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.
Thank you for your time.

  • 1
    \$\begingroup\$ In your final equation, arctan should be cos(arctan. \$\endgroup\$
    – user80875
    Jun 2, 2016 at 21:15
  • \$\begingroup\$ Then calculating that out correctly should give you the correct answer. You made a calculation error. \$\endgroup\$
    – user80875
    Jun 2, 2016 at 21:18
  • \$\begingroup\$ Also, there is another way to do it by which you can solve it without a calculator. \$\endgroup\$
    – user80875
    Jun 2, 2016 at 21:21
  • \$\begingroup\$ @CharlesCowie What confuses me is putting an equal sign between \$I\$ and \$\frac{U}{Z}\$ (their effective values). Can I really do that? \$\endgroup\$
    – A6SE
    Jun 2, 2016 at 21:22
  • \$\begingroup\$ Yes you can do that. \$\endgroup\$
    – user80875
    Jun 2, 2016 at 21:23

2 Answers 2


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.