The sentences given with blanks are to be filled with an appropriate word(s). Four alternatives are suggested for each question. For each question, choose the correct alternative and click the button corresponding to it.
They abandoned their comrades ______the wolves.
Direction: Select the one which is different from the other three responses.
49  33 = 16
62  46 = 16
83  67 = 16
70  55 = 15
All options have a difference of 16 except the D option.
Hence, option D is the answer.
Amit rows a boat 9 kilometres in 2 hours downstream and returns upstream in 6 hours. The speed of the boat (in kmph) is:
Downstream rate = 9/2 = 4.5 kmph
Upstream rate = 9/6 = 1.5 kmph
The speed of the boat = (4.5 – 1.5) kmph = 3 kmph
Directions: In the following question, out of the four alternatives, select the word opposite in meaning to the given word.
Turgid
Turgid = swollen and distended or congested; pompous
Bloated = inflated
Humble = plain, simple
Puffy = inflated
Tumescent = swollen
Hence, humble is the correct answer.
Direction: Study the following information carefully and answer the given questions:
The following pie chart shows the percentage distribution of total number of Apples (Dry + Wet) sold in different days of a week:
The pie chart2 shows the percentage distribution of the total number of Apples (Wet) sold on different days of the week.
Find the ratio of Apples (Dry + Wet) sold on Tuesday to that of the Apples (Wet) sold on Thursday?
Required ratio = 13% of 7000: 23% of 4500 = 182:207
Directions: In each of the questions below, some statements are given followed by some conclusions. You must consider the statements to be true even if they seem to be at variance with commonly known facts. You must decide which of the following conclusions logically follows from the given statements. Give answer.
Statements:
Some jacket is shirt.
Some shirts are trouser.
No shoes are tshirt.
All trousers are shoes.
All pants are tshirt.
Conclusions:
I. All shirts being tshirt is a possibility.
II. Some shoes are trouser.
III. Some jackets are tshirt.
IV. Some shirts being tshirt is a possibility.
From the above diagram
• Some shoes are trouser.
• Some shirts being tshirt is a possibility.
Hence option B is correct.
Direction: Which of the following is the MOST SIMILAR in meaning to the given word?
Remnant
‘Remnant’ means ‘remainder’ which is the same as ‘residue’.
Direction: Read the information carefully and give the answer of the following questions
Total number of children = 2000
If twoninths of the children who play football are females, then the number of male football children is approximately what percent of the total number of children who play cricket?
57.4% (If 2/9 th of children play football are female, then
male children = 1  2/9 = 7/9th of children play football;
Let’s take z% of male football children equal to cricket children, then
⇒ z% of cricket children = 7/9 th of football;
⇒ z% of (23% of 2000) = 7/9 th of (17% of 2000);
⇒ z% of 23 = 7/9 th of 17
⇒ z = 7× 17 × 100 = 57.4%;)
Shortcut : Z% of 23 = 7/9 of 17
(7x17x100)/ (23x9) =57.4%
Required percentage = (7/9)×17×100/ 23 = 57.4%
Direction: Study the following information carefully and answer the given questions.
Seven people, P, Q, R, S, T, W and X sitting in a straight line facing north, not necessarily in the same order. R sits at one of the extreme ends of the line. T has as many people sitting on his right, as on his left. S sits third to the left of X. Q sits on the immediate left of W. Q does not sit at any of the extreme ends of the line.
If all the people are made to sit in alphabetical order from right to left, the positions of how many people will remain unchanged?
From the given conditions, we can conclude:
After arranging in alphabetical order,
Hence, only Q′ s position will remain unchanged.
A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in total from both as interest. The rate of interest per annum is.
Let the rate % = R
According to the question,
= 2200
100R + 120R = 2200
220R = 2200
R = 10%
Hence required rate % = 10%
A 10 KVA, 400 V, 3 phase, star connected synchronous generator has an armature resistance of 0.8 Ω per phase and synchronous reactance of 1.5 Ω per phase. The voltage regulation at a power factor 0.6 lagging is _______.
R_{a} = 0.80X_{s} = 1.5Ω
V_{ph} = 230.94V
E_{g} = 255.19V
Voltage Regulation =
=
V.R. = 10.50%
How many essential prime implicants are there in the given kmap?
