Wattless Current

Setting up the extreme case. Section 7.13 showed that the average power in a general AC circuit is $P_{avg}=V_{rms}I_{rms}\cos\phi$. Consider the special case of a purely inductive (or purely capacitive) circuit, where $\phi=90^\circ$ exactly. Since $\cos90^\circ=0$, the formula gives

exactly zero average power -- even though $V_{rms}$ and $I_{rms}$ are both genuinely nonzero, and a real, measurable current is flowing through the circuit at every instant. This current, which flows but delivers no NET power to the circuit on average, is called a wattless current (sometimes the idle current or the reactive current).

Why the power is zero despite a real current flowing. Looking at the instantaneous power $p=vi$ for this case makes the reason clear: with the current $90^\circ$ out of phase with the voltage, $p$ is positive for exactly one quarter of every cycle (energy genuinely flowing INTO the inductor's magnetic field, or the capacitor's electric field, from the source) and NEGATIVE for the following quarter (that same stored energy flowing back OUT, returning to the source) -- and this in-and-out exchange repeats, with the positive and negative halves of every cycle exactly cancelling. Averaged over a whole cycle, exactly as much energy is returned to the source as was taken from it, so the net energy transfer -- and hence the average power -- is zero, even though the instantaneous power is very much nonzero at almost every instant along the way.

Which part of the current is 'wattless'. In a general LCR circuit (not purely $L$ or $C$), the current phasor $I$ can be resolved into two perpendicular components relative to the voltage phasor: one component, $I\cos\phi$, IN PHASE with the voltage, which is the part that genuinely delivers power ($P_{avg}=V_{rms}(I_{rms}\cos\phi)$, exactly recovering Section 7.13's formula); and a second component, $I\sin\phi$, at $90^\circ$ to the voltage, which is the wattless (reactive) component -- it contributes nothing to the average power delivered, however large it is, purely because of this $90^\circ$ phase relationship. …