What is the Difference Between Active and Reactive Power in A.C. Circuits?

[zorro]

Homework Statement

A capacitor of rating 50 VAR, a heater of rating 1000 W and impedance rating 500 VA at a power factor of 0.8 lagging are connected in parallel to a 230V, 50 Hz power-supply system.Find

a) The system-active power
b) The system-reactive power
c) The system-apparent power
d) The system-power factor.

The Attempt at a Solution

I don't understand what VAR and VA stand for. What are active and reactive powers?

 

[gneill]

VA is volt amperes, a power unit. VA is used to designate apparent power. VAR is reactive power. Real power is given in watts.

Look up "power triangle".

 

[zorro]

ok I understood their meanings.
How do I find out each of them?

 

[gneill]

ok I understood their meanings.
How do I find out each of them?

Good question. I'm not sure what the designation "impedance rating 500 VA" means. Does your text give any enlightenment? If it were an apparent power for a given power factor I could understand it, but it's associated with a "1000W" heater. The real power should not be greater than the apparent power (which forms the hypotenuse for the power triangle).

 

[zorro]

I just know the answers
1400W, 250VAR, 1422VA, 0.98

 

[gneill]

Okay, it's not a "heater of rating 1000W and impedance rating 500VA". It's a heater of rating 1000W, and a separate impedance with an apparent power of 500VA when the power factor is 0.8; They are describing two separate components. That should clear things up nicely, right?

 

[zorro]

Yeah cleared up a little.
I want to ask one thing - the power which we calculate using P = VIcos(phi), where V and I are r.m.s. voltages, is it the apparent power?

 

[gneill]

Yeah cleared up a little.
I want to ask one thing - the power which we calculate using P = VIcos(phi), where V and I are r.m.s. voltages, is it the apparent power?

The apparent power (in VA) is on the hypotenuse of the power triangle. Real power (in watts) is the horizontal leg, and reactive power (in VAR) the vertical leg.

VA² = W² + VAR²

 

[zorro]

You did not get my question.
The average power in a complete cycle is V_r.m.s.I_r.m.s.cosΦ
Is it equal to the apparent power?

EDIT: Got it. That is true power.

Thanks for your help!

 

[gneill]

Ah. I get you now. The average (or mean) power is the real component of the power triangle, not the apparent power.