AB and are essential prime implicants because only these two have at least single 1′ ' which can have only one pairing.
The various resistance of a wheatstone bridge are shown in figure. The battery has an internal emf of 8 V with negligible internal resistance. If the sensitivity of the galvanometer is 12 mm/µA with an internal resistance of 150 Ω.
What will be the sensitivity of the bridge in terms of deflection per unit change in resistance.
Resistance of unknown resistor required for balance,
Therefore, the deviation in unknown resistance from the balance condition,
ΔR = 1650  1600
= 50Ω
So now Thevenin source emf,
ε_{0} = 0.039V Now the internal resistance of bridge looking into terminal C & d−
= 481.95Ω
Hence the current through the galvanometer
then deflection of galvanometer
θ = Si. Ig
= 12 × 61.7mm
θ =740.4mm
Therefore, sensitivity of bridge
The circuit shown below has C = 20 μF and L = 5 μH. Initial voltage across the capacitor is V_{s} = 230 V. The load current is constant at 300 A . Let ‘A’ be the conduction time period of the auxiliary thyristor and ‘B’ is the circuit turn off time of the main thyristor. Than A + B will be _________.
The above circuit is class 'B' commutation.
So, the conduction time of auxiliary thyristor is =
A = 31.4 \musec
The peak value of resonant current,
Voltage across the main thyristor, when it gets turned off will be
Circuit turn off time for main thyristor is
So, A + B = 31.4 + 11.62
=
In the circuit shown in figure the switch is closed at time (t=0). The rate of change of current
across inductor will be 
At (t = 0^{+}) as the circuit was unenergized at (t = 0^{ }) ⇒
Inductor L is open circuit (O.C)
Capacitor C is short circuit (S.C)
In the figure shown below,
Assume V_{th} = 0.3 V, the transistor is operating in
Since the device is NMOS,
V_{th }= 0.3 V
V_{S} = 0.4 V
V_{D} = 2 V
V_{G} = 0 V
For cutoff region, V_{GS} < V_{th}
Since, VGS = –0.4 V
So, the device is in the cutoff region.
If a finite length of wire having a total length QR = 80 m, a current of 10 A is flowing in it.
A point P from the origin point or reference point is away at a distance of 4 mm. If the point P is making an angle of 75° with the point Q in clockwise direction & 60° with the point R in anticlockwise direction. What will be the magnetic field density B, at point P.
B =
where the sign of angle is tve, is being taken, a is the distance from a reference point.
S_{0} B at point P
B =
= 2.5 x 10^{4}(0.966 + 0.866)
= 0.458 x 10^{3}Wb/m^{2}
B = 0.458 mWb/m^{2}
A chopper circuit shown in the figure has an input voltage of V_{s} = 5 V. Duty ratio is 0.6 . The load resistance is 10 Ω. The chopper is operating at 25 kHz switching frequency. If L = 150 μH and C = 220 μF. Find the ratio of peak value of capacitor voltage to peak value of inductor current?
The given circuit is a boost regulator.
For the boost regulator,
When the switch is ON.
Peak value of inductor current is
The ripple voltage at capacitor is
Peak voltage at capacitor,
Required ratio is = =
= 3.56
Consider the circuit given in the figure below, if voltage V_{AB} and V_{CD} in the circuit are 200 V each, then the value of the capacitor in the circuit, C is approximately ________ (in μF).
Let assume angle between V_{AB} and V_{BC} is θ
So, taking magnitude both sides
By solving, θ = 120∘
Current through inductor h_{L}=10A
since, we know that inductor current lags 90^{∘} by its voltage and for capacitor its current leads 90^{∘} by its voltage
C =
C =
A uniform volume charge density ρ_{v}= 5 µC/m^{3} is present in the spherical shell of 0.9 < r < 1 m and ρ^{v }= 0 elsewhere. The charge present in the shell approximately is
Q =
=
=
Q =
=
Q = 5.675µC
A 15 MVA, 11 kV, alternator has positive, negative and zero sequence reactance are 0.3 p.u., 0.2 p.u. and 0.05 p.u. respectively. Find the value of resistance to be added in the generator neutral so that the fault current of LG fault does not exceed 1.5 times of rated line current.
For LG fault Current
R_{n}= 0.641 p.u.
