Part 12 - Power in AC Circuits

In this part we’ll define Power for AC circuits, we’ve had all the tools now for a while, but just never applied them.

Power in AC circuits can be divided into three categories:

Resistive load $(\theta = 0)$ .
Inductive load $(Z = \omega L \angle 90^{\circ})$ .
Capacitive load $(Z = \dfrac{1}{\omega C} \angle -90^{\circ})$ .

Resistive Load

When we have a resistive load, meaning no phase shift. Then our Voltage and Current are,

\begin{align*} V(t) & = V_m \cos(\omega t), \newline I(t) & = I_m \cos(\omega t). \end{align*}

Which, naturally means our power is,

P(t) = V_m I_m \cos^2(\omega t).

This is the so-called, instantaneous power.

Our average power is,

P_{\text{avg}} = \dfrac{V_m I_m}{2}.

Recall that the Root-mean-square is,

\begin{align*} V_{\text{rms}} & = \dfrac{V_m}{\sqrt{2}}, \newline I_{\text{rms}} & = \dfrac{I_m}{\sqrt{2}}. \end{align*}

This also means,

P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}}.

Inductive Load

In an inductive load, the voltage leads the current by $90^\circ$ ,

\begin{align*} V(t) & = V_m \cos(\omega t), \newline I(t) & = I_m \cos(\omega t - 90^\circ). \end{align*}

Or, we can write $I(t)$ as,

I(t) = I_m \sin(\omega t).

Which means, our instantaneous power is,

P(t) = V_m I_m \cos(\omega t) \sin(\omega t).

Using one of the trigonometry identities,

P(t) = \dfrac{V_m I_m}{2} \sin(2\omega t).

Our average power then? Well, the period of our power, $P$ , is half the period of $V$ or $I$ .

This means that when either $V(t)$ , or $I(t)$ are equal to 0, so is $P$ .

Therefore, the average power will also be zero!

P_{\text{avg}} = 0

Capacitive Load

In a capacitive load, the current now leads by $90^\circ$ ,

\begin{align*} V(t) & = V_m \cos(\omega t), \newline I(t) & = I_m \cos(\omega t + 90^\circ). \end{align*}

We can again rewrite $I(t)$ as,

I(t) = -I_m \sin(\omega t).

Which means our instantaneous power is,

P(t) = -V_m I_m \cos(\omega t) \sin(\omega t).

Or using the same trigonometry identity,

P(t) = -\dfrac{V_m I_m}{2} \sin(2\omega t).

Since we’ve only flipped our sign, the average power behaves the same,

P_{\text{avg}} = 0.

Example 1 (Average power)

Let’s try to understand the average power with an example:

If we have,

\begin{align*} Z_{L} & = 8 - j11, \newline I & = 5 \angle 20^\circ. \end{align*}

What’s the average power in $Z_{L}$ ?

We know that the reactive part is going to be zero,

P_{\text{avg}, Z_{\text{reactive}}} = 0.

The resistive part,

P_{\text{avg}, Z_{\text{resistive}}} = \dfrac{V_m I_m}{2}.

Which we can rewrite with Ohm’s law,

\begin{align*} P_{\text{avg}, Z_{\text{resistive}}} & = \dfrac{I_m^2 R}{2} \newline & = \dfrac{5^2 \cdot 8}{2} = \boxed{100 \ \mathrm{W}}. \end{align*}

Power in AC circuits - general

Let’s now write this as generally as we can,

\begin{align*} V(t) & = V_m \cos(\omega t + \theta_V), \newline I(t) & = I_m \cos(\omega t + \phi_I). \end{align*}

Which means our instantaneous power,

P(t) = V_m I_m \cos(\omega t + \theta_V) \cos(\omega t + \phi_I).

We can rewrite this as,

P(t) = \dfrac{1}{2} V_m I_m \cos(\theta_V - \phi_I) + \dfrac{1}{2} V_m I_m \cos(2 \omega t + \theta_V + \phi_I).

As we can see, this has one constant part and one periodic part.

This means our average power is,

P_{\text{avg}} = \dfrac{V_m I_m}{2} \cos(\theta_V - \phi_I).

Or using RMS,

P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos(\theta_V - \phi_I).

Different types of power in AC

Since we always deal with a complex part and a real part in AC, we can also define different types of power:

Definition 1 (Types of power)

Real Power: $P = V_{\text{rms}} I_{\text{rms}} \cos(\theta_V - \phi_I) \ [\mathrm{W}]$ $P = V_{rms} I_{rms} cos (θ_{V} - ϕ_{I}) [W]$
- Unit is in Watts, if this is purely resistive, meaning no phase shift, $P = V_{\text{rms}} I_{\text{rms}}$ .
Reactive Power: $Q = V_{\text{rms}} I_{\text{rms}} \sin(\theta_V - \phi_I) \ [\mathrm{VAR}]$ $Q = V_{rms} I_{rms} sin (θ_{V} - ϕ_{I}) [VAR]$
- Unit is in Volts Amperes Reactive, if purely resistive, $Q = 0$ .
Complex Power: $S = P + jQ$ $S = P + j Q$ or in polar form, $S = V_{\text{rms}} I_{\text{rms}} \angle (\theta_V - \phi_I) \ [\mathrm{VA}]$ $S = V_{rms} I_{rms} ∠ (θ_{V} - ϕ_{I}) [VA]$
- Unit is in Volt Ampere.
Apparent Power: $|S| = V_{\text{rms}} I_{\text{rms}} \ [\mathrm{VA}]$ $∣ S ∣ = V_{rms} I_{rms} [VA]$ .
- Unit is in Volt Ampere.

This also means that,

P^2 + Q^2 = (V_{\text{rms}} I_{\text{rms}})^2.

Power Factor

The last thing we’ll talk about is the so-called power factor.

Definition 2 (Power factor)

We define the power factor as,

\text{PF} = \cos(\theta_V - \phi_I) \leq 1.

We call the angle the power angle,

\text{Power Angle} = \theta_V - \phi_I.

Which means we can define the power factor as,

\text{PF} = \dfrac{P}{|S|}.

The power factor and the power angle say a lot about the type of circuit we have.

An inductive load will always have a positive power angle, on the other hand, a capacitive load will have a negative power angle!

Power Relationships

Now, with all these tools, let’s write down some relationships. There’s a lot.

P = V_{\text{rms}} I_{\text{rms}} = I_{\text{rms}}^2 R = \dfrac{V_{\text{rms}}^2}{R}

Note, here $X$ is equal to the reactance,

Q = I_{\text{rms}}^2 X = \dfrac{V_{\text{rms}}^2}{X}

\begin{align*} S & = P + j Q \newline & = V_{\text{rms}} I_{\text{rms}} \angle (\theta_V - \phi_I) \newline & = \dfrac{V_m I_m}{2} \angle (\theta_V - \phi_I). \end{align*}