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 and 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:

V(t) = V_m cos(\omega t) \newline I(t) = I_m cos(\omega t)

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_{avg} = \dfrac{V_m I_m}{2}

Recall that the Root-mean-square is:

V_{rms} = \dfrac{V_m}{\sqrt{2}}

I_{rms} = \dfrac{I_m}{\sqrt{2}}

This also means:

P_{avg} = V_{rms} I_{rms}

Inductive Load

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

V(t) = V_m cos(\omega t) \newline I(t) = I_m cos(\omega t - 90^\circ)

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_{avg} = 0

Capacitive Load

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

V(t) = V_m cos(\omega t) \newline I(t) = I_m cos(\omega t + 90^\circ)

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 signed, the average power behaves the same

P_{avg} = 0

Example 1 (Average power)

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

If we have:

Z_{L} = 8 - j11 \newline I = 5 \angle 20^\circ

What’s average power in $Z_{L}$ ?

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

P_{avg:\ Z_{reactive}} = 0

The resistive part:

P_{avg:\ Z_{resistive}} = \dfrac{V_m I_m}{2}

Which we can rewrite with Ohm’s law:

P_{avg:\ Z_{resistive}} = \dfrac{I_m^2 R}{2}

P_{avg:\ Z_{resistive}} = \dfrac{(5^2) \cdot\ 8}{2} = \boxed{100 W}

Power in AC circuits - general

Let’s now write this as general we can:

V(t) = V_m cos(\omega t + \theta_V) \newline I(t) = I_m cos(\omega t + \phi_I)

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_{avg} = \dfrac{V_m I_m}{2} cos(\theta_V - \phi_I)

Or using RMS:

P_{avg} = V_{rms} I_{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:

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

This also means that:

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

Power Factor

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

We define the power factor as:

PF = cos(\theta_V - \phi_I) \leq 1

We call the angle, for the power angle:

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

Which means we can define the power factor as:

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_{rms} I_{rms} = I_{rms}^2 R = \dfrac{V_{rms}^2}{R}

Note, here $X$ is equal to the reactance

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

S = P + j Q

S = V_{rms} I_{rms} \angle \theta_V - \phi_I

S = \dfrac{V_m I_m}{2} \angle \theta_V - \phi_I