R = 5.17 Ω
Consider a circuit shown in figure below, switch is closed at t = 0 in the circuit for V = 10 V, R = 5 Ω L = 3 H. Find the value of maximum current ‘I’ in the circuit. (in Amperes)
Expression for current 11 for t≥0 is
Where τ=LR
Similarly, for current I_{2}
Current 1 = 1_{1 }I_{2}
The value of current at this time is
I = 0.5
A causal system has input x[n] = δ[n] + 14δ[n − 1] − 18δ[n − 2]
and output y[n] = δ[n] − 34δ[n − 1]
The impulse response of this system is
⇒
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., j0.2 p.u. and j0.3 p.u. respectively. If prefault 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, Z_{n} = j0.02 p.u.)
Similarly, the equivalent negative sequence reactance is
For a single line to ground fault, fault current is
Given an assembly language program in 8085 microprocessor,
LXI B, 2384 H
LOOP: DCX B
MOV A, C
ORA B
JNZ LOO
HLT
If the system clock frequency is 2 MHz. The time required for complete execution of the program is _________ msec.
Number of Tstates required = Count x Number of T states inside loop + Number of T states outside loop Number of T states when jump is invalid.
[Here, count in decimal (2384)16=(9092)10]
=(9092 × 24 )+ 16 − 3 = 218221 Tstates
Clock frequency is 2MHz
T=1 / 2 ×10^{6 }= 0.5μsec
Total execution time =218221 × 0.5 × 10 − 6 = 0.10911sec
Hence, the time required for execution of the program is 109.11 msec.
Let x(t) be a signal that has a rational Laplace transform with exactly 2 poles located at S = –1 & S = –3. If g(t) equal to e^{2t }x(t) & G(ω) converge. The conclusion about g(t) is
One pole in RHS & another pole in LHS to make the system converging i.e. stable, the signal must be noncausal.
Hence Doublesided signal.
If the crystal frequency of 8085 microprocessor is 6MHz then total time (in millisecond) required to execute the program, before HLT instruction is _____msec.
T_{Total} = ⅓ x 60 = 20ߎsec = 0.02msec
Consider a feedback system with characteristics equation .
The point of intersection of the asymptotes of the root locl with real axls on
Characteristic equation is equation as: 1+G(s)H(s) = 0
On comparing this characteristics equation with the given equation
The point of intersection of the asymptotes = centroid(o)
In a current cumulated chopper, peak commutating current is twice that of the maximum possible load current. The source voltage is 250 V dc and main SCR take off time is 50 μs. Voltage is 250 V dc and main SCR taken off time is 50 μs. If maximum load current is 250 A, then peak capacitor voltage is………V.
Peak Capacitor Voltage V_{c} = V_{s} + Io
Peak Current = I_{p }+ 2I_{o} = 2(250) = 500
= 2
= 0.5
V_{CP} 250 + 250(0.5) = 375V
Which of the following is incorrect:
For inverting amplifier,
(Av)1= −R_{F} / R_{L}
if R_{1} > R_{F} then it will be less than 1
for NonInverting amplifier,
(AV)_{NI }= (1 + R_{F} / R_{L})
when R_{F }= 0, AV = 1
it can be one but cannot be less than one in any case.
So, option B is an incorrect statement and correct Answer.
A singlephase transformer, with secondary voltage of 250 V, 50 Hz delivers power to R load of 15 Ω through a half wavecontrolled rectifier circuit. For firing angle of 45°, the form factor is____
Average value of output voltage
RMS value of output voltage,
Vorms =168.55V
Form factor, FF = = 1.75
Time period of the function X(n) is e^{jm2} will be
x(n) = ejm2 = cosπn^{2} + jsinπn^{2}
x(n) = (1)^{r2} + 0
x(n) = (1)^{n2}
x(o0 = (1)0^{2} = 1
x(1) = (1)^{2} = 1
x(2) = (1)2^{2} = 1
x(3) = (1)3^{2} = 1
Hence time period will be 2
A 200 V, unsaturated shunt motor has an armature resistance (including brushes and interpolar winding) of 0.05 Ω and field resistance of 200 Ω.
The value of resistance in Ω to be added in the field circuits increases the speed from 1500 RPM to 1800 RPM. When the supply current is 300 A.
