Final bank
Electrical Engineering Quiz
Test your knowledge in electrical engineering with this comprehensive quiz! Covering topics from power systems to transformers, this quiz is designed for both students and professionals looking to challenge themselves.
Key features of the quiz:
- 30 carefully crafted questions
- Multiple choice format for easy answering
- Detailed problem-solving scenarios
The voltage across a component is measured as 80 Vr.m.s. And the current through it is 4 A r.m.s. If the current leads the voltage by 20-the angle between the current and the voltage.what is the apparent power in the component?
109 VA
116 VA
301 VA
320 VA
The voltage across a component is measured as 80 Vr.m.s. And the current through it is 4 A r.m.s. If the current leads the voltage by 20-the angle between the current and the voltage. What is the reactive power in the component?
109 VA
116 VA
301 VA
320 VA
The voltage across a component is measured as 80 Vr.m.s. And the current through it is 4 A r.m.s. If the current leads the voltage by 20-the angle between the current and the voltage.what is the active power component?
109 VA
116 VA
301 VA
320 VA
Balanced positive sequence star connected voltage source with Eab=480 angle 0^o V is applied to a balanced delta connected load with Zdelta =30 angle 40^o ohm. The line impedance between the source and load Zl =1 angle 85^o ohm for each phase. Calculate: Line currents.
Il=25.83 angle -73.78^o
Il=15.83 angle 73.78^o A,
Il=25.83 angle -63.78^o A,
A balanced positive sequence star connected voltage source with Eab=480 angle 0^o V is applied to a balanced delta connected load with Zdelta =30 angle 40^o ohm. The line impedance between the source and load Zl =1 angle 85^o ohm for each phase. Calculate: Delta load currents.
Iph=10.9 angle -43.78^o A
Iph=14.9 angle -43.78^o A,
Iph=14.9 angle -13.78^o A,
A balanced positive sequence star connected voltage source with Eab=480 angle 0^o V is applied to a balanced delta connected load with Zdelta =30 angle 40^o ohm. The line impedance between the source and load Zl =1 angle 85^o ohm for each phase. Calculate: The voltage at the load terminals.
Vph= 407.3 angle -3.78^o V
Vph= 447.3 angle 3.78^o V
Vph= 447.3 angle -3.78^o V
A balanced abc sequence Y-connected source with Van =100 angle 10^o V is connected to delta - connected load (8+j4) ohm/phase. Calculate the phase and line currents
) Iph= 19.36 angle 13.43^o A, Il= 33.53 angle -16.57^o A
Iph= 10.36 angle 13.43^o A, Il= 33.53 angle -16.57^o A
Iph= 19.36 angle 13.43^o A, Il= 10.53 angle 16.57^o A
A three phase, 6 poles, star connected synchronous generator revolves at 1000 rpm. The flux per pole is 0.05 wb. Calculate the voltage generated if the machine constant equals 532.8 V/HZ.Wb.
Eline= 307 V
Eline= 237 V
Eline= 2307 V
The effective resistance of 2200 V, 50 HZ, 440 KVA 1-phase alternator is 0.5 ohm. On short circuit of field current of 40 A gives full load current of 200 A. The emf on open circuit with the same field current is 1160 V. Calculate the synchronous impedance and reactance.
) Zs= 5.8 ohm. Xs=5.77 ohm.
) Zs= 1.8 ohm. Xs=1.77 ohm.
Zs= 5.8 ohm. Xs=1.77 ohm.
200 kVA, 480 V, 50 Hz, Y-connected synchronous generator with a rated field current of 5 A was tested and the following data were obtained : 1-VT,OC = 540 V at the rated IF. (line voltage) 2-IL,SC = 300 A at the rated IF. ((line current) 3-When a DC voltage of 10 V was applied to two of the terminals, a current of 25 A was measured. Find the generator�s model at the rated conditions (i.e., the armature resistance and the approximate synchronous reactance).
Ra= 0.2 ohm. Zs=1.08 ohm. Xs= 1.02 ohm.
Ra= 0.2 ohm. Zs=1.08 ohm. Xs= 2.02 ohm.
Ra= 0.2 ohm. Zs=0.08 ohm. Xs= 1.02 ohm.
