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

Question

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?

Answer 1

Your mistake is your premise that the real and reactive powers are independent of whether the elements are in series or parallel. That is false. The expressions you start with are for the two elements in parallel with a voltage source, and your expression for the impedance that you arrive at is also for the two elements in parallel.

There are some simple limiting cases where it's easy to see that the powers for parallel and series elements must me different. Consider large inductances at high frequencies. In this situaion, the inductor looks like an open circuit. Put the inductor in parallel, and the apparent power will look like that of a plain resistor. Put the inductor in series however, and the power approaches 0.