Description : The per unit impedance ofa circuit element is 0.15. If the base kV and base MVA are halved, then the new value of the per unit impedance of the circuit element will be?
Last Answer : The per unit impedance ofa circuit element is 0.15. If the base kV and base MVA are halved, then the new value of the per unit impedance of the circuit element will be 0.30.
Description : A transmission line has 1 P.0 impedance on a base of 11 KV, 100 MVA. On a base of 55 KV. it will have a P.0 impedance of
Last Answer : A transmission line has 1 P.0 impedance on a base of 11 KV, 100 MVA. On a base of 55 KV. it will have a P.0 impedance of 0.02 P.U
Description : A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j0.1, j0.1 and j0.04 respectively. The neutral of ... the fault is (A) 513.8 V (B) 889.9 V (C) 1112.0 V (D) 642.2 V
Last Answer : A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j0.1, ... line. The voltage of the alternator neutral with respect to ground during the fault is 642.2 V
Description : A 3-phase circuit breaker is rated at 1250 A, 2000 MVA, 33 kV, 4s. Its making current capacity will be (a) 35 kA (b) 89 kA (c) 79 kA (d) 69 kA
Last Answer : 89
Description : A three-phase, 33 kV oil circuit breaker is rated 1200 A, 2000 MVA, 3 s. The symmetrical breaking current is (A) 1200 A (B) 3600 A (C) 35 kA (D) 104.8 kA
Last Answer : A three-phase, 33 kV oil circuit breaker is rated 1200 A, 2000 MVA, 3 s. The symmetrical breaking current is 35 kA
Description : A 3-phase star-connected 75 MVA, 25 kV 3-phase synchronous generator has a synchronous reactance of 1.0 p.u. The per unit value to a 90 MVA base and 30 kV is (A) 2/15 p.u. (B) 5/6 p.u. (C) 1.0 p.u. (D) 1.2 p.u.
Last Answer : A 3-phase star-connected 75 MVA, 25 kV 3-phase synchronous generator has a synchronous reactance of 1.0 p.u. The per unit value to a 90 MVA base and 30 kV is 5/6 p.u.
Description : A 60 Hz, 4 pole turbo generator rated 100 MVA, 13.8 KV has an inertia constant of 10 MJ/MVA. Find the stored energy in the rotor at synchronous speed. (A) 10 MJ (B) 100 J (C) 1000 J (D) 1000 MJ
Last Answer : c
Description : If we give 2334 A, 540V on Primary side of 1.125 MVA step up transformer, then what will be the Secondary Current, If Secondary Voltage=11 KV?
Last Answer : A. As we know the Voltage & current relation for transformer-V1/V2 = I2/I1 We Know, VI= 540 V; V2=11KV or 11000 V; I1= 2334 Amps. By putting these value on Relation540/11000= I2/2334 So,I2 = 114.5 Amps
Last Answer : As we know the Voltage & current relation for transformer-V1/V2 = I2/I1 We Know, VI= 540 V; V2=11KV or 11000 V; I1= 2334 Amps. By putting these value on Relation540/11000= I2/2334 So,I2 = 114.5 Amps
Description : Which of the following statements is incorrect ? (a) Lightning arrestors are used before the switchgear (b) Shunt reactors are used as compensation reactors (c) The peak short current is (1.8 xV2) ... component (d) The MVA at fault is equal to base MVA divided by per unit equivalent fault reactance
Last Answer : (a) Lightning arrestors are used before the switchgear
Description : A 50 Hz, 4-pole, 500 MVA, 22 kV turbo-generator is delivering rated megavolt amperes at 0.8 power factor. Suddenly a fault occurs reducing in electric power output by 40%. Neglect losses and assume constant power input to the ... the time of fault will be (A) 1.528 (B) 1.018 (C) 0.848 (D) 0.509
Last Answer : A 50 Hz, 4-pole, 500 MVA, 22 kV turbo-generator is delivering rated megavolt amperes at 0.8 power factor. Suddenly a fault occurs reducing in electric power output by 40%. Neglect losses and assume ... the shaft. The accelerating torque in the generator in MNm at the time of fault will be 1.018
Description : A 500 MVA, 11 KV synchronous generator has 0.2 p.u. synchronous reactance. The p.u. synchronous reactance on the base values of 100 MVA and 22 KV is A) 0.16 B) 0.01 C) 4.0 D) 0.25
Last Answer : A 500 MVA, 11 KV synchronous generator has 0.2 p.u. synchronous reactance. The p.u. synchronous reactance on the base values of 100 MVA and 22 KV is 0.01