A 2000/200, 20 KVA single phase transformer has 66 turns in the secondary. Calculate: 1) Primary turns. 2) Primary and secondary full load current. Neglect losses
Np= 660 turns. Ip= 10 A, Isec=100 A.
Np= 620 turns. Ip= 10 A, Isec=100 A.
Np= 260 turns. Ip= 10 A, Isec=100 A.
-An ideal 25 KVA single phase transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3000 V, 50 HZ supply. Calculate: Primary and secondary currents on full load.
Ip = 2.33 A, Isec= 104.125 A
Ip = 8.33 A, Isec= 104.125 A
Ip = 3.33 A, Isec= 104.125 A
-An ideal 25 KVA transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3000 V, 50 HZ supply. Calculate: Secondary emf.
E2=140 V.
E2=220 V.
E2=240 V.
-An ideal 25 KVA transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3000 V, 50 HZ supply.Calculate: The maximum core flux
Max.Flux= 27 mWb
Max.Flux= 7 mWb
Max.Flux= 17 mWb
A single phase 2200/250 V, 50 HZ transformer has a net core area of 36 cm2 and maximum flux density of 6 Wb/m2. Calculate the number of turns of primary and secondary
Np= 458 turns. Nsec= 52 turns.
Np= 220 turns. Nsec= 25 turns.
Np= 159 turns. Nsec= 52 turns.
-A transformer takes a current of 0.6 A and absorbs 64 W when the primary is connected to its normal supply of 200 V, 50 HZ. The secondary is being open circuit. Find the magnetizing and ion loss currents.
Im= 0.507 A, Iw= 0.32 A
Im= 0.32 A, Iw= 0.507 A
Im= 0.506 A, Iw= 0.12 A
A 230/2300 V transformer takes no load current of 5 A at 0.25 power factor lagging. Find: 1) The core losses. 2) Magnetizing current.
Piron= 207.5 W, Im= 1.25 A
Piron= 260.5 W, Im= 4.84 A
Piron= 287.5 W, Im= 4.84 A
A 4000/400 V, 10 KVA transformer has primary and secondary winding resistances of 13 ohm and 0.15 ohm respectively. The leakage reactance referred to primary is 45 ohm. The magnetizing impedance referred to primary is 6 k ohm and the resistance corresponding to the core loss is 12 K ohm. Determine the input current. 1) When secondary terminals are open circuit. 2) When secondary load current is 25 A at power factor 0.8 lagging-Use simplified equivalent circuit.
I1 = 0.745 angle -26.6^o A 2) I1= 0.18 angle -42.89^o A
I1 = 0.745 angle -63.4^o A 2) I1= 3.18 angle -42.89^o A
I1 = 0.745 angle 63.4^o A 2) I1= 1.18 angle 42.89^o A
A 10 KVA, 2000/400 V single phase transformer has the following data: R1=5 ohm. X1=12 ohm. R2=0.2 ohm. X2=0.48 ohm. Determine the secondary terminal voltage at full load 0.8 power factor lagging when the primary supply is 2000 V.
V2= 350 V
V2= 22.4 V
) V2= 377.6 V
The primary and secondary windings of a 40 KVA, 6600/250 V single phase transformer have resistances of 10 ohm and 0.02 ohm respectively. The leakage reactance of the transformer referred to primary side is 35 ohm. Calculate the percentage voltage regulation of the transformer when supplying full load at a power factor 0.8
%V.R=3.8 %
%V.R=1.8 %
%V.R=0.8 %
-A single-phase transformer is rated at 10 KVA, 230/100 V. The iron loss of the transformer is 180 W. And its copper losses corresponding to full load is 240 W. calculate:1) Efficiency of the transformer at full load unity power factor. 2) The load corresponding to max. Efficiency unity power factor and calculate max. efficiency.
) Efficiency = 85.97 %, 2) m=0.87, Max. efficiency= 96.03 %
) Efficiency = 95.97 %, 2) m=1.15, Max. efficiency= 98.03 %.
Efficiency = 95.97 %, 2) m=0.87, Max. efficiency= 96.03 %
A 600 KVA, single-phase transformer when working at unity has an efficiency of 92% at full load and also at half full load. Determine the efficiency when it operates at unity power factor and 60% of full load
Efficiency= 90.3 %.
