Complex power

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elec123

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Hi guys,

My lecturer sucks so could any brights sparks give us a hand with this problem....

A consumer has three identical loads, each with an impedance of 9+j12 ohms, connected in parallel across the 240V (rms) 50Hz mains.

Could you explain how I can find:

1. the complex power supplied by the mains.
2. the real (average) power supplied by the mains
3. the reactive power supplied by the mains
4. the magnitude of the current drawn from the mains
5. the power factor of the load and whether it is leading or lagging

Any help would be grateful coz all i've got in front of me is a load of formulae not having a clue what they all mean.....

Thanks

pop

New Member
First of all find total impedance of the circuit:

Z=[(1/(9+j12))+(1/(9+j12))+(1/(9+j12))]^-1

Once you got Z, then you can find current supplied by the voltage source (240 V).

I=240/Z;

1. Complex Power: S=VI'
where I' means conjugate of I

2. What you will get when you do S=VI' is S=P+jQ.

P is your real or average power

3. Q is your reactive power

4. Magnitude of current:

I=240/Z as before and you will get I=Re{I}+j*Im{I}
where Re means real part
Im means imaginary part.

To get magnitude do following

[((Re{I})^2)+(Im{I})^2)]^0.5

5. Power factor: To get power factor draw a triangle with:
P=horizontal part and Q=vertical part
Now, P is always positive but Q can be either negative or postive.

tan (theta)=Q/P

power factor=cos(theta)

is theta>0, then lagging, if theta<0 then it is leading

elec123

New Member
thanks a lot you've really cleared things up for me.......

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