Description : The inertia constant of a 100 MVA, 11 kV water wheel generator is 4. The energy stored in the rotor at the synchronous speed is: A) 400 Mega Joule B) 400 Kilo Joule C) 25 Mega Joule D) 25 Kilo Joule
Last Answer : The inertia constant of a 100 MVA, 11 kV water wheel generator is 4. The energy stored in the rotor at the synchronous speed is: 400 Mega Joule
Description : A generator is connected through a 20 MVA, 13.8/138 kV step down transformer, to a transmission line. At the receiving end of the line a load is supplied through a step down transformer of 10 MVA, 138/69 kV rating. A 0.72 ... load (in per unit) in generator will be (A) 36 (B) 1.44 (C) 0.72 (D) 0.18
Last Answer : A generator is connected through a 20 MVA, 13.8/138 kV step down transformer, to a transmission line. At the receiving end of the line a load is supplied through a step down transformer of 10 MVA, 138/ ... MVA and 69 kV in load circuit, the value of the load (in per unit) in generator will be 36
Description : A 100 MVA, 11 KV, 3-phase, 50 Hz, 8—pole synchronous generator has an inertia constant H equal to 4 seconds. The stored energy in the rotor of the generator at synchronous speed will be A) 100 MJ B) 400 MJ C) 800 MJ D) 12.5 MJ
Last Answer : 800
Description : A 50 Hz, 4 pole turboalternator rated at 20 MVA. 13.2 KV has an inertia constant H = 4 KW sec/KVA. The K.E. stored in the rotor at synchronous speed is
Last Answer : A 50 Hz, 4 pole turboalternator rated at 20 MVA. 13.2 KV has an inertia constant H = 4 KW sec/KVA. The K.E. stored in the rotor at synchronous speed is 80 MJ
Description : A single phase transformer rated at 3000 kVA, 69 kV 4.16 kV, 60 Hz has a total internal impedance Zp of 127 ohm, referred to the primary side. Calculate the primary current if the secondary is accidentally short circuited. (A) 43.5 A (B) 543 A (C) 9006 A (D) 721 A
Last Answer : 543A
Description : For a given base voltage and base volt-amperes, the per unit impedance value is x. For the doubled base values of both voltage and volt-amperes, the per unit impedance will be (a) 2x (b) 0.25x (c) 0.5x (d) no change
Last Answer : Answer is 2x Base impedance =(base voltage) ²/base current
Description : If the base of right rectangular prism remains constant and the measures of the lateral edges are halved, then its volume will be reduced by : -Maths 9th
Last Answer : answer:
Description : The p.u. parameter for a 500 MVA machine on its own base are: inertia, M = 20 p.u. ; reactance, X = 2 p.u. The p.u. values of inertia and reactance on 100 MVA common base, respectively, are (A) 4, 0.4 (B) 100, 10 (C) 4, 10 (D) 100, 0.4
Last Answer : The p.u. parameter for a 500 MVA machine on its own base are: inertia, M = 20 p.u. ; reactance, X = 2 p.u. The p.u. values of inertia and reactance on 100 MVA common base, respectively, are 100, 0.4
Description : The base of a right triangular prism is an equilateral triangle. If the height is halved and each side of the base is doubled, find the ratio of the -Maths 9th
Last Answer : 1 : 2 Let each side of the base of the original prism be a units and the height of the prism be h units. Then Required ratio = Vol. of original prismVol. of new prismVol. of original ... )2×h3√4×(2a)2×h234×(a)2×h34×(2a)2×h2 = 2a2h4a2h2a2h4a2h = 1 : 2.
Description : For which of the following ratings of the transformer differential protection is recommended ? (a) above 30 kVA. (b) equal to and above 5 MVA (c) equal to and above 25 MVA (d) none of the above
Last Answer : (b) equal to and above 5 MVA
Description : The forced oil and air circulation method is usual one for transformers of capacities (a) upto 5 MVA (b) upto 10 MVA (c) upto 20 MVA (d) 30 MVA upwards
Last Answer : (d) 30 MVA upwards
Description : As per chart on energy consumption the supply unit of furnace oil is ……… a. kWh c. kV b. kg. d. Watts
Last Answer : kg
Description : As per pie chart on energy consumption the supply unit of the electricity is……….. a. kWh c. kV b. kg. d. Watts
Last Answer : kwh
Description : 99. As per chart on energy consumption the supply unit of furnace oil is ……… a. kWh c. kV b. kg. d. Watts
Last Answer : kg.