Efficiency= 96.3 %.
Efficiency= 86.3 %.
A 10 kVA, single-phase transformer has a turns ratio of 300/25. The primary is connected to a 1500 V, 50 Hz supply. Find the secondary voltage and the values of currents in the two windings on full load. Find also the maximum value of magnetic flux in the transformer core.
V2=115 V, I1=20/3 A, I2=80 A, Max.flux =0.0225 Wb
V2=110 V, I1=20/3 A, I2=80 A, Max.flux =0.0225 Wb
V2=125 V, I1=20/3 A, I2=80 A, Max.flux =0.0225 Wb
The no-load current of a transformer is 5 A at 0.3 power factor lagging when supplied at 230 V, 50 Hz. The number of turns on the primary winding is 200, calculate: a- The components of no-load current. b- The core iron loss.
Iw= 1.5 A, Im= 4.77 A, Piron= 345 W.
Iw= 0.5 A, Im= 4.77 A, Piron= 345 W.
Iw= 4.77 A, Im= 1.5 A, Piron= 345 W.
-A 2200/200 V, transformer takes 1 A at the high voltage side on no-load at a power factor of 0.25 lagging. The transformer supplies a load of 20 kW at a power factor of 0.8 power factor lag, calculate: A) The iron loss. B) Primary current and its power factor.
Piron = 50 W, I1 =12.2 angle-39.6^o�A, P.F= 0.766 lag
Piron = 550 W, I1 =12.2 angle -39.6^o�A, P.F= 0.766 lag
Piron = 550 W, I1 =8.2 angle-39.6^o�A, P.F= 0.766 lead
The equivalent circuit for a 200/400 V step up transformer has the following parameters referred to the low voltage side, Req1 = 0.15 ohm, Xeq1= 0.37 ohm, Ro= 600 ohm,Xo= 300 ohm. The transformer is supplying a load at 10A of 0.8 Power factor lagging, calculate: a-The primary current. b-The secondary terminal voltage.
I1= 20.67 A, V2= 396.3 V
I1= 10.67 A, V2= 386.3 V
I1= 20.67 A, V2= 386.3 V
A 5 kVA, 2300/230 V, 50 Hz transformer has an iron loss 40 W and a cupper loss of 112 W. Calculate the efficiency of the transformer at 0.8 power factor lag for the loading of 1.25 and 2.5 kVA.
Efficiency= 95.51 % (for 1.25 KVA, Efficiency=96.71% (for 2.5 KVA)
Efficiency= 85.51 % (for 1.25 KVA, Efficiency=96.71% (for 2.5 KVA)
Efficiency= 90.51 % (for 1.25 KVA, Efficiency=95.51% (for 2.5 KVA)
A 40 kVA, single-phase transformer has an iron loss of 400W and full load cupper loss 800W. Find the load at which maximum efficiency is achieved at 0.8 power factor lagging and calculate the value of maximum efficiency.
M=0.107, Corresponding output=28.28 KW, Corresponding efficiency= 97.25 %
m=1.414, Corresponding output=45.28 KW, Corresponding efficiency= 90.25 %
m=0.707, Corresponding output=28.28 KW, Corresponding efficiency= 97.25 %
A Y-connected, balanced three-phase load consisting of three impedances of 20 angle 30^o ohms each is supplied with the balanced line to-neutral voltages 220 angle 0^o : A. Calculate the phase currents in each line. B. Calculate the line-to-line phasor voltages. C. Calculate the total active and reactive power supplied to the load
Iph= 11 angle -30^o A, Vab = 381.05 angle 30^o V, Pt= 6287.35 W, Qt=3630 VAR
Iph= 8 angle -30^o A, Vab = 381.05 angle 30^o V, Pt= 6287.35 W, Qt=3630 VAR
Ph= 11 angle 0^o A, Vab = 281.05 angle 30^o V, Pt= 6287.35 W, Qt=3630 VAR
-A three phase 50 km transmission line a resistance of 0.24 ohm/km and an inductance 1.0186 mH/km. If the line is supplying 3500 kw , 33 kv , 0.82 lagging p.f load. Find sending end voltage.
