Is there an RMS value for power delivered to an inductor?

Question

My textbook defines power delivered to an inductor as:

$$P= V_{L\rm\ peak}I_{\rm peak} \cos ( \omega t) \sin( \omega t)$$ where $\omega$ is angular frequency.

but makes no mention of $P_{RMS}$. It simply says that $P_{av}$ is zero (which makes sense since it's defined as the product of two circular functions).

However, when we covered power, current, and voltage delivered to a resistor in an AC circuit, we used RMS values for current and voltage, and an average value for power. This made sense since power delivered to a resistor is a function of a squared sinusoidal function, so average was adequate.

In this section (inductors in AC circuits), only instantaneous power was discussed. This seemed odd to me. In previous sections the book discussed how taking an average of a sinusoidal function just returns zero, which is why we use RMS values instead. That makes perfect sense, so why not apply that approach here? Do we not care about RMS power? if so, why not?

They did say that the average power is given by $I_{rms}r$ where $r$ is internal resistance, assuming internal resistance is substantial. I'm curious about cases where internal resistance is negligible.

Answer 1

$P_{\mathrm{RMS}}$ is a bit of a misnomer. "RMS" is short for "root-mean-square", and is applied in the reverse order in which it is read (i.e. square the quantity, average it, then take the square root). What you really have is average power, $P_{\mathrm{av}}$, which can be computed using the RMS voltage and RMS current if current and voltage are proportional and in phase with one another.

Working with complex impedance, you have $v=iz$, with $v$ the amplitude of the voltage (assumed real) and $i$ the complex amplitude of the current. The instantaneous power is given by $$P = \mathcal{R}\left\{i \mathrm{e}^{i\omega t} \right\} \mathcal{R}\left\{v \mathrm{e}^{i\omega t} \right\}.$$ You can show that the cycle averaged power is given by $$P_{\mathrm{av}} = \frac{1}{2} \mathcal{R}\left\{i v^* \right\},$$ and, if $\phi=\,$$\operatorname{Arg}$$(z) \equiv\,$$ \operatorname{arctan2}$$\left(\mathcal{I}\{z\},\, \mathcal{R}\{z\}\right) $ then \begin{align} P_{\mathrm{av}} &= \frac{1}{2} |i| |v| \cos\phi\\ &= i_{\mathrm{RMS}}v_{\mathrm{RMS}}\cos\phi, \end{align} where $i_{\mathrm{RMS}}$ root-mean-square current, which is the real amplitude of the current divided by $\sqrt{2}$ for sine waves.

Answer 2

With circuit components, we're typically interested in the amount of energy that's dissipated over time. If we have an average power $\langle P\rangle$, then the energy dissipated over time $t$ is simply $\langle P\rangle t$. RMS power is useless here: there's no way to go directly from RMS power to figure out how much energy is dissipated.

RMS voltage and current are useful because the average power dissipated in a resistive component is given by: $$ \langle P\rangle={V_{RMS}^2\over R}=I_{RMS}^2R$$

Inductive components have $\langle P\rangle=0$, and so the energy dissipated over any time interval is zero. This doesn't mean that the average is "insufficient," but instead that inductive components don't dissipate energy. The power put into a perfect inductor is stored in its magnetic field, and can be recovered without losses.