Description : 95. As per pie chart on energy consumption the supply unit of the electricity is……….. a. kWh c. kV b. kg. d. Watts
Description : In a simple DC circuit, the resistance is held constant while the applied voltage is halved. Current flow, therefore, will ______________. A. double B. remain the same C. be divided by two D. be divided by four
Last Answer : Answer: C
Description : If the inertia constant H of a machine of 200 MVA is 2 p.u., its value corresponding to 400 MVA will be (a) 4 p.u. (b) 2 p.u. (c) 1.0 p.u. (d) 0.5 p.u.
Last Answer : If the inertia constant H of a machine of 200 MVA is 2 p.u., its value corresponding to 400 MVA will be 1.0 p.u.
Description : Which of the following precautions should be taken when troubleshooting various power circuits using a common solenoid-type mechanical voltage tester (wiggins)? A. Never use this tester on circuits greater ... voltage expected in the circuit in order to prevent damage from an off-scale reading.
Last Answer : Answer: A
Description : If a cable of homogeneous insulation has a maximum stress of 10 kV/mm, then the dielectric strength of insulation should be (a) 5 kV/mm (b) 10 kV/mm (a) 15 kV/mm (d) 30 kV/mm
Last Answer : (b) 10 kV/mm
Description : Compare BJT common base configuration with common collector configuration on the basis of (i) Current gain (ii) Voltage gain (iii) Input impedance (iv) Output impedance
Last Answer : Parameter Common Base Common Collector Current gain Low (About 1) High (1+β) Voltage gain High 1 Input impedance Low High Output impedance High Low
Description : Two transformers of identical voltages but different capacities are operated in parallel, for satisfactory load sharing: a) Impedances should be equal b) Per unit impedance must be equal c) per unit impedance and X/R ratio must be equal d) Impedance and X/R ratio must be equal
Last Answer : Ans: C
Description : Two transformers operating in parallel will share the load depending upon their (A) Rating. (B) Leakage reactance. (C) Efficiency. (D) Per-unit impedance.
Last Answer : Ans: A Transformers having higher kVA rating will share more load.
Description : Air blast circuit breakers for 400 kV power system are designed to operate in (a) 100 microsecond (b) 50 millisecond (c) 0.5 sec (d) 0.1 sec
Last Answer : (b) 50 millisecond
Description : A centrifugal pump has the following specifications: Power = 4 H.P.; Speed = 800 rpm Head = 8 metres Flow = 1000 litres/minutes. If its speed is halved, then the new head will be __________ metres. (A) 2 (B) 4 (C) 8 (D) 5.5
Last Answer : (A) 2
Description : Draw a circuit diagram of R.C. series circuit. Draw impedance triangle and power triangle for same circuit.
Last Answer : Draw a circuit diagram of R.C. series circuit. Draw impedance triangle and power triangle for same circuit.
Description : Draw impedance triangle and voltage triangle for RL series circuit
Last Answer : impedance triangle and voltage triangle for RL series circuit
Description : A negative feedback loop returning a voltage to the input raised the input impedance and making the circuit a better
Last Answer : A negative feedback loop returning a voltage to the input raised the input impedance and making the circuit a better voltage sensor
Last Answer : In an R-C-L series circuit, during resonance, the impedance will be minimum.
Description : In a series resonant circuit, the impedance of the circuit is?
Last Answer : In a series resonant circuit, the impedance of the circuit is minimum.
Description : I start off red, then I am halved, and turn blue. I am then halved once again, and turn pink. What am I ? -Riddles
Last Answer : Australian notes (money) $20 note is red $10 note is blue And a $5 note is pink!
Description : If in a cylinder, radius is doubled and height is halved, then find its curved surface area. -Maths 9th
Last Answer : Let r and h be radius and height of the cyclinder, then C.S.A. = 2πrh Now, radius is doubled and height is halved. ∴ New radius = 2r and new height = h / 2 New C.S.A. = 2π × 2r × h / 2 = 2πrh .
Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th
Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3
Description : In a cylinder, radius is doubled and height is halved, then curved surface area will be -Maths 9th
Last Answer : The curved surface area will remain same. So, there is no change in the curved surface area of cylinder . Hence the curved surface area will remain same.
Description : If the distance between two charges is halved, then the force between them becomes (a) Half (b) Double (c) Four times (d) One-fourth
Last Answer : Ans:(c)
Description : If the length of a simple pendulum is halved then its period of oscillation is - (1) doubled (2) halved (3) increased by a factor √ 2 (4) decreased by a factor √ 2
Last Answer : (4) decreased by a factor √ 2