20.476 angle 1.306^o KV
19.05 angle 1.306^o KV.
19.05 angle -34.915^o KV.
20.476 angle -34.915^o KV.
-A three phase 50 km transmission line a resistance of 0.24 ohm/km and an inductance 1.0186 mH/km. If the line is supplying 3500 kw , 33 kv , 0.82 lagging p.f load. Find sending end current.
74.67 angle 34.915^o Amp.
74.67 angle -34.915^o Amp.
20.47 angle 36.87^o Amp.
20.47 angle -36.87^o Amp.
1-A three phase 50 km transmission line a resistance of 0.24 ohm/km and an inductance 1.0186 mH/km. If the line is supplying 3500 kw , 33 kv , 0.82 lagging p.f load. Find sending end p.f.
0.807 leading.
0.807 lagging.
0.76 leading.
0.76 lagging.
-A three phase 50 km transmission line a resistance of 0.24 ohm/km and an inductance 1.0186 mH/km. If the line is supplying 3500 kw , 33 kv , 0.82 lagging p.f load. Find Efficiency.
92.56 %.
93.41 %.
94.58 %.
90.25 %.
A three phase 50 km transmission line a resistance of 0.24 ohm/km and an inductance 1.0186 mH/km. If the line is supplying 3500 kw , 33 kv , 0.82 lagging p.f load. Find Voltage regulation.
8.9 %.
7.47 %.
5.9 %.
10.33 %.
A three phase 50 km transmission line a resistance of 0.24 ohm/km and an inductance 1.0186 mH/km. If the line is supplying 3500 kw , 33 kv , 0.82 lagging p.f load. Find Power loss.
265.5 KW.
230.65 KW.
200.5 KW.
41.25 KW.
A three phase 160 km transmission line has the following parameters per phase per km : r=0.1557 ohm/km. l=1.32 mH/km. c=0.008756 micro F/km. If the line is supplying 50 MVA , 132 kv ,0.8 lagging p.f. Using pi-model. find: Sending end voltage and current.
) Vs= 88.54 angle 5.63^o KV ,Is= 197.45 angle -28.79^o Amp.
Vs= 76.21 angle 1.25^o KV ,Is= 218.7 angle -36.87^o Amp.
Vs= 88.54 angle 5.63^o KV ,Is= 218.7 angle -36.87^o Amp
Vs= 76.21 angle 1.25^o KV ,Is= 197.45 angle -28.79^o Amp
-A three phase 160 km transmission line has the following parameters per phase per km : r=0.1557 ohm/km. l=1.32 mH/km. c=0.008756 micro F/km. If the line is supplying 50 MVA , 132 kv ,0.8 lagging p.f. Using pi-model. find: Sending end p.f.
) 0.825 lagging.
0.825 leading.
) 0.787 laggig.
0.787 leading.
A three phase 160 km transmission line has the following parameters per phase per km : r=0.1557 ohm/km. l=1.32 mH/km. c=0.008756 micro F/km. If the line is supplying 50 MVA , 132 kv ,0.8 lagging p.f. Using pi-model. find: Efficiency of T.L.
89.8 %
95.32 %
86.52 %
92.45 %
A three phase 160 km transmission line has the following parameters per phase per km : r=0.1557 ohm/km. l=1.32 mH/km. c=0.008756 micro F/km. If the line is supplying 50 MVA , 132 kv ,0.8 lagging p.f. Using pi-model. find: Voltage regulation.
7.8 %.
9.63 %
15.8 %
17.9%
-A three phase 160 km transmission line has the following parameters per phase per km : r=0.1557 ohm/km. l=1.32 mH/km. c=0.008756 micro F/km. If the line is supplying 50 MVA , 132 kv ,0.8 lagging p.f. Using pi-model. find: Power loss.
5.86 MW.
2.05 MW.
3.27 MW.
4.44 MW.
Three phase T.L has the following parameters: A= 0.93 angle 1.5^o. B= 115 angle 77^o. Determine c and D
C= 0.93 angle 1.5^o. ,D= 7.52*10^-3 angle -74^o.