As, armature current is nearly equal to supply current. The value of back emf will not change much
As we known,
is nearly constant
R_{enva} = 240  200 = 40Ω
Two resistors R1 and R2 are connected in series. The voltage across the series combination is V. A voltmeter V1 is used to measure the voltage across the resistor R1 and a voltmeter V2 is used to measure the voltage across the resistor R2.
The voltmeter V1 is ±X% and the voltmeter V2 is +y% accurate. The voltmeter V1 reads 10 V and V2 reads 5 V. The absolute error of two voltmeters are the same and absolute error in total voltage V is 0.2 V.
The values of X and Y (in %) respectively are:
The voltmeter V1 is ±x% and voltmeter V2 is +y% accurate
V1 = 10V, V2 = 5
Absolute error,
∨1 = x100 × 10 = 0.1x
V2 = y100 × 5 = 0.05y
Absolute error of two voltmeter are same
0.1x = 0.05y
y = 2x −(i)
Total voltage, V = V1 + V2
Absolute error in total voltage,
V = 0.1 x + 0.05y
0.2 = 0.1 x + 0.05y
Put equation (i) in above expression
0.2 = 0.1 x + (0.05)2x
⇒x=1%, y=2%
Independent trials consisting of a rolling of a fair die are performed, the probability that 2 appears before 3 or 5 is __________ (Up to three decimal)
(0.2, 0.4)
Probability =
=
=
= 3 / 8
= 0.375
In the circuit shown all the values are up on 100MV, 400KV base (LL) If a (solid 3 phase vault) occurs at the cantro of the transmission line at an instant to = 9.735msec. What is the magnitude of initial DC offset current? (in kA) (Nominal system frequency f = 50Hz). The voltage in phase a is V = V_{m} sin ωt.
Magnitude of DC offset current =
At switching instant Ioc =
= 23.57 pu
I_{DC}= 3.402KA
In a short circuit test on a 132 KV 3–phase system, the breaker gave the following results: Pf of the fault 0.4, recovery voltage 0.95 of full line voltage, the breaking current is symmetrical and the restricting transient had a natural frequency of 16 KHz. What would be the rate of rise of restriking voltage.(assume that the fault is grounded)
The peak value of phase voltage
= V_{mph} = 132 √3 × 2 = 107.77KV
Active Recovery voltage
= K_{1}K_{2}V_{mph}sinϕ
K_{1 }= 0.95
K_{2 }= 1 (grounded fault)
cosϕ = 0.4 ⇒ ϕ = cos^{−1}(0.4)=66.42^{∘}
Sinϕ = 0.916
Active recovery voltage
= 0.95 × 1 × 107.77 × 0.916
= 93.85KV
Restriking Voltage
= 93.85k(1 − cosωt)
RRRV = d / dt [93.85 × 10^{3}(1 − cosωt)]
RRRV =93.85 ×103 × ω. sinwt (RRRV) average = 2π(93.85 × 10^{3} × ω)
=2/π × 93.85 × 10^{3 }× 2πf
= 4f x 93.85 x 10^{3}
= 4 x 16 x 93.85 x 10^{6}
= 4f x 93.85 x 10^{–3} KV/μsec
Average RRRV = 6 KV/μsec
The transformation ratio of an auto transformer for a 30 kW, 400, 3ϕ induction motor is______. If the starting torque is to be 75% of full load torque. Assume the full load slip to the 5% and the short circuit current to be 6 times the full load current.
As we know,
x^{2} = 0.41667
x = 0.645
In resonant pulse commutation, C = 30 μF and L = 5 μH. Initial voltage across capacitor = V_{S }= V_{C} = 240 V. If the constant load current of 240 A, then circuit then of time for main thyristor is______μsec.
Circuit turn off time,
Where, V_{ab} = V_{s}cosωt
V^{ab} is voltage across the main thyristor when it gets turned off.
