“Nobody praised her ________ nature, so she decided to mend her ways to become ________.”
The words that best fill the blanks in the above sentence are
‘complaisant’- willing to please or oblige others
‘complacent’- self-satisfied, unconcerned
Thus, the sequence in option 3 must be used to make the sentence contextually sensible.
“All the development programmes of the government have been ________ over the years by the perpetual rise in the population of the country.”
‘vitiate’- to make faulty or defective, hurt
‘ratify’- to approve and sanction formally
The sentence implies that the ever-increasing population of the country is posing a hindrance to all the developmental programmes initiated by the government.
Hence, option 1 is the correct answer.
D = ABC + BCA + CAB where A, B and C are decimal digits, then D is divisible by
D = ABC + BCA + CAB
= (100 A + 10 B + C) + (100 B + 10 C + A) + (100 C + 10 A + B)
= 111 (A + B + C)
So, D is a multiple of 111.
111 is divisible by 37 but not divisible by 29. Thus, D is divisible by 37 but not divisible by 29.
In an examination 100 questions of 1 mark each are given. After the examination, 20 questions are deleted from evaluation, leaving 80 questions with a total of 100 marks. Student A had answered 4 of the deleted questions correctly and got 40 marks, whereas student B had answered 10 of the deleted questions correctly and got 35 marks. In this situation,
In an examination 100 questions of 1 mark each are given.
After the examination, 20 questions are deleted from evaluation.
Now, total number of questions considered for evaluation = 80
Marks for each correct question = 100/80 = 1.25
Student A:
Total marks secured = 40
Number of questions answered correctly = 40/1.25 = 32
Student A had answered 4 of the deleted questions correctly
Number of questions answered correctly before the deletion of 20 questions = 32 + 4 = 36
Marks as per 100 question paper = 36
Student B:
Total marks secured = 35
Number of questions answered correctly = 35/1.25 = 28
Student A had answered 10 of the deleted questions correctly
Number of questions answered correctly before the deletion of 20 questions = 28 + 10 = 38
Marks as per 100 question paper = 38
Conclusion: B lost more than A.
Two solutions of X and Y containing ingredients A, B and C in proportions a : b : c and c : b: a, respectively are mixed. For the resultant mixture to have A, B and C in equal proportion. It is necessary that
The ratio of ingredients A, B and C = a : b : c
Let x be the factor.
The quantity of A in the solution X = ax
The quantity of B in the solution X = bx
The quantity of C in the solution X = cx
Solution - Y:
The ratio of ingredients A, B and C = c : b: a
Let x be the factor.
The quantity of A in the solution Y = cx
The quantity of B in the solution Y = bx
The quantity of C in the solution Y = ax
Mixture Solution:
The quantity of A in the mixture solution = (a + c)x
The quantity of B in the mixture solution = 2bx
The quantity of C in the mixture solution = (a + c)x
The ratio of ingredients A, B and C = (a + c)x : 2bx : (a + c)x = (a + c) : 2b : (a + c)
It is given that, the resultant mixture to have A, B and C in equal proportion.
⇒ 2b = a + c
In the given diagram of Testbook team, contents are represented in the parallelogram, technical in the rectangle and DTP in the triangle. Out of the total number of the people, the percentage of content shall be in the range of ________.
Total number of people in content = total number of people in rectangle
Total number of people in rectangle = 18 + 50 + 25 = 93
Total number of people in Testbook team = 152
Hence the answer the is 61 to 75
How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 where n is an odd integer less than 60?
Since m and n are both positive and their sum is equal to
Hence the value of both there fractions must be less than
Therefore n-can take values > 48 and odd
N = {49, 51, 53, 55, 57, 59}
Now we have to check which values of ‘n’ will give integral values of ‘m’.
Equation:
For m to be on integer (n - 48) should be a multiple of 12n
The above condition is satisfied for
N = 49, 51, 57
Corresponding values of m
N = 49
M = 12 × 49
N = 51
Most successful professional football players began playing football as youth, though a significant number learned the sport later in life. No professional football player who ignored the advice of his coaches, however, has ever become successful. If the statements are true, which of the following must also be true?
According to the second sentence of the passage, a necessary condition for being a successful professional football player is not ignoring the advice of coaches.
So all of the successful professional football players mentioned in the passage’s first sentence, including those who played the sport as youths, do not ignore the advice of coaches. That’s what correct answer states.
The diagram shows a circle with one equilateral triangle inside and one equilateral triangle outside. The ratio of areas of big triangle to the small triangle is n :1. Find the value of n.
Rotating the smaller triangle by 180°, we get the following figure.
