What is the purpose of Q-point in transistor?

MADHUSHREE SINHA

• Nov 9th, 2015

It is a DC operating point between saturation and cutoff. It must be selected in a way that it does not go into saturation or cutoff.

bala

• Oct 24th, 2015

It is more important for a transistor because according to this value, it may operate as an amplifier or a switch and most importantly the point should not vary with temperature hence we normally prefer voltage divider bias.

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A.Rajasekharreddy

• Sep 13th, 2016

Brush less alternator is nothing but stationary armature, Brush alternator is a rotating armature.

Shan

• Aug 24th, 2016

Is there a exciting stator in brushed generator? If not what is the reason?

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Gaurav

• Jan 13th, 2017

Capacitor causes a lag in the phase, resulting in a good starting torque.

manoj yadav

• Sep 1st, 2016

Capacitors provide phase difference between voltage and current and give starting torque

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What is the frequency of DC Voltage?

Gaj

• Oct 30th, 2017

Frequency is no. of cycles per second i.e. rate of change w.r.t time. DC has constant amplitude it means no change w.r.t. time. So it has Zero frequency.

Anup Kumar

• Sep 13th, 2017

Zero

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Answered by: NagaseshaReddy.M

• Sep 10th, 2007

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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.

Raghu

• Feb 9th, 2018

In transformer there is two losses
1. Iron loss
2. Copper loss
The iron loss is depands on Voltage
And the Copper loss is depands on current.

Umar Naik

• Nov 26th, 2017

There are cu losses and iron losses which are not dependent on power factor thats why we rated it in KVA.

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Answered by: joe

• Dec 23rd, 2006

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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.

Dharmender

• Feb 7th, 2018

Connected Load is 96kw &125kvaDG set will have run stand by unit for Hotel is suitable plz give your technical supporting answer

Mayilsamy Kasipandi

• Jul 11th, 2017

The Generator produce the power is exactly what we withdraw as load. 100 kw genset can produce 1 kw to 115kw power depending upon the connected/withdrawing load. Normally the generator can produce 115% of designed rating.

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KaranSorout  

 Profile Answers by KaranSorout Questions by KaranSorout

• Member Since Jul-2006 | Jul 13th, 2006

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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 ?3V_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 ? load current (per phase)

And,

I = ?3V_s/R       {as line voltage of the supply is directly applied to the phase of the delta load}

So,

            P_d = (?3V_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  

RUP KAMAL KUTUM

• Feb 3rd, 2018

Well first you assumed voltage per phase as Vs and Line voltage as sqrt(3) x Vs: Which is for star connection. Then, you assumed the same thing for Delta Connection(where line and phase voltages are t...

Anbumozhi R

• Dec 3rd, 2017

Power is Same. Whether it is star to Delta conversion or Delta to star conversion. Power is 1:1

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MOHAMMED MUZAMMIL

• Apr 1st, 2018

So net voltage in the primary winding will be (Vin-0) more due to which primary winding draws more current which leads to burning of winding. ... But when a particular High DC voltage is applied to th...

phani prasad

• Sep 13th, 2017

Windings get burned

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