# Using Phasors in AC Circuits

concept

Dealing with sines and cosines is mathematically a pain in the ass. Which is unfortunate because in AC it's mostly what we do.
Luckily it's possible to use a new mathematical tool called "phasors" to remove the trigonometry from our problems and replace it with normal multiplication, division, addition and subtraction.
Phasors are a way of representing sine waves as well as complex numbers and operating on them without needing to do any complex algebra or trigonometry.
Before we go into how to use phasors we'll need to do a quick refresher on complex numbers.

fact

A complex number is a point on a 2D graph written \(x + iy\) where \(x\) is the horizontal coordinate and \(y\) is the vertical coordinate.
\(x + iy\) is the same as point (x, y) in geometry.

fact

The \(i\) in complex numbers happens to be:
\(i = \sqrt{-1}\)

fact

In a complex number the \(x\) is called the "Real" part and the \(y\) is called the "Imaginary" part (hence the use of the letter \(i\) to denote the imaginary part).

fact

Two complex numbers can be added together by adding their real and imaginary parts separately.

example

If \(Z_1 = 1 + 2i\) and \(Z_2 = -3 + i\) find \(Z_1 + Z_2\)

The real part of \(Z_1 + Z_2\) is going to be the real part of \(Z_1\) added to the real part of \(Z_2\). So the real part of \(Z_1 + Z_2\) is \(-2\). Likewise the imaginary part of \(Z_1 + Z_2\) is going to be \(2 + 1 = 3\) So \(Z_1 + Z_2 = -2 + 3i\) It's useful when doing multiplication or division to write complex numbers in their polar form instead, this uses polar coordinates to mark the point rather than rectangular coordinates like we used above.

fact

Polar form is given by \(Z = re^{i\theta}\)
To convert \(Z = x + iy\) into polar form we use the formulas:
\(r = \sqrt{x^2 + y^2}\)
\(\theta = \tan^{-1}(\frac{y}{x})\)

example

Convert \(1 + i\) into polar form

We just use the above formulas to get: \(r = \sqrt{2}\) \(\theta = \tan^{-1}(1) = \frac{\pi}{4}\) So \(i + i = \sqrt{2}e^{i\frac{\pi}{4}}\)fact

To convert a polar form to a rectangular form use the following formulas:
For \(Z = Ae^{i\theta}\):
\(Z = A\cos(\theta) + iA\sin(\theta)\)

fact

To multiply two complex numbers use their polar forms and the following formulas:
\(Z_3 = Z_1Z_2\)
\(r_{Z3} = r_{Z1}\cdot r_{Z2}\)
\(\theta_{Z3} = \theta_{Z1}+\theta_{Z2}\)

fact

To divide two complex numbers use their polar forms and the following formulas:
\(Z_3 = \frac{Z_1}{Z_2}\)
\(r_{Z3} = \frac{r_{Z1}}{r_{Z2}}\)
\(\theta_{Z3} = \theta_{Z1}-\theta_{Z2}\)

Okay, that's enough of the complex numbers for us to get into phasors.
A phasor is really just a complex number in polar form. We can use it to represent a thing called "Impedance" which replaces resistance in AC circuits.
We can also use phasors to represent sine waves so we can convert trigonometry problems into simpler algebra problems

fact

To write the polar complex number \(re^{i\theta}\) as a phasor we use the slightly different form:
\(r \angle \theta\)

fact

If all of the AC sources in a circuit have the same frequency then the voltage and current everywhere in the circuit has that one frequency.

fact

If all of the AC frequencies in a circuit are the same we can write all the voltages and currents as phasors.

fact

To write a cosine wave \(A\cos(\omega t + \theta)\) as a phasor it's simply:
\(A \angle \theta\)

fact

To write a sine wave \(A\sin(\omega t + \theta)\) as a phasor we write:
\(A \angle (\theta - 90^\circ)\)

example

Find \(12\angle 30^\circ \cdot 2\angle 11^\circ\)

\(12 \cdot 2 = 24\) \(30 + 11 = 41\) So \(12\angle 30^\circ \cdot 2\angle 11^\circ = 24\angle 41^\circ\)example

Find \(\frac{12\angle 30^\circ}{2\angle 11^\circ}\)

\(\frac{12}{2} = 6\) \(30 - 11 = 19\) so \(\frac{12\angle 30^\circ}{2\angle 11^\circ} = 6\angle 19^\circ\)fact

To add two phasors we must first change them into rectangular form, add them like we normally add two complex numbers, then turn the result back into phasor form.

example

Find \(5\angle 30^\circ + 2\angle 45^\circ\)

\(5\angle 30^\circ = 5\cos(30^\circ) + 5i\sin(30^\circ) \approx 4.33 + 2.5i\) \(2\angle 45^\circ = 2\cos(45^\circ) + 2i\sin(45^\circ) \approx 1.41 + 1.41i\) So \(5\angle 30^\circ + 2\angle 45^\circ \approx 5.74 + 3.91i\) Now we return it to phasor form: \(5.74 + 3.91i = 6.95\angle 34.26^\circ\)
practice problems