V_{ab} = 240cos24.09^{o}
V_{ab} = 219.09V
Consider the circuit shown below:
The value of capacitance C required to make circuit power factor unity is:
The equivalent impedance seeing from source side is
Y_{eq} = (0.035  j0.0136) + jwC
For unit power factor, the imaginary part should b zero
j(wC  0.0136) = 0
Let x (t) = rect (t−1/2) (where rect (x)=1 for −1/2 ≤ x ≤ 1/2 and zero otherwise). Then if
sinc(x) = sin(πx) / πx, the Fourier Transform of x(t) + x(−t) will be given by
x (t) = rect (t1/2)
y (t) = x (t) + x (t) = rect (t1/2) + rect (t1/2) = rect(t/2)
Fourier Transform of y (t) is
y(ω) = ∫−1^{1I}⋅e^{−jω}dt
y(ω) = [e^{−jω}/−jω]−11
y(ω) = e^{jω}−e^{−jω}/jω
y(ω) = 2/ω[e^{jω}−e^{−jω}/2j
y(ω) = 2/ωsinω
y(w) = −2sin(n−ω/n)/(π−ω/n)
y(ω) = 2sin(ω/n)
But none of the four options match this Answer The result can be modified as follows.
y(ω) = 2/ωsinω
y(ω) = 2/ω⋅2sin(ω/2)⋅cos(ω/2)
y(ω) = 2sin(n−ω/2π)/(Π−ω/2π)⋅cos(ω/2)
y(ω) = 2sinc(ω/2π)⋅cos(ω/2)
which matches option C.
A 100 MVA 50 Hz two alternator operates at no load at 3000 rpm. A Load of 27 mw is suddenly applied to the machine and the steam values to the turbine commence to open after 0.6 sec due to time – log in the governor system. Calculate the frequency to which the generator voltage drops before the steam flow to increase to meet the steam flow to increase to meet the new load. Inertia constant H = 4.5 KW – sec / KVA.
We know the swing equation
= sync honours speed
Integrate on both side
W_{r }= rotor speed
So ds/dt = 9.42t
Now at t = 0.6
W_{r } w_{s}= 5.652
Wr = w_{s}  5.652
= 308.348 rad/sec
f’ = 49.1 Hz
In the given circuit, the silicon transistor has β = 75 and a collector voltage V_{C}=9V.
Then the ratio of R_{B} and R_{C} is
Correct answer is 105.1
Given β = 75
V_{C} = 9 V
We redraw the given circuit as
From the circuit, we have
15 − R_{C}(I_{C} + I_{B}) = V_{C}
15−V_{C }= R_{C}(I_{C }+ I_{B})
R_{C}(I_{C }+ I_{B}) = 6V……….(1)
Applying Kirchoff's law across BE Junction,
V_{C }− R_{B}I_{B }− V_{BC }= 0
or 9 − 0.7 = R_{B}I_{B}
or R_{B}I_{B }= 8.3......(2)
Also,I_{C }= βI_{B }= 75l_{B}…….(3)
Solving equations (1),(2), and (3), we get
R_{C}(75 + 1)I_{B }= 6……….(4)
R_{B}⋅I_{B }= 8.3……….(5)
Hence, dividing equation (5) by (4), we obtain the desired ratio as
R_{B}/R_{C }× (1/76)= 8.3/6
R_{B}/R_{C} = {76 x 8.3} / 6 = 105.13
The output characteristics for a BJT are given below.
Determine the hybrid parameter h_{fe}, if V_{CEQ }= 6V and I_{CEQ} = 3.5mA
Firstly, the given Qpoint is plotted on the transfer characteristics.
Keeping V_{CE} Constant, the change in ib is chosen along the straight line at V_{CEQ} = 6V
Next, the corresponding change in i_{c} by drawing horizontal lines intersecting at I_{B} to the vertical axis.
Thus, h_{fe} is computed as,
⇒ h_{fe} =
⇒ h_{fe} = = 200
Find the resistance and inductance of the coil if the arms of an ac Maxwell Bridge are given as: AB and BC are nonreactive resistors of value 100 Ω each. Let DA is a standard variable reactor and value of L1 of resistance 32.7Ω and CD comprises a standard variable resistor R connected in series with a coil of unknown impedance. Balance was obtained with L1=47.8mH and R=1.36Ω?
Figure shows Ac Bridge; in the figure we observe the products of
the resistances of opposite arms are equal
32.7 × 100 = (1.36 + R4)100
32.7 = 1.36 + R4
R4 = 32.7 − 1.36 = 31.34Ω
L1 × 100 = L4 × 100
L4 = L1 = 47.8 mH
A 3 phase full converter charges a battery from a three phase supply of 230V, 50Hz. The battery emf is 200V and its internal resistance is 0.5Ω. On account of inductance connected with the battery charging current is constant at 20A. Compute the supply power factor.
