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".
Inductive loads create lagging power triangles and capacitive loads create leading triangles.
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 \(Q = \sqrt{9600} = 97.98\)VARexample
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