AC Circuit Analysis

The good news is that every technique we have learnt to date for DC circuit analysis carries over to AC circuit analysis, all we have to do is simply treat resistance as impedance and voltage as a phasor and recall the reactance formulas for capacitors and inductors, that's it!

And so this page will serve as a treasure trove of example problems and questions to build familiarity with AC Circuit Analysis by reusing the techniques we have already learnt albeit in AC form.

Quick Tricks

But before we get stuck in, let's go over a few tricks of the trade regarding quick conversions to polar coordinates:

This is the case because j refers to any point directly above the x-axis at an angle of pi/2 counter clockwise. Similarly -j refers to a point directly below the x-axis at an angle of -pi/2.

It is easier to multiply and divide with the polar form vs the rectangular form. For multiplication in polar form we simply multiply the magnitudes together and add the phases together. For division we divide the magnitudes and subtract the phases.

It is easier to add and subtract using rectangular form, we just add/subtract the real and imaginary components.

Examples

1. a) Determine the impedance of the circuit in phasor form:

For impedance, we just need to sum together the resistive and reactive components. We can elegantly compute these in a single expression as follows:
To determine the phasor form of impedance we can first determine the magnitude via Pythagoras' Theorem:

b) Determine the current flowing through the circuit

We can re-use Ohm's Law to determine the current as follows:

c) Determine the voltage across the capacitor and resistor

d) Determine the power dissipated in the resistor

e) Determine the reactive power of the capacitor

2. a) Determine the impedance of the circuit shown below:

As above let's break this down in terms of resistive and reactive pieces:
Let's then determine the phasor form as this will be easier to work with when calculating the current and voltages as they involve a multiplication and division which is a simpler operation to complete in phasor form.

b) Determine the current flowing through the circuit

c) Determine the voltage across each component

3. Determine the Thevenin Equivalent of the circuit shown below:

Recall from DC Circuit Analysis that the Thevenin Equivalent Circuit consists of a voltage source in series with a Thevenin Resistance. It turns out for AC Circuit Analysis this works by using the Thevenin Impedance. So let's begin by determining the Thevenin Impedance by zeroing the source and calculating the impedance measured between terminals X and Y. Recall that we zero a voltage source with a short circuit as a short circuit has 0V across it nullifying its effect. We can then calculate the resistance of the following circuit looking in from terminals X and Y.
We are then looking at an inductor in series with a capacitor in parallel with a resistor:
Hmm... That denominator won't play nicely for division so let's convert it into its phasor form for ease of division.
Let's now substitute this back in!
Let's convert this back into polar coordinates for simpler addition.
Okay let's now determine the Thevenin Voltage which is simply the voltage across terminals X and Y. The Thevenin Voltage is then just the voltage across the capacitor as no current will travel through the inductor as terminals X and Y are open so we effectively have a simple RC circuit to analyse which is nice.
Knowing the Thevenin Voltage and Thevenin Resistance we can now draw the Thevenin Equivalent Circuit:

4. Determine the nodal voltage at the node connecting R1, C1, L1 and R2.

The question gives us a prompt to use "nodal analysis" so let's follow the prompt :) As you may recall from DC analysis we begin by labelling all of the currents, nodes and voltages, so let's do that now.
From here we can write a nodal equation at node Va:
We can then plug in the values and solve for va:
Wait what just happened there? That was some slight of hand! What I did was use the trick of converting polar coordinates to rectangular coordinates and recalling -1/j = j and that 1/j=-j

Questions

1. a) Determine the impedance of the circuit in phasor form:

Answer

2000+j37.699

b) Determine the current flowing through the circuit

Answer

999.823 < -0.019uA

c) Determine the voltage across the resistor and inductor

Answer

vr = 1.998 < 0.019V, vl = 0.038 < 0.019V

d) Determine the power dissipated in the resistor

Answer

1.999mW

e) Determine the reactive power of the inductor

Answer

376.857 < pi/2 uVAR

2. a) Determine the impedance of the circuit shown below:

Answer

2000-j0.019

b) Determine the current flowing through the circuit

Answer

1 < 0.0000095 mA

c) Determine the voltage across each component

Answer

vc1=314 < -1.571 uA, vl1=295 < 1.571 uA, vr1=vr2=1.999V

3. Determine the Norton Equivalent of the circuit shown below:

4. Determine the nodal voltage at the node connecting R1, C1, L1 and R2.

Answer

4.914 < 0.121 V

Lab:

It is time to start simulating AC circuits! I would say simulating these circuits is even more important to build intuition DC with the increased complexity so take the time to play around with these!!

Let's start simple with our old mate the voltage divider:

We can define a 2Vpk-pk 1kHz sine wave with the setup as below:

And simulate for 5ms (5 periods, 1/1kHz=1ms)

Let's then measure the voltage across the voltage source to see our beautiful sine wave!

There it is! Now isn't that such a nice image? So, let's now take a look at the voltages across the resistors starting with R2:

Here we can compare the vs and vR2 waveforms to see that most of vs is dropped across R2 which is expected as it is the larger resistance. Let's check vR1 on the same graph (perform another left click but this time hold and drag from the left to the right terminal of R1 to calculate the voltage across it):

Very cool! So now if you imagine adding vr1 and vr2 you would get vs and notice all sine waves are also in phase.

Let's now check the current:

Note the current begins swinging the other way as LTSpice determined the incorrect direction of the current.

Let's now try replacing R2 with a capacitor:

Let's now try plotting the voltage source and the voltage across the capacitor in the same curve by probing both each node and a single left click in turn:

Notice one major difference between the purely resistive circuit and this one? The difference is the phase of the voltage across the capacitor. We can say that there is a voltage lag which occurs due to the charging of the capacitor and this manifests itself as a phase shift compared to the phase of the voltage source.

The same exercise can be performed for an inductor which I encourage you to do as well as connect up multiple voltage and current sources together to get a feel for how it works!

Ready for the next module? Return to the course home.