TAFE SA Associate Degree in Biomedical Engineering

Associate Degree in Electronic Engineering

Associate Degree in Electrical Engineering
ELECTRONIC CIRCUITS

LECTURE 5
Circuit analysis – Mesh and Nodal analysis
1 Voltage and current sources

Voltage source Current source

1.1 Voltage source

An ideal voltage source will have a constant output voltage for all output current. Actual voltage sources include the
internal source resistance, which can drop a small voltage under load. A practical voltage source is drawn as an ideal
source in series with the source resistance. When the internal resistance is zero, the source reduces to an ideal one.

1.2 Current source

A current voltage source will have a constant output current for all output voltage. Practical current sources have
internal source resistance, which takes some of the current. A practical current source is drawn as an ideal source
with a parallel source resistance. When the source resistance is infinite, the current source is ideal.

1.3 Source conversion

Any voltage source with an internal resistance can be converted to an equivalent current source and vice-versa by
𝑉𝑆
applying Ohm’s law to the source. I.e., 𝑉𝑠 = 𝐼𝑆 × 𝑅𝑠 and 𝐼𝑆 = 𝑅𝑠
. The source resistance, RS, is the same for both.

2 Thevenin’s theorem
Thevenin’s theorem states that any two-terminal, resistive circuit can be replaced with a simple equivalent circuit of
a voltage source when viewed from two output terminals.

The Thevenin’s equivalent voltage is defined as the open circuit voltage and can be determined by measuring the
open circuit (unloaded) voltage between the output terminals of the system.
Lecture 5 Page 1 of 6
EC Lecture 5 2020s2 - Circuit Analysis.docx (Revised: 20/10/20)
TAFE SA Associate Degree in Biomedical Engineering
Associate Degree in Electronic Engineering
Associate Degree in Electrical Engineering
ELECTRONIC CIRCUITS
The Thevenin’s equivalent resistance can be determined by removing the voltage/current source, then measure the
total resistance between the two output terminals of the system.

• To remove a voltage source replace it by a short circuit such that the potential difference of the voltage
source is zero.
• To remove a current source replaces it by an open circuit such that the current is zero.

Example: determine the Thevenin’s equivalent circuit between terminal A and B of the following circuits.

(a) (b) (c)

3 Norton theorem
Norton’s theorem states that any two-terminal, resistive circuit can be replaced with a simple equivalent circuit
when viewed from two output terminals.

The Norton equivalent current is defined as the short circuit current and can be determined by measuring the short
circuit current between the output terminals of the system.

The determination of Norton equivalent resistance is the same as Thevenin’s equivalent resistance; i.e. remove the
voltage/current source and then measure the total resistance between the two output terminals of the system.

Example: determine the Norton’s equivalent circuit between terminal A and B of circuits a, b and c in the previous
example.

4 Mesh and Nodal analysis

4.1 DC Mesh Analysis
Kirchhoff’s current law is commonly used to solve for the current in each branch of a given network. While the
methods used were relatively simple, branch-current analysis is awkward to use because it generally involves solving
Lecture 5 Page 2 of 6
EC Lecture 5 2020s2 - Circuit Analysis.docx (Revised: 20/10/20)
TAFE SA Associate Degree in Biomedical Engineering
Associate Degree in Electronic Engineering
Associate Degree in Electrical Engineering
ELECTRONIC CIRCUITS
several simultaneous linear equations. It is not difficult to see that the number of equations may be prohibitively
large even for a relatively simple circuit.

A better approach and one which is used extensively in analysing linear bilateral networks is called mesh (or loop)
analysis. While the technique is similar to branch-current analysis, the number of simultaneous linear equations
tends to be less. The principal difference between mesh analysis and branch-current analysis is that we simply need
to apply Kirchhoff’s voltage law around closed loops without the need for applying Kirchhoff’s current law.

The steps used in solving a circuit using mesh analysis are as follows:

1. Arbitrarily assign a clockwise current to each interior closed loop in the network. Although the assigned
current may be in any direction, a clock-wise direction is used to make later work simpler.
2. Note: Convert each current source in the network to its equivalent voltage source. This step, although not
absolutely necessary, makes further calculations easier to understand.
3. Using the assigned loop currents, indicate the voltage polarities across all resistors in the circuit. For a
resistor which is common to two loops, the polarities of the voltage drop due to each loop current should be
indicated on the appropriate side of the component.
4. Applying Kirchhoff’s voltage law, write the loop equations for each loop in the network. Do not forget that
resistors which are common to two loops will have two voltage drops, one due to each loop.
5. Solve the resultant simultaneous linear equations.
6. Branch currents are determined by algebraically combining the loop currents which are common to the
branch.

Example:

Find the current in each

branch for the circuit
Step 1: Loop currents are assigned as shown in the
circuit. These currents are designated I1 and I2.

