EE301 – AC Source Transformation and Nodal Analysis

Learning Objectives 1. Construct equivalent circuits by converting an AC voltage source and a resistor to an AC current source and a resistor 2. Apply Nodal Analysis to an AC Circuit DC Source transformation Source transformation is the process of replacing a voltage source vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa.

v viRi or s ss sR

AC Source transformation A voltage source with impedance Z in series is the same as a current source with an impedance Z in parallel.

1 Example: Convert the voltage source to a current source

Solution:

2 Example: Using source transformations, determine the voltage drop VR across the 3 ohm resistor.

Solution:

1 EE301 – AC Source Transformation and Nodal Analysis

3 Example: Using source transformations, determine the voltage drop VR across the 10 ohm resistor.

Solution:

AC Nodal Analysis AC Nodal Analysis is exactly the same procedurally as DC Nodal Analysis. The only difference is that the numbers are now complex.

Given a circuit with n nodes: Step 1. Select a reference node. Step 2. Assign voltages Va, Vb, Vc …. to the remaining nodes, identifying any known voltages. Step 3. Assume a direction for the current passing through each resistor adjacent to a node with an unknown voltage and express the branch currents in terms of node voltages. Step 4. Apply KCL to each node. Step 5. Solve the resulting simultaneous equations to obtain the unknown node voltages. So, let’s take it slow by considering the circuit below:

6 Ω 8 Ω 12 Ω 10∡30V 20 Ω 7∡20V

2 EE301 – AC Source Transformation and Nodal Analysis

We first select a reference node, and assign labels to the remaining nodes. The circuit above has four nodes. Let’s select the reference node (ground) to be the bottom node (Step 1 complete!), and let’s label the remaining three nodes as a, b, and c, as shown below: a b c

6 Ω 8 Ω 12 Ω 10∡30V 20 Ω 7∡20V

Now, each of the labeled nodes, a, b, and c, will have a voltage associated with it: VVVabc , , . Now…we are not quite done with Step 2; we have to identify any know voltages. Look at the figure above: Are any

of the quantities VVVabc , , known?

VVUNKVabc10 30, , 720 Step 2 complete! Next, for Step 3, we assume a direction for the current passing through each resistor adjacent to a node with an unknown voltage (node b). Stated another way, we are going to write the branch currents. We

have chosen to label the currents as iii123,, in the direction shown below.

i1 a b i3 c

6 Ω 8 Ω i2 12 Ω 10∡30V 20 Ω 7∡20V

But we’re not done with Step 3: we now have to express each of the branch currents in terms of node voltages. This is done using Ohm’s Law, but you have to be careful about the polarities. The way I have labeled the direction of i1 assumes that the voltage drop across the 4 resistor has it’s positive polarity

at node b. Stated another way, I am assuming that Vb is greater than Va .

So, expressing the currents as iii123,, in terms of the node voltages, we have: VV V10 30 V i ba b 1 68jj 68  V  0 V i b b 2 jj20 20 VV V720 i bc b 3 12 12 3 EE301 – AC Source Transformation and Nodal Analysis

Step 3 complete! Next, we apply KCL to node b.

KCLb : i123 i i 0

VV10 30 V  V7 20 KCL : bb b 0 b 68jj  20  12 

Step 4 complete!

This is one equation with one unknown. At this point the Nodal Analysis is essentially finished and the algebra begins. You can solve this with some simple algebra. It is left to the student to compute the value of Vb.

Now that we know Vb we can determine any other quantity in the circuit (such as the currents iii123,,).

That’s it!! That’s all there is conceptually to AC Nodal Analysis!!

4 Example: Use Nodal Analysis to determine the voltage at node b in the circuit below.

Solution: b

10 Ω 12 Ω 5∡15V 15 Ω 3∡10A

Step 1&2: Select a reference node and assign voltages Va, Vb, Vc to the remaining nodes, identifying any known voltages. i1 i3 a b c

10 Ω i2 12 Ω 5∡15V 15 Ω 3∡10A

VVUNKVUNKabc515, ,

4 EE301 – AC Source Transformation and Nodal Analysis

Note that, strictly speaking we don’t know the voltage at node c however a KCL equation at this node would be trivial and is not necessary.

So we only need one KCL equation and node b:

KCLb : i123 i i 0

Step 3 is complete.

In this case, since i3 is a current source and constant, we can simply use the constant and there is no need to express it in terms of voltage and resistance. Note the polarity of i3 is opposite the red arrow so it is negative.

VV V515 V i ba b 1 10 10 V  0 V i b b 2 jj15 15

iA3 310

Now express the KCL equation in terms of voltage and resistance: VV10 30 V KCL: b b 13 10 A 0 b 10j 15

Step 4 is complete and this is the end of the Nodal Analysis. From here on the problem is strictly algebra.

5 Example: In light of the previous example, determine IUNK in the circuit above, assuming IUNK points from node c to node b.

Solution:

5