From symmetry we can say R is the midpoint of BC.
So P and Q are also mid points of AB and AC
From the midpoint theorem:
Since two equilateral tringles are also similar triangle we know
The following hydrograph depicts the rainfall intensity of an area for the time period of 8 hours. Also, infiltration curve is plotted on the same graph for the given land characteristics. Hydrograph region below the infiltration curve represents infiltrated portion of rainfall. Calculate the excess of rainfall occurred.
The region above the infiltration curve represents the excess of rainfall.
∴ Excess rainfall = Area of the hydrograph above infiltration curve.
Area of hydrograph above infiltration curve (A) is given by
∴ Excess rainfall is 11 mm.
Which of the following expression represents the given logic circuit?
= A̅ B̅ C + ABC
⇒ Y = (A̅ B̅ + AB) C
⇒ Y = C(A⊙B) = C(A̅⊙B̅)
In figure, 
the balancing component of current is _____ A
We see that both current shown in the figure enter their respective winding at dots. so, for ampere turn balance
A Transformer is rated at 100 KVA. At full load its copper loss is 1200 W and its iron loss is 850 W. Calculate the load (in KVA) at which maximum efficiency will occur.
Two transformers operating in parallel having impedance of 6 % and 4 %. If the total load supplied by the combinations is 100 A, then the current supplied by 6 % impedance transformer _______ A.
This question is based on load sharing
By using current division rule
However, if the impedance is in PU then their PU value should be converted on a common base to ensure that the ohmic ratio remain unchanged.
Let x(t) = 1 – 2t. Then what is the value of 
where δ(t + 1) is unit impulse at t = -1.
From the property of unit pulse function,
what is the inverse Laplace transform of the product F1(s) . F2(s)?
F(s) = F1(s) . F2(s)
By using partial fractions,
By applying inverse Laplace transform,
Latching current for an SCR, inserted in between a dc voltage source of 300 V and the load, is 150 mA. Compute the minimum width of gate pulse current required to turn-on this SCR in case the load of L = 0.3 H in series with R = 30 Ω
1 – e-100t = 0.015
-e-100t = 0.985
-100t = -0.01511
t = 151.136 μsec
A step-up chopper shown in figure is to deliver 3A
In to the 10 Ω load. The battery voltage is 12 V, L = 20 μH, C = 100 μF and chopper frequency is 50 kHz. Determine the on-time of the chopper ________ μs
Calculation:
Vo = IoRL = 3 × 10 = 30 V
Vb = Vo (1 – D)
On time = TON = DT = 0.6 × 20 × 10-6 = 12 μs
When the transistor is switched on, current ramps up in the inductor and energy is stored.
When the transistor is switched off, the inductor voltage reverses and acts together with the battery voltage to forward bias the diode, transferring energy to the capacitor.
When the transistor is switched on again, the load current is maintained by the capacitor, energy is stored in the inductor and the cycle can start again.
The value of the load voltage is increased by increasing the duty cycle or the on-time of the transistor.
During the transistor on-period, assuming an ideal inductor and transistor
Vb = VL = Ldi/dt
Vb/s = (sL) i(s)
i(s) = Vb/s2L
i(t) = (Vb/L)t = kt
Over an on-time of TON, the change of battery current will be
Δi = k TON
During the off-period
Vo = Vb + VL = Vb + L(di/dt) = Vb + L(Δi/Δt)
= Vb + L(Vb/L)TON /Toff = Vb(1 + TON /Toff)
Let the duty cycle
D = TON T, and Toff = T – TON = T(1 - D). Now
Vo = Vb (1 – DT/T(1 – D))
which simplifies to
Vo = Vb/(1 - D)
In an ideal circuit, VbIb = V0I0. Therefore, from eq.
Ib = (Vo/Vb)Ib = Io/(1 – D)
Ib = Ib + Δi/2
Ibo = Ib – Δi/2
When the steady variation of battery current has been reached, the variation will be between lo and I1, as shown in.
For the circuit, shown in the figure below, the values of i0 and i1 respectively are
By applying voltage division in the input loop,
i0 will be zero, as there is no closed path.
By applying KCL in the right-side circuit,
As 10 Ω and 5 Ω are connected in parallel, voltage across them are equal.
⇒ 10 i1 = 5 i2 ⇒ i2 = 2i1 ----(2)
From equation (1) and (2)
i1 + 2i1 = -3 ⇒ i1 = -1 A
For the two – port circuit shown in the figure, the S – domain expression for the parameter d is where
The s-domain circuit for the given circuit is
When V2 = 0
By applying KVL in the second loop
sI2(s) + 4 (I1(s) + I2(s)) = 0
⇒ 4 I1 (s) + (s + 4) I2(s) = 0
Match List-I (Transfer function of the system) with List-II (Type and order of the system) and select the correct answer using the codes given below the Lists:
Type of the system indicates the number of poles at origin.
