Electrical Engineering Interview Questions

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Answered by: NagaseshaReddy.M

kVA is the unit for apparent power. Apparent power consists of active and reactive power. Active power is the share of the apparent power which transmits energy from the source (generator) to the user. Reactive power is the share of the apparent power which represents a useless oscillation of energy from the source to the user and back again.

It occurs when on account of some »inertia« in the system there is a phase shift between voltage and current. This means that the current does not change polarity synchronous with the voltage. But the heat generated in a winding as well as the eddy current losses generated in a transformer core depend on the current only, regardless of whether it aligns with the voltage or not.

Therefore the heat is always proportional to the square of the current amplitude, irrespective of the phase angle (the shift between voltage and current). So a transformer has to be rated (and selected) by apparent power.

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Answered by: sinhasunilk

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Member Since Jul-2010 | Answered On : Jul 3rd, 2010

Transformer can only operate with A.C supply. Voltage will induce in the secondary winding of transformer only when there is a change in flux. The induced voltage in the secondary winding is no. of turns in secondary x rate of change of flux. In D.C circuit the rate opf change of flux is Zero, hence induced voltage in secondary winding is Zero and no Back MMF produce in the magnetic circuit of core of transformer. This will result in very high current drawal in the primary winding and will result in burning of primary winding of transformer. We can not think of applying D.C source in transformer.  

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Answered by: KaranSorout

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Member Since Jul-2006 | Answered On : Jul 13th, 2006

The power ratio between Star to Delta is 1:3

Explanation:

Let’s suppose V_s be the supply voltage per phase.

So the line voltage of the supply will be Ö3V_s.

Now assume any type of load; for simplicity I’m assuming it a only resistive load.

And let it be ‘R’ per phase.

For Delta connected load:

Calculation for per phase power; P_D= I²R

Where I à load current (per phase)

And,

I = Ö3V_s/R       {as line voltage of the supply is directly applied to the phase of the delta load}

So,

            P_d = (Ö3V_s/R )²R = 3V_s²/R  watts per phase.

For 3 phases:

            P_3D = 3P_d = 3*3V_s²/R = 9V_s²/R watts.

Now for Star connected load:

P_S = I²R = (V_s/R)²R = V_s²/R watts

For 3 phases: P_3S = 3P_S = 3 V_s²/R watts

Conclusion:

P_3S / P_{3D  =}  3V_s²/R / 9V_s²/R = 1/3

It may be better concluded by drowing figures. If any variations, can connect to me through karan.sorout@gmail.com  

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Answered by: joe

100 kw is the full load rating of the generator..... ..that means it can supply 100 kw without exceeding in specified temperature limits .that doesnt mean that it wont be able to supply a load more than 100 kw,it can but not for a longer period of time.after sometime the generator will burn out.if a load of 50 kw is connected to generator the generator will generate only 50 kw& not more than that.the generator will generate power only according to the requirement.

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Answered by: cvk

The series capacitor is conected to compensate for the line inductance and thus decrease the line reactance so that more power can be tranferred through the line (EV/X sin(delta)) and thus the system stability can be increased.

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Answered by: rahulr30

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Member Since May-2008 | Answered On : May 30th, 2008

Power factor is the ratio of Active ppwer to apparent power in sysytem,It is an index to see how much of the energy is actually being utilized to do useful work

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Answered by: abbasanwarali

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Member Since Oct-2008 | Answered On : Oct 16th, 2008

Actually there is no volt during normal condition. In case of any fault resistor will act as a low resistance path. because above mentioned components are high impedence components. so the huge fault (volt) will be grounded and the equipment remains safe.

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Answered by: csbalu

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Member Since Aug-2009 | Answered On : Aug 29th, 2009

When there is a sudden loss of load, The Generator terminal voltage increases due to the fact that the IX drop in the generator winding decreases. The EMF generated being the same the reduced reactance loss leads to a sudden rise in generator voltage..but the AVR (Automatic Voltage Regulator) will act instantly to bring this voltage to the set value by reducing the excitation.
There is practically no effect of this on the transformer barring the reduction in I2R (copper Losses) and reduced IX drop.