# The Power Triangle in AC Circuits

concept
In the previous topic we covered real and reactive power in AC circuits. In this topic we'll cover power across a load that is both resistive and reactive in nature. For this we use a concept known as the power triangle.
fact
Apparent Power ($$S$$) is measured in Volt Amps (VA) and is given by the usual formula for power: $$S = VI$$ Where both $$V$$ and $$I$$ are rms values.
fact
We visualise this apparent power as a right-angled triangle, this shows it as consisting of the sum of a real power component ($$P$$) and a reactive power component ($$Q$$). The angle $$\theta$$ is given by: $$\theta = \theta_v - \theta_i$$. Where $$\theta_v$$ is the phase angle of the voltage and $$\theta_i$$ is the phase angle of the current.
fact
If $$0^\circ \lt \theta \lt 90^\circ$$ we call the circuit "lagging". If $$0^\circ \gt \theta \gt -90^\circ$$ we call the circuit "leading".
fact
Using the triangle shows us useful relationships between the three power values: $$S^2 = P^2 + Q^2$$ $$P = S\cos(\theta)$$
fact
The value $$\cos(\theta)$$ is called the "Power Factor" (pf). The higher the power factor, the better.
The power factor is a relationship between the apparent power and the real power of a circuit. High power factors mean that almost all the apparent power comes from real power, which is what we want.
Because $$\cos(\theta) = -\cos(\theta)$$ the power factor cannot tell you whether the load is lagging or leading.
example

Find the reactive power across a load with $$S = 100$$VA and $$P = 20$$W

Using our pythagorean rule we see that: $$Q^2 = S^2 - P^2 = 10000 - 400 = 9600$$ So $$Q = \sqrt{9600} = 97.98$$VAR
example

Find the power factor of the circuit with $$Q = 40$$VAR and $$P = 100$$W

First we find $$\theta = \tan^{-1}(\frac{Q}{P}) = \tan^{-1}(0.4)$$ $$\theta = 21.8^\circ$$ Now pf$$= \cos(\theta) = \cos(21.8^\circ)$$ pf$$= 0.93$$
practice problems