Order of the system indicates the total number of poles.
The feedback system shown below is stable for all values of K given by
Characteristic equation,
1 + G(s)H(s) = 0
⇒ s(s + 1)(s + 6) = 0
⇒ s3 + 7s2 + 6s + K = 0
Routh array:
For the system to be stable, there should be no sign change in the first column of Routh array.
A 60 μA meter has a resistance of 100 ohms. If the meter has to measure 100 mA, then the value of shunt resistance and the current through shunt respectively are
Given that, current through meter (Im) = 60 μA
Total required current (I) = 100 mA
Meter resistance (Rm) = 100 Ω
Current flows though shunt resistance
Ish = I - Im
= 100 × 10‑3 – 60 × 10-6
= (100 – 0.06) × 10-3
= 99.94 mA
In the measurement of power by 3 voltmeter method connected as shown below. The expression for power consumed in the load is given by
Power consumed in the load,
PL = V3 IL cos θ
V1 = ILR
The output of the 4 to 1 MUX shown in the below figure is
Find VDS under DC condition of the given circuit
Given rd = 15 KΩ, IDSS = 20 mA, Vp = -10V
Given:
VGS = -6V
VDS = VDD – IDRD = 40 – 3.2 × 3 = 30.4 V
What is the output voltage Vo (in Volt)of the given circuit?
Voltage at A point VA = 4 × 4 = 16 V
By using virtual short circuit condition VA = VB = 16 V
Voltage droop across feedback resistance V4kΩ = 4 × 4 = 16 V
Output voltage Vo = 16 + 16 = 32 V
A 30 kV, 50 Hz, 60 MVZ generator has the positive negative and zero sequence reactance of 0.2 pu, 0.15 pu and 0.05 pu respectively. The generator ground with a reactance of 0.05 pu. The fault current magnitude for a single line to ground fault is ________(in pu)
Given that, Z1 = 0.2 pu
Z2 = 0.15 pu
Z0 = 0.05 pu
Zn = 0.05 pu
For a G fault, fault current is given by
The distribution system shown in the figure is to be protected by over current system of protection. For proper fault discrimination, directional over current relays will be required at which locations?
Directional relays are not required at supply side and at radial feeders as the power flow unidirectional but not bidirectional.
so, the directional over-current relays will be required at 2 and 3 only.
In a four-bus system the total self-admittance of bus 2 is Y22 = -j14 p.u. A transmission line having a series reactance of j 0.2 p.u. and shunt capacitance of j 0.5 p.u. is connected between buses 2 and 3. The magnitude value of y22 modified is________
y22(old) = -j 14 p.u
Z23 = j 0.2 p.u
y’23 = -j5 p.u
y22 = -j 14 – j5 + j 0.25
= -j 18.75 p.u.
A fair die is rolled two times independently. Given that the outcome on the first roll is 1, the expected value of the sum of the two outcomes is
Let X is a random variable represents an event of rolling a die and P(X = xi) represents the probability of getting sum as xi.
While rolling a die, the probability of getting different numbers as follows.
A curve is given by x2 + 2y2 = 16 is revolved around x axis. The volume of solid generated is
Consider a function f(x, y, z) = x2yz + 3xy2. The greatest rate of increase of function f at point (2, 1, -1) is
Greatest rate of increase of function f is directional derivative at that point.
∇f = (2xyz + 3y2) i + (x2yz + 6xy) j + x2y k
At the given point (2, 1, -1), ∇f = -i + 8 j + 4 k
Greatest rate of increase
If a 3 × 3 square matrix A has the eigen values 3, 4 and 5, then the determinant of adjoint of matrix A is
Given that, Eigen values of a matrix A = 3, 4 and 5
From the properties of Eigen values, determinant of a matrix is equal to the product of their Eigen values.
⇒ |A| = 3 × 4 × 5 = 60
|Adj A| = |A|n-1 = |A|2 = 602 = 3600
In a lot of 600 solenoids 30 are defective. Find the probability that at least one is defective in a random sample of 24 solenoids.
