 # Amateur Extra: Power Factor in an RL Circuit

Ah, another fine math problem from the Amateur Extra exam (don’t you just love these?) [E5D16]

## What is the power factor of an RL circuit having a 30 degree phase angle between the voltage and the current?

A. 1.73
B. 0.5
C. 0.866
D. 0.577

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This question is a bit tricky, because we have to calculate the numeric value of the power factor due to the phase difference.  Luckily we can throw out A, because its greater than 1 (that can’t happen!)

But, we’re not actually given any values for the resistance or inductance in the circuit! How are we supposed to do this?

Simple.  All you need to remember, is how to plot the values for an RLC circuit on a set of coordinates.  We’re given the 30 degree phase difference.  So draw a line from the origin, one unit long, at an angle of 30 degrees from the X axis.  OK?

Now, this starts to look like a trig problem.  Let’s just remember that the power factor is defined as the actual resistance of the circuit (not the impedance!) divided by the total impedance.  It then represents the actual power dissipated by an equivalent resistance.  So, we need to find the value of the triangle leg along the X axis.  (Yes, I realize I need to update this with an actual image of a graph. Remind me to do this because I will forget!)

Therefore, we use the definition of a cosine.  (Remember your trig?) $\cos \theta = O / H$

where O is the “opposite” side of the triangle vs. the angle, and H is the hypotenuse.  In our case, this looks like this: $\cos 30^{\circ} = O / 1$

We can get rid of the 1, leaving us with only this, and since the cosine of 30 degrees is 0.866: $O = \cos 30^{\circ} = .866$

Therefore C. 0.866 is our answer!

### 1 thought on “Amateur Extra: Power Factor in an RL Circuit”

1. Oh please put an image to illustrate.

Thank you