# 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:

**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

**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:

**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:

**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.

**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

**b)** Determine the current flowing through the circuit

#### Answer

**c)** Determine the voltage across the resistor and inductor

#### Answer

**d)** Determine the power dissipated in the resistor

#### Answer

**e)** Determine the reactive power of the inductor

#### Answer

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

#### Answer

**b)** Determine the current flowing through the circuit

#### Answer

**c) **Determine the voltage across each component

#### Answer

**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

## 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!