_______ lag.
We know that
√3v_{s}s_{s}cosϕ = v_{o}l_{o}…(i)
Battery terminal voltage Vo is
V_{0 }= E + I_{O}r
= 200 + 20 × 0.5
= 210V
I_{O} = 2OA (given)
We know that supply current is question square wave of amplitude
I_{o} = 20A and for 120^{∘} over every half cycle
Using (i)
√3 × 230 × 16.33cosϕ = 210 × 20
cosϕ = 0.646lag
A nonsalient pole, 3phase, synchronous machine with negligible resistance is on an infinite bus. Its synchronizing power coefficient will be maximum when δ = _______ degrees.
Synchronizing power coefficient at any,
This will be maximum when δ=0.
The current coil in a wattmeter is connected in series with an ammeter & an inductive load. The ammeter reading is 5.5 A. A voltmeter & the voltage coil are connected across 200 Hz supply. The reading of voltmeter & wattmeter is 220V & 25 watt respectively. The inductance of the voltage circuit is 12 mH & its resistance R = 2400 Ω. If the voltage drop across the ammeter & the current coil are negligible, then what will be the approximate percentage error in the wattmeter reading.
Inductive reactance of pressure coil =2π × 200 × 12 × 10^{−13} = 15Ω
Resistance of pressure coil =2400Ω
Phase angle of pressure coil
Since,
tanβ = 152 / 400 = 6.25 × 10^{−3}
So, β = tan^{−1}(6.25 × 10^{−3})
=0.36^{∘} or β = 0.36^{∘}
We know that for inductive load,
Reading of the wattmeter αcosβ.cos(ϕ−β)
True power αcosϕ
Where cos ϕ load p.f.
Therefore,
True power
= cosϕ / cosβ.cos(ϕ − β) × Re ading of wattmeter
Here true power = 1^{2}R
= (5.5)^{2}R
Which is equal to = 1^{2}Zcosϕ
Impedance
Z = 220 / 5.5 = 40Ω
So, true power = (5.5)^{2} × 40cosϕ
Reading of wattmeter is given as = 25 watt
So,
(5.5)^{2} × 40 cosϕ = cosϕ / cosβ⋅cos(ϕ − β) × 5
Or,cosβ⋅cos(ϕ − β) = 25(5.5)^{2 }× 40 = 0.0206
since β = 0.36^{∘}
So, cos(0.36)⋅cos(ϕ − 0.36^{∘}) = 0.0206
or cos(ϕ − β) ≃ 0.0206
(ϕ−β) ≃ cos^{−1}(0.0206)
ϕ−β ≃ 88.81^{∘}
so, ϕ ≃ 88.81 + 0.36^{∘} ≃ 89.17^{∘}
therefore, percentage error
=tanϕ⋅tanβ × 100
=tan(89.17)⋅tan(0.36) × 100
=43.37%
A synchronous motor is drawing 40 A from 440 V, 3ϕ supply at unity p.f. with a field current of 0.9 A. The motor has synchronous reactance of 1.4 Ω. If mechanical load on motor remains the same. Find the % change in the field current which would result in 0.85 p.f. leading is _________? (Assuming linear magnetization)
Since it is given that mechanical load is constant i.e. P= constant
V/cosϕ = constant
I Cos ϕ = constant
I_{1}Cosϕ_{1} = I_{2}Cosϕ_{2}
40 × 1 = 1_{2}×0.85
I_{2 }= 47.05A
When power factor is 0.85 leading motor takes the current 1 = 47.05
∠31.78∘
In that condition the generated emf is
E_{f} = 254.03∠0^{∘ }− (47.05 < 31.78)(1.4="" />< />^{∘})
E_{f} = 254.03 − 65.87 < />^{∘}
E_{f} = 294.09∨
Since given that linear magnetization
E_{f }∝ I_{f}
E_{f2} / E_{f1 }= I_{f2 }/ I_{f1}
294.09 / 260.12 = I_{f2} / 0.9
l_{f2} = 1.017A
% change in the field current
= 1.017 − 0.9 / 0.9 × 100
= 13% increases
Two identical 1 − ϕ transformer of 15kVA, gave the following results, when tested by back to back method.