Probability that solenoid is defective 
The probability that solenoid is not defective = 1 – 0.05 = 0.95
Since 24 samples are there, the probability of zero defectives will be
The probability that at least one is defective = 1 - the probability of zero defectives
= 1 – 0.292 = 0.708
The answer is in between 0.65 and 0.75
What is the value of the departure angle of root locus from the complex pole at s = -2 + j for
Angle of departure is given by,
θD = 180° + ∠G(s)H(s) (or) 180° - ϕ
ϕ = (∠ poles - ∠ zeros) of G(s)H(s)
θD = 180° - 90 – (π – tan-1 (2))
= -90 + tan-1(2) = -26.56
The steady state error due to a unit-step disturbance input D(s) in figure is
We need to find ess due to D(s).
So, R(s) = 0, the reduced block diagram will be
The state space representation of a system is given by
Then the impulse response of the system at t = 1 sec is ___
Transfer function for a given state space representation is, C[sI - A]-1 B + D
Impulse response = L-1[T/F] = 
At t = 1sec, impulse response = 1
A single-phase half wave converter is operated from 220 V, 50 Hz source and the load resistance R = 12 Ω, For the firing angle delay 60° determine The RMS Voltage.
A DC series motor chopper drive has the following parameters:
Battery voltage = 100 V
Ra + Rf = 0.1 Ω
Motor velocity constant (kV) = 10 mV/A -rad/sec
Chopper frequency = 130 Hz
Calculate the armature speed with an average armature current of 100 A and a chopper on-time of 6 ms.
Vavg = Vb Ton f = 100 × 6 × 10-3 × 130 = 78 V
E = Va – Iavg × R = 78 – (100 × 0.1) = 68 V
E = kV Iavg ω
A three – phase half-controlled thyristor converter has a highly inductive load of 10 Ω, and a supply of 240 V at 50 Hz. Determine average power (in kW) for firing angle delay of 30°.
A 230 V. dc single phase bridge inverter is connected to load R = 12 Ω and L = 0.04 H Determine the RMS value of third harmonic for quasi-square wave output with an on -period of 0.5 of a cycle
For quasi – square wave
2d = 0.5 × 180° = 90°
d = 45°
RMS value of fundamental voltage
Load impedance at third harmonic frequency is
RMS value of third harmonic
Consider the AC bridge shown below. If ωRC = 1 and 
is approximately equal to
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°
The circuit shown in the figure below uses ideal positive edge-triggered synchronous J-K flipflops with outputs Q0Q1Q2. If the initial state of initial state of output is Q0 = 0, Q1 = 1, Q2 = 0 just before the arrival of the first clock pulse, the state of the output just before the arrival of third clock pulse is
J0 = K0 = 1
J1 = K1 = Q0
J2 = K2 = Q0Q1
Initial value of Q0Q1Q2 = 010
The state just before the third clock pulse is the state at the end of second clock pulse.
At the end of second clock pulse
= Q0Q1Q2 = 001
A 3 phase, 6 pole star connected alternator revolves at 1000 rpm. The stator has 90 slots and 8 conductor per slot. The flux per pole is 0.05 wb. Calculate phase voltage generated by the machine if the winding factor is 0.96.
Pole = 6, conductor/slot = 8, flux = 0.05
Winding factor KW = 0.96, speed = 1000 RPM
Number of conductor z = slot × conductor/slot = 90 × 8 = 720
Eph = 4.44 Kc Kd f N = 4.44 × 0.96 × 0.05 × 50 × 120 = 1278.72 V
A 6 pole, 3ϕ , 440 V – 50 Hz induction motor has the rotor impedance of [0.2 + j1.6] Ω, The ratio of the stator of rotor turn is 1.8 and the full load slip is 0.06. The full load torque is ________ N – m
A DC shunt motor takes an armature current of 110 A at 480 V. The armature resistance is 0.2 Ω, the machine has 6 poles, and armature is lap connected with 860 conductors. The flux per pole is 0.05 wb, calculated torque developed (N-M) by the armature
Back emf Eb = V – IaRa = 480 – 110 × 0.2 = 458 V
where Z = 860
For lap winding A = P = 6
Flux ϕ = 0.05
A 400 MVA, 33 kV synchronous generator is tested for O.C.C and S.C.C the following data are obtained from these characteristics: If = 1080 A, VCC = 33 kV, ISC = 12.3 kA calculate Xs in PU
The sampled signal has z-transform 
Which of the following time-domain signals does it represents?
