# Why can't I get correct reactive and real power in what seems to be easy task

As the title says Why can't I get correct real and reactive power

Source with voltage $$\U = 100\angle{30} \$$ is plugged on to impedance $$\ Z = 3 + 4j \$$. Determine real, reactive and apparent power.

what I did was:

I find apparent power: $$\ \frac{U^2}{Z} = 2000\angle{6.87 } W \$$

And to find real power i tried to multiply cos(6.87) with 2000 and I get $$\ 1985.65 \$$ but the answer supposed to be $$\ 1200 \$$. What have I did wrong isn't the real power equals cosφ * apparent power?

The angle of 30 degrees associated with your voltage supply is a complete red herring. Your voltage supply is 100 volts period because your load cares not one bit about the phase angle. All the load sees is 100 volts RMS.

So, without the safety net of a calculator, I can see the load impedance is 5 ohm and this takes 20 amps. That 20 amps flows through the resistive element of the load of 3 ohms and therefore dissipates 400 x 3 watts.

• Just one question, so do I in this kind of tasks ignore angles of voltage and just take in the consideration angles of current and impedance? Jul 28 '19 at 14:46
• You choose a reference phase angle and that is usually taken as zero degrees. Then align whatever voltage or current signals with that new reference to make it easier to solve. Given that there is only one voltage source, it’s an easy and simple task but, if the problem had multiple sources all at different phase angles then it’s a little arbitrary which one to choose but, after making that choice and resetting the angle for that signal, you must move the other sources by that same angle. Jul 28 '19 at 15:03

As stated previously ignore angle of voltage. Voltage is the reference and is 100 ∠0

Magnitude of Z, |Z| = 5

Angle of Z = arctan (4/3) = 53.1 degrees.

So 53.1 is the angle between the voltage and current which is equal to phi.

Apparent power = (V^2)/5 = (100^2)/5 = 2000VA

Real power = 2000 * cos 53.1 = 1200W

Reactive power = 2000 * sin 53.1 = 1600VAR

Now check results:-

Apparent power = Sqrt ((real power^2) + (reactive power^2))

Apparent power = Sqrt ((1200^2) + (1600^2))

Apparent power = 2000VA

All's good!