W_{1 }= 1000 watt (Primary)
W_{2 }= 1500watt( Secondary )
Calculate the percentage efficiency of the transformer at 1/4^{th} load 0.8 P.F. lag
In the back to back test the primary wattmeter gives the iron loss.
Each transformer has iron loss
=W_{1} / 2 = 1000 / 2 = 500W and secondary wattmeter reading gives the full load copper loss.
Each transformer has full load copper loss
=W_{2} / 2 = 1500 / 2 = 750W
Full load loss at 1/4^{th } of load is
=
Efficiency η=
η=0.8458×100%
η=84.58%
A long transmission line has R = 6 Ω, G = 240 mS L = 2 H and C = 80 mF is operating at 50 Hz, 400 kV. If this line is terminated with 5 Ω, the voltage across the load as ______ kV.
Z_{c} =
=
= 5Ω
Z_{c} = Z_{L}
Hence, V_{L }= V_{S }= 400 kV
The value of is Where [x] is the greatest integer function.
[x] = 0; 0 ≤ x ≤ 1
[x] = 1; 1 ≤ x ≤ 2
[x] = 2; 2 ≤ x ≤ 3
So.
= 3 /2 + 5
= 6.5
The Cauchy’s mean value of the functions e^{x} & e^{−x }between (0, 1) is
f(x) = e^{x}, g(x) = e^{−x}
a = 0, b = 1
Cauchy's mean value a < c="" />< b="" />
f′(x) = e^{x} & g(x) = −e^{−x}
⇒ −e^{2c} =
⇒ −e^{2c }= −e
⇒ 2c = 1
⇒ c = 0.5
The daily variation of load on a 100 KVA transformer is as follows:
8:00 AM to 1 PM – 65 kW, 45 KVAR
1 PM to 6 PM – 80 kW, 50 KVAR
6 PM to 1 AM – 30 kW, 30 KVAR
1 AM to 8 AM – No load
This transformer has no load area loss of 270 W and a full load ohmic loss of 1200 W. The allday efficiency of the transformer is
No load core losses are constant throughout 24 hrs. core losses = 270 × 24 = 6.48 kW
8 AM to 1 PM
kW output = 65 kW, kWh output = 325 kWh
Cu loss is proportional to S^{2}
Culoss = 749.86 x 5 = 3.75kWh
1PM to 6PM
KWh output = x 5 = 400kWh
Culoss = 1068 x 5 = 5.34kWh
Culoss = 216 x 7 = 1.512kWh
= 97.96%
The output of a continuoustime, linear timeinvariant system is represented by T{x(t)} where x(t) is the input signal. A signal z(t) is called eigensignal of the system T, when T{z(t)} = Y z(t), where Y is a complex number, in general, and is called an eigenvalue of T. Assume the impulse response of the system T is real and even. Then which of the following statements is TRUE?
Given impulse response is real and even Thus H(jω^{0}) will also be real and even Therefore H(jω_{0}) = H(−jω^{0})
If cos(t) is input then
Output will be =
If sin(t) is input then
Output will be
So sin(t) and cos(t) are Eigen signal with same Eigen values.
The opencircuit and shortcircuit characteristic of a singlephase synchronous generator are shown on the same graph below.
A field current of 60 A produces the rated voltage. If the effective armature resistance of the generator is 0.3 Ω, determine the synchronous reactance of the machine [in Ohm].
From the given figure, at the field current, I^{f }= 60A
Opencircuit voltage, V_{OC }^{ }= 500V
Shortcircuit armature, I_{SC} = 275A
Hence, the synchronous impedance,
Synchronous reactance,
Find the inverse Z transform of X(z)
A three phase to ground fault occurs at " ' in the network show in figure. Generator G_{1} : E_{1 }= 1 pu. ,x_{1} = x_{2} = 0.8 pu. x0 = 0.5 pu.
Generator G_{2} : E_{2} = 1pu,x_{1} = x_{2} = 0.6pu
X_{0} = 0.3pu
Transformer T_{1} : x_{1} = x_{2} = X_{0} = 0.22pu
T_{2} : x_{1} = x_{2} = x_{0} = 0.15pu
Each line:
x_{1} = x_{2} = x_{0} = 0.67pu, x_{2} = 0.02pu
The current flowing through ground is :
Three phase to ground fault occurs so we require only positive sequence network
Fault current
For the given system, determine the overall response Y(s) of the system corresponding to both inputs.
