# Calculating true power of a predominantly inductive load?

I'm stuck at this problem, which doesn't seem to be so difficult, but, I guess there is some basic stuff that still confuses me. So, if we have a load represented with complex impedance $Z=30+j40 \ \Omega$ powered with a sinusoidal signal generator of voltage $U=200 \ V$, how can we find a true power on that load?

What I did, knowing that true power is: $P=U\cdot I\cdot \cos \phi$, where $\phi$ is phase difference between voltage and current through the load and equals $\phi=\cos\big(\arctan\frac{X}{R}\big)$ $\big(X$ - reactance, $R$- resistance $\big)$, since I don't know the current, I expressed it as: $I=\frac{U}{Z}$ and then calculated true power as: $$P=\frac{U^2}{Z}\cos\bigg(\arctan\frac{X}{R}\bigg)=742 \ W$$ but, I think I'm completely missing some parts regarding complex values of impedance and current.

• In your final equation, arctan should be cos(arctan. – Charles Cowie Jun 2 '16 at 21:15
• Then calculating that out correctly should give you the correct answer. You made a calculation error. – Charles Cowie Jun 2 '16 at 21:18
• Also, there is another way to do it by which you can solve it without a calculator. – Charles Cowie Jun 2 '16 at 21:21
• @CharlesCowie What confuses me is putting an equal sign between $I$ and $\frac{U}{Z}$ (their effective values). Can I really do that? – A6SE Jun 2 '16 at 21:22
• Yes you can do that. – Charles Cowie Jun 2 '16 at 21:23

In the text below I will use $\mathbf{Z}$, $\mathbf{U}$, $\mathbf{I}$, $\mathbf{S}$ as complex numbers and $Z$, $U$, $I$, $S$ as their magnitudes.

Note that $$Z\cos\left(\arctan\frac{X}{R}\right) = \operatorname{Re}(\mathbf{Z})$$ so you can simplify the expression $$\frac{U^2}{Z}\cos\left(\arctan\frac{X}{R}\right) = \frac{U^2}{ZZ}Z\cos\left(\arctan\frac{X}{R}\right) = \frac{U^2}{Z^2}\operatorname{Re}(\mathbf{Z})$$

In fact it's much easier to avoid $\cos(\varphi)$ at all from the beginning, since the magnitude of the current is just $$I = \frac{U}{Z}$$ and, since the current is in phase with voltage at the active resistance, the active (real) power is just $$P = I^2 R = \frac{U^2}{Z^2}\operatorname{Re}(\mathbf{Z})$$ which is the same formula as above.

Also, there is a concept of complex power, usually denoted by $\mathbf{S}$. The complex power is defined as $$\mathbf{S} \equiv \mathbf{U}\mathbf{I}^*$$ where both $U$ and $I$ are complex numbers and $\mathbf{I}^*$ means the complex conjugate. The equivalent formulas are $$\mathbf{S} = \frac{U^2}{\mathbf{Z}^*} = I^2 \mathbf{Z}$$ Real power $P = \operatorname{Re}(\mathbf{S})$

So the alternative way to solve your problem is $$P = \operatorname{Re}\left(\frac{U^2}{\mathbf{Z}^*}\right)$$

Solution without a calculator:

The impedance 30 + J40 can, by inspection, be represented by a 3-4-5 triangle, so Z = 50. I = U/Z = 200/50 = 4. Power = I squared R. 16 X 30 = 480 watts.