# power in an ac circuit - difference between in series and parallel

In AC-circuits, we have different kinds of power: active power $P$, reactive power $Q$ and apparent power $|S|$.

Let's say we have a circuit with a resistor $R$ and an inductor $L$. My understanding is that we then have $$P = \frac{U_\text{rms}^2}{R}, \,\,\,\,Q = \frac{U_\text{rms}^2}{\omega L}$$ independent of whether the elements are in series or parallel. Is this correct? If not, how do these powers look like when having the elements in series vs. when having them parallel?

If my current understanding is correct, I get the following problem. I think we have $$S = P + iQ = \frac{U_{\text{rms}}^2}{R} + i\frac{U_\text{rms}^2}{wL} = \frac{U_\text{rms}^2}{R} - \frac{U_\text{rms}^2}{iwL} = U_\text{rms}^2 \left(\frac{iwL+R}{iwLR}\right)$$ and $$S = \frac{U_\text{rms}^2}{Z}$$

So it seems that the impedance $Z$ would be also independent of whether we put the elements in series or parallel which is clearly false.

Where is my mistake?

• Your question may perhaps be better suited to the electrical engineering stack exchange, but I think that a problem might be that the definition of $U_{rms}$ isn't made clear here. The expressions for S looks correct for a parallel circuit. In that case both R and L have the same voltage across them, which is taken to be $U_{rms}$. For a series circuit, though, the resistor and inductor will not have the same voltage across them at every instant in time.
– user93237
Mar 9 '16 at 19:56
• Ah sure, I see it now. Thank you! I will try to derive the expressions for a series circuit and see if I understand everything then.
– Marc
Mar 9 '16 at 20:37