When multiple inputs are present in a linear system, output is computed by treating each input independently. The overall response of the system is obtained by adding the outputs corresponding to each input.
Assuming x_{2 } (s)= o
The block diagram can be reduced to,
Reducing the inner feedback loop,
The block diagram is reduced to,
Thus,
= output due to X_{1}(s) acting alone
⇒
Assuming X_{1}(s)=0
The block diagram is reduced to,
thus. Y_{x2}(s)= output due to X_{2}(s) acting alone
⇒
Hence, the response of the system,
⇒
Find the inverse fourier transform of X (t) =
Given
X(w) =
We know that, X(w) =
=
=
In sinusoidal pulse width modulation, the lowestorder harmonic in the modulated signal is the 15^{th} harmonic. Determine the number of pulses per half cycle of the output signal.
In sinusoidal PWM, all the harmonics less than or equal to (2p – 1) are eliminated, where p is the number of pulses per half cycle. Since the lowestorder harmonic is 15th, the harmonics till 13th are eliminated.
Thus,
2p − 1 = 13
⇒ 2p = 13 + 1 = 14
⇒ p = 14/2 = 7
Hence, there are 7 pulses per half cycle in the output signal.
A 300V DC shunt motor is drawing a full load current of 45A. The armature resistance and shunt field resistance are 0.6Ω and 160Ω respectively. If the torque remains same throughout, by how much should the flux be reduced to increase the speed by 40%?
T ∝ ∅I_{a}
since torque is constant and flux and speed are changing. SOt
∅_{1}I_{a1 }=∅ _{2}I_{a2}
I_{a2} = ∅_{1} / ∅_{2}I_{a1}
Let
∅_{1} / ∅_{2} = x, then
I_{a2} = 45x as I_{a1} = 45A
E_{b1} = 300 − (0.6 × 45) = 273V
E_{b2} = 300 − (0.6 × 45x) = (300 − 27x)V
It is given that N_{2 }= 1.4N_{1} since speed is increased by 40% Now, we know E_{b }∝ N∅
⇒ E_{b1} / E_{b2} = N1∅_{1} / N_{2}∅_{2}
382.2 = 300x − 27x^{2}=0
27x^{2} − 300x + 382.2=0
x = 9.64,1.4685
For x = 9.64, the increase in speed criteria is not satisfied. Percentage change in flux
= ∅_{1}−∅_{2} / ∅1
= 1 − ∅_{2} / ∅_{1 }= 1 − 11.4685 = 31.9%
A 50 cycle, 4 pole turbo generator of rating 30MVA, 11.2KV has an inertia constant of 10KW sec/KVA. If the acceleration computed is constant for 10 cycles and input less the rotational losses is 26800 and electric power developed is 16MW. Find the change in torque angle [in Electrical degree] in that period. Assume that the generator is connected to a large system and has no accelerating torque before the 10 cycle period commences.
GH = 30 × 10 = 300MJ
Pm = 26800hp = 26800 × 0.746 = 20,000KW
Pe = 16000KW
Pa = Pm − Pe = 20,000 − 16,000 = 4000KW
Pa = Md^{2}δ/dt^{2}
M = GH/180f = 300/(180 × 50) = 1/30
4 = 1/30(d^{2}δ/d^{1})
d^{2}δ/dt^{2 }= 120%sec^{2 }= 120 × π/180 = 2.09rad/sec^{2}
Multiply by 2(dδ/dt). dt on both sides
Integrating the above equation
Further integration
constant for 10 cycle which means 10 cycles = 0.2 sec δ = 1.045(0.2)^{2} = 0.0418rad = 1.8^{∘}
Find the value of average output voltage if a 230V, 50Hz, singlepulse SCR is feeding a RL load with α = 40 ° and β = 210 °?
Vo = Vm/2π x (cosαcosβ)
Where Vm = √2Vs
Vm = 84 V
Consider a circuit as shown in fig below
Equivalent resistance across the terminals A & B is_____Ω.
⇒ Req = 1 + ¾
⇒ 7/4 Ω
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