- A 800 kVA, single phase transformer has a voltage ratio of 6600/5000 volt. If Emf per turn = 8 Volt, calculate number of turns on high voltage and low voltage side.
- A 3φ 2000 kVA transformer has a voltage ratio of 6600/415 volt, and has pu value of resistance and leakage reactance are 0.02 and 0.1 respectively. Calculate full load regulation if power factor is 0.8 lagging. Given Vr = 2, Vx = 10.
% regulation = Vr cosφ + Vx sinφ = 2 X 0.8 + 10 X 0.6 = 7.6 %
- A balanced 3 φ load of 150 kW at 1000 volt and p.f 0.866, is supplied from a 2000 V source having 3 phase mains connected in delta. Calculate the primary phase current.
√3.VL.IL.cosφ = 150000 or IL = 100 Amp
Secondary phase current = IL/√3 = 57.7 Amp
Turns ratio = 1000/2000 = 1/2
Primary phase current = 57.76/2 = 28.85 Amp
i.e. option B
- 2 transformers are connected in Scott connection operating from 440 volt 3 φ supplying 2 1-φ transformers on 200 volt each. If total output is 150 kVA. Then secondary to primary turns ratio of the teaser transformer is:
- The basic function of a transformer is to change
- Tappings of a step down transformer are provided at
In H.V side winding current loading is comparatively smaller than L.V side winding and high voltage winding placed at the outer level.
- In a transformer, zero voltage regulation at full load is
- In case of distribution transformer which connection is applied-
For distribution system, a neutral wire is very much required. In star connection only neutralwire is provided where line voltage is √3X(phase voltage), line current is phase current. So, in case of distribution transformer delta - star connection is applied.
- A Buchholz relay is used for
Buchholz relay protects the transformer from all internal faults.
- For large capacity transformer which core is used?
Stepped core is used in large capacity transformers to have optimum advantage of space, cost and material.
- Transformers cause transfer of energy from one circuit to another without change in
- Transformers with rating 100 MVA which cooling is preferred?
Natural radiation is preferred when transformer rating is 5KVA.
Oil filled and self-cooled is preferred when transformer rating is 5KVA to 100 MVA.
Forced cooling with air blast is preferred when transformer rating is above 100 MVA.
- The oil used in the transformers serves as a
The oil used in the transformer serves in two ways.
1. It acts as a coolant.
2. It acts as an insulator.
It must be a good insulator with dielectric strength of 200KV ⁄ cm. It should not absorb moisture and dust. It must be free from acids, alcoholics and sulfur. The oil is generally used for transformer are
1. Liquid silicon.
4. Mineral oils.
- Leakage flux in a transformer depends upon
The flux set up by in the primary & seconary which is set up tn the core linking with its own turns & not linking with the other is known as leakage flux.The effect of leakage flux is to develop into their respective windings emfs of self-inductance which are proportional to load current ,and, are therefore,equivalent in effect to the addition of an inductive coil in series with each winding.So, leakage flux in a transformer depends upon load current.
- A 40KVA transformer has a core loss of 400W & full load copper loss of 80W. At Maximum efficiency percent of full load is
Pi = iron loss(core loss) = 400 Watt, Pc = 800 Watt(copper loss), maximum efficiency occurs at fraction x of full load such that x = √(Pi ⁄ Pc) = √(400/800)=0.707X100 = 70.7 % .
- AT(ampere turn) in magnetic circuit is analogous to which quantity in electric circuit
The unit of magneto-motive force (MMF) is the ampere-turn (AT). MMF in magnetic circuit is analogous to EMF (electromotive force) in electrical circuit. Similar to the way that electromotive force (EMF) drives a current of electrical charge in electrical circuits, magneto-motive force (MMF) drives magnetic flux through magnetic circuits. The unit of electromotive force (EMF) is voltage.
- A 2400V/240 V ,150KVA single phase transformer has a coreloss of 1.8KW at rated voltage .Its equivalent resistance is 1.1 percent .Then the transformer efficiency at 0.8 power factor &fullload is-
We know, copper loss = resistance per unit X rated KVA
So here, copper loss = 0.011X150 = 1.65 kW
- A single phase transformer when supplied from 220 V, 50 Hz, eddy current loss is 50 W. If transformer is supplied voltage 430 V,50 Hz then eddy current loss will be -
Pe(eddy current loss)=Kef^2Bm^2(f->frequency,Bm->magnetic flux density) and V=4.44 fBmAn, so Pe1/Pe2=(f1*Bm1/f2Bm2)=(v1/v2)^2 or,Pe2=Pe1(v2/v1)^2=50*(430/220)^2=191.01 Watts.