Let the signal represents ramp function
r(t) = t u(t)
signal is sampled at t = T
s(n) = r(nT)
= nT u(nT) = nT u(n)
By applying z-transform
For the Fourier transform X(jω) shown in figure, evaluate the value of 
The power supplied by the 40 V source in the circuit shown in the figure is _____(in W)
By applying Delta to star conversion as shown below:
Now the given circuit becomes,
Power supplied by 40 V source = 40 × 0.5 = 20 W
For the circuit shown in the figure below, the maximum average power delivered to the load of Vg = 80 cos 5000 t V is _____(in W)
Maximum average power delivered to the load, when 
R = 40 Ω
XL = jωL = j (5 × 103) (8 × 10-3) = 40 j Ω
Now the Thevenin circuit will be
|Irms| = 1 A
Average power delivered = (1)2 (20) = 20 W
In the voltage divider circuit shown in figure the no-load values of Vo is 4V. When the load resistance RL is attached across the terminal a and b, Vo drops to 3V. The value of RL in ohms is ______
Circuit under no-load condition
By voltage division
⇒ 5R2 = 40 + R2 ⇒ R2 = 10 Ω
The circuit with load RL,
Let the parallel combination of R2 and RL is Req.
By voltage division,
Find the value of DC current IE in (mA) for the given circuit shown in figure
VBE = 0.7, re = 25mV/IE, B = Very large
The Thevenin equivalent circuit for DC model is
Since B is very large, therefore IB is small and can be ignored.
The switch in the circuit shown in the figure has been in position 1 for a long time. At t = 0 the switch moves instantaneously to position 2. Find the value of R (in ohms) so that 10% of the initial energy stored in the 10 mH inductor is dissipated in R in 10 μs.
Given figure shows an electromagnet to lift a load of W gram. Assume infinite permeability for iron. The ampere turns of the exciting coil to maintain an equilibrium at an air gap of x mm would be given by
Force due to weight = Wg
Force due to electromagnet = 2Fe
Magnetic field in air gap,
Bg = μ0Hg
Ampere turn, NI = Hg. x
A solenoid is 100 cm long, 5 cm in diameter and consists of 1200 turns. The cylindrical core has a relative permeability of 60. This coil is co-axial with another solenoid 120 cm long, 6 cm diameter and 1000 turns. The mutual inductance between these two co-axial coils would be ____ (in mH)
Given that, μr = 60
N1 = 1200
N2 = 1000
d1 = 5 cm ⇒ r1 = 2.5 cm
l = 100 cm
An alternator rated at 6.6 kV protected by balanced circulating current system has its neutral grounded through current system has its neutral grounded through a resistance of 10 Ω. The protective relay is set to operate when there is out of balance current of 1 A in pilot wires which are connected to secondary winding of 800/5 ratio current transformer. The percentage of winding which remains unprotected is______
Let x% be the percentage of winding which remains unprotected
The voltage of unprotected winding 
The industry load or a three phase 2200 volts supply line is 1200 kW at a lagging power factor of 0.6. In order to raise the line power factor to unity (neglecting all the losses) What is the kVA rating of a synchronous condenser installed in parallel with the load and the kVA rating of an alternator required to supply the new total load?
Reactive power absorbed by load
Q = P tan ϕ
= 1200 × tan (cos-1 (0.6))
= 1600 kVAR
At unity power factor, reactive power requirement = 0
Rating of synchronous condenser = 1600 kVAR
Rating of alternator = 1200 kVA
Evaluate 
where C is the circle: |z| = 4
[Sum of residues at poles]
Poles: z = 0, 3
Both the poles are inside the |z| = 4
Residue at z = 0,
The value of 
and C is the curve x2 = 4y and y2 = 4x is _____.
The given integral is in the form of M dx + N dy
By using Green’s theorem,
In the differential equation 
is solved using the Euler’s method with step size h = 0.1, then y(1.2) is equal to (round off to 2 places of decimal)
Given that, y
y(1) = 2 ⇒ y0 = 2, x0 = 1
h = 0.1
According to Euler’s method, y
First iteration:
y0 = 2, x0 = 1, h = 0.1
y
y
Second iteration:
y1 = 2.22, x1 = x0 + h = 1 + 0.1 = 1.1, h = 0.1
y
y
Consider the 2 × 2 matrix 
What is the maximum value of x, for which both the Eigen values of matrix are real and positive?
⇒ (2 – λ) (4 – λ) – 3x = 0
⇒ λ2 – 6λ +8 – 3x = 0
Roots of the above equation are Eigen values and let the Eigen values be λ1 and λ2
From the properties of Eigen values,
For the Eigen values to be positive, λ1λ2 > 0
⇒ 8 – 3x > 0
⇒ x < 8/3
For roots to be real, 
⇒ 36 – 32 + 12x > 0
⇒ x > -1/3
The range of for which Eigen values of matrix are real and positive is: 
Maximum value of x = 8/3 = 2.67
The particular integral of 
Particular integral is,
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