# Power in AC Circuits

concept

Our usual definition of power \(V\cdot I\) from DC circuits still holds in AC.
However because we deal with trigonometric functions and we have energy stored and released through capacitors and inductors we develop a few tools to make our power analysis simpler.
There are a few concepts in AC power that might seem more difficult than they should be, but they really do make analysing circuits a whole lot easier.

It's convenient to use a value known as the "root mean square" (rms) of an AC signal. This makes power calculations more straightforward. It's also how most AC signal magnitudes are given in the real world. For instance your house outlet is likely given by either 110V or 240V, both of these are rms values, not the peak values of the sine wave.

fact

The Root Mean Square (rms) value of an AC signal \(A(t)\) is given by:
\(A_{rms} = \sqrt{\frac{1}{T_2-T_1}\int_{T_1}^{T_2}[A(t)]^2 dt}\)

fact

For a sine wave \(V_{rms} = \frac{V_m}{\sqrt{2}}\)
Where \(V_m\) is the magnitude of the sine wave.

fact

Whenever you see \(V\) or \(I\) without the \(_m\) subscripts it's referring to rms values.

fact

*Instantaneous Power*is the term given to the usual power definition, it is a function of time: \(p(t) = v(t)i(t)\) W(watts) Where \(p(t)\) is the instantaneous power. \(v(t)\) is the voltage. \(i(t)\) is the current.

example

Find the instantaneous power at \(t = 1\) when \(v = 2\sin(t)\) and \(i = \sin(t+45^\circ)\)

We'll just plug in \(t=1\) into both \(v(t)\) and \(i(t)\) then multiply the results to get \(p\). \(v(1) = 2\sin(1)\) \(v(1) = 1.68\)V \(i(1) = \sin(1 + 45^\circ) = \sin(1 + \frac{\pi}{4})\) \(i(1) = 0.98\)A \(p(1) = v(1)i(1) = 1.65\)W Remember that the inside of the sine is in radians. Forgetting to keep track of what's in radians and what's in degrees, and what setting your calculator is in, is one of the most common source of problems.fact

In AC circuits the power delivered can be negative at some points, this represents energy being delivered back into the source by capacitors and/or inductors.

fact

*Active Power*(\(P\)) is a measure of how much useful energy is delivered to a load and is measured in Watts. The active power (P) (also termed

*real power*) is the average of the power waveform. If \(P \gt 0\) then power flows into the load. If \(P \lt 0\) then power flows from the load and into the source.

fact

For a purely resistive load \(v(t)\) and \(i(t)\) are completely in phase:
\(P = \frac{V_mI_m}{2}\)
Where \(V_m\) is the magnitude of the voltage waveform and \(I_m\) is the magnitude of the current waveform.

derivation

Let \(v(t) = V_m\sin(\omega t)\), \(i(t) = I_m\sin(\omega t)\).
Now \(p(t) = v(t)\cdot i(t) = V_mI_m\frac{1}{2}(1-\cos(2\omega t))\)
Now the real power is the average value of \(p(t)\).
Since \(\cos(2\omega t)\) has an average of \(0\):
\(P = V_mI_m\frac{1}{2}\)

fact

Using rms values for a resistive load:
\(P = VI = I^2R = \frac{V^2}{R}\)

fact

*Reactive Power*(\(Q\)) is the power that travels into and then back out of the load and is measured in VAR (volt amp reactive).

fact

For a purely reactive load (no resistance):
\(Q = VI = I^2X = \frac{V^2}{X}\)
Where \(X\) is the reactance of the load.

fact

In AC circuits ideal capacitors and inductors never dissipate any real power (\(P = 0\)) because their voltage is always \(90^\circ\) out of phase with the current.

\(Q\) is positive for inductive loads and negative for capacitive loads.
Because reactive power flows back and forth its average value is 0 and it therefore contributes nothing to the average power to the load.

fact

For a load with both resistive and reactive components:
\(P + jQ = \frac{V^2}{Z} = I^2Z\)
Where \(Z\) is the impedance of the load.

example

A load has a voltage across it of 120V rms and an impedance of \(3 + 4j\). Find the real power across it.

\(Z = 3 + 4j = 5\angle 53.13^\circ\) \(\begin{align} P + jQ = \frac{V^2}{Z} & = 2880\angle -53.13^\circ \\ & = 1728 - j2304 \\ \end{align}\) So we can see that \(P = 1728\)W and \(Q = -2304\)VAR Therefore the real power across the load is \(1728\)W
practice problems