C= 7.52*10^-3 angle -74^o. ,D= 115 angle 77^o C) C= 7.52*10^-3 angle -74^o. ,D= 0.93 angle 1.5^
C= 7.52*10^-3 angle -74^o. ,D= 0.93 angle 1.5^o
C= 115 angle 77^o. ,D= 0.93 angle 1.5^o.
Three phase T.L has the following parameters: A= 0.93 angle 1.5^o. B= 115 angle 77^o. Determine the sending end voltage if the load at receiving end being 250MW, 275 KV, 0.8 lag power factor
Vs= 211.9 angle -36.87^o KV
Vs= 211.9 angle 14.34^o KV
Vs= 158.77 angle 14.34^o KV.
Vs= 158.77 angle -36.87^o KV.
-A three phase transmission line 200 Km long has the following constants r=0.15 ohm/Km xl=0.5 ohm/Km y=2*10^-6 mho/Km voltage at recieving end is 132 KV, the line is delivering 50 MVA at 0.85 p.f lagging. Using T-model .calculate sending end voltage
95.2 angle 10.25^o KV.
92.64 angle 9.56^o KV.
100.25 angle 10.25^o KV
76.2 angle 31.79^o KV.
A three phase transmission line 200 Km long has the following constants r=0.15 ohm/Km xl=0.5 ohm/Km y=2*10^-6 mho/Km voltage at recieving end is 132 KV, the line is delivering 50 MVA at 0.85 p.f lagging. Using T-model .calculate sending end current.
218.7 angle 31.79^o Amp.
218.7 angle -31.79^o Amp
200.1 angle 23.97^o Amp.
) 200.1 angle -23.97^o Amp.
-A three phase transmission line 200 Km long has the following constants r=0.15 ohm/Km xl=0.5 ohm/Km y=2*10^-6 mho/Km voltage at recieving end is 132 KV, the line is delivering 50 MVA at 0.85 p.f lagging. Using T-model .calculate line efficiency
91.6 %.
95.2 %
89.9 %
) 94.1 %
5-Which of the following rule is used to determine the direction of rotation of D.C motor?
Coloumb’s Law
Lenz’s Law
Fleming’s Right-hand Rule
Fleming’s Left-hand Rule
In which of the following application DC series motor is used?
Centrifugal Pump
Motor Operation in DC and AC
Water pump drive
Starter for car
Nowadays DC motor is widely used in
Electric Traction
Air compressor
Centrifugal Pump
Machine shop
Which of the following DC motor have the tendency of load instablity?
Cumulative compound motor
Shunt Motor
Series motor
Differentially compound motor
Which DC motor is preferred for Elevator?
Seperate excited motor
Series motor
Shunt Motor
Compound motor
Which DC motor is preferred for constant speed?
Series motor
Compound motor
Shunt motor
Differential motor
Which DC motor is preferred for crane and hoist?
Series motor
Cumulative compound motor
Shunt motor
Differentially compound motor
The speed of synchronous motor:
increases as the load increases
Decreases as the load decreases
Always remains constant
None of the above
In which of the following motors the stator and rotor magnetic fields rotate at the same speed:
Universal motor
Synchronous motor
Induction motor
Reluctance motor
An induction motor is analogous to
Auto-transformer
Two windings transformer with secondary short circuited
Two windings transformer with secondary short circuited C) Two windings transformer with secondary open circuited
Synchronous motor
Induction motors have the advantage of
Less Maintenace
) Less cost
Simple in construction
All of the above
The electric motor generally used in computer printer driver is
Reluctance motor
hysteresis motor
universal motor
Stepper motor
The power of the Motors israted in:
KW.
KVA
KVAr.
KWh.
A switched reluctance motor differs from a VR stepper motor in the sense that it
has rotor poles of ferromagnetic material
rotates continuously
Is designed for open-loop operation only
has lower efficiency.
Which of the following applications make use of a universal motor
Portable tools
Lathe machines
Oil expeller
Floor polishing machine
A universal motor is one which-A universal motor is one which
Is available universally
Can be marketed internationally
Can be operated either on dc or ac supply
Runs at dangerously high speed on no-load
Speed of the universal motor is
Dependent on frequency of supply
Proportional to frequency of supply
Independent of frequency of supply
None of the above
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