Step 2: Voltage polarities are assigned according to the loop currents. Notice that the resistor R2 has two different
voltage polarities due to the different loop currents.

Step 3: The loop equations are written by applying Kirchhoff’s voltage law in each of the loops. The equations are as
follows:

Loop 1: 6V - (2Ω)I1 - (2Ω)I1 + (2Ω)I2 - 4V = 0

Loop 2: 4V - (2Ω) I2 + (2Ω)I1 - (4Ω)I2 + 2V = 0

Note that the voltage across R2 due to the currents I1 and I2 is indicated as two separated terms, where one term
represents a voltage drop in the direction of I1 and the other term represents a voltage rise in the same direction.
The magnitude and polarity of the voltage across R2 is determined by the actual size and directions of the loop
currents. The above loop equations may be simplified as follows:

Loop 1: (4Ω)I1 - (2Ω)I2 = 2V

Lecture 5 Page 3 of 6
EC Lecture 5 2020s2 - Circuit Analysis.docx (Revised: 20/10/20)
TAFE SA Associate Degree in Biomedical Engineering
Associate Degree in Electronic Engineering
Associate Degree in Electrical Engineering
ELECTRONIC CIRCUITS
Loop 2: -(2Ω)I1 + (6Ω)I2 = 6V

Step 4: Using simultaneous equations or determinants solve to give I1 = 1.20A and I2 = 1.40A.

Step 5: Current across R2 (in downwards direction) = I1 - I2 = -0.20A

If the circuit being analysed contains current sources, the procedure is a bit more complicated. The circuit may be
simplified by converting the current source(s) to voltage sources (using Thevenin’s theorem) and then solving the
resulting network using the procedure shown in the previous example. Alternatively, you may not wish to alter the
circuit, in which case the current source will provide one of the loop currents.

4.2 DC Nodal Analysis

With nodal analysis we will apply Kirchhoff’s current law to determine the potential difference (voltage) at any node
with respect to some arbitrary reference point in a network. Once the potentials of all nodes are known, it is a
simple matter to determine other quantities such as current and power within the network.

The steps used in solving a circuit using nodal analysis are as follows:

1. Arbitrarily assign a reference node within the circuit and indicate this node as ground. The reference node is
usually located at the bottom of the circuit, although it may be located anywhere.
Note: Convert each voltage source in the network to its equivalent current source. This step, although not
absolutely necessary, makes further calculations easier to understand.
2. Arbitrarily assign voltages (V1, V2,... , Vn) to the remaining nodes in the circuit. (Remember that you have
already assigned a reference node, so these voltages will all be with respect to the chosen reference.)
3. Arbitrarily assign a current direction to each branch in which there is no current source. Using the assigned
current directions, indicate the corresponding polarities of the voltage drops on all resistors.
4. With the exception of the reference node (ground), apply Kirchhoff’s current law at each of the nodes. If a
circuit has a total of n+1 nodes (including the reference node), there will be ‘n’ simultaneous linear
equations.
5. Rewrite each of the arbitrarily assigned currents in terms of the potential difference across a known
resistance.
6. Solve the resulting simultaneous linear equations for the voltages (V1, V2,... , Vn)

Example:

Determine the voltage at node A for A

R1 R2
the circuit

V1 R3 V2

Step 1: Reference node assigned at bottom of circuit.

Step 2: Only one node voltage to assign (VA).

Step 3: Currents across R1 and R2 (I1 and I2 respectively) flow into node A. Current across R3 (I3) flows out of node A.

Lecture 5 Page 4 of 6
EC Lecture 5 2020s2 - Circuit Analysis.docx (Revised: 20/10/20)
TAFE SA Associate Degree in Biomedical Engineering
Associate Degree in Electronic Engineering
Associate Degree in Electrical Engineering
ELECTRONIC CIRCUITS
Step 4: We now apply Kirchhoff’s current law at the node labelled as VA:
Node VA: I1 + I2 = I3

Step 5: The currents are rewritten in terms of the voltages across the resistors as follows:
I1 = (V1 – VA) / R1
I2 = (V2 – VA) / R3
I3 = VA / R3
The nodal equations become
((V1 – VA) / R1) + ((V2 – VA) / R3) = VA / R3

Step 6: Solve for VA.

Example:

Determine the nodal

voltages for the
circuit

Step 1: Assign reference node at bottom of

circuit.

Step 2: Convert the voltage sources into equivalent current sources.

Steps 3 and 4: Arbitrarily assign node voltages and branch currents. Indicate the voltage polarities across all resistors
according to the assumed current directions.

Step 5: We now apply Kirchhoff’s current law at the nodes labelled as V1 and V2:
Node V1: I1 + I2 = 2A
Node V2: I3 + I4 = I2 + 3A

Step 6: The currents are rewritten in terms of the voltages across the resistors as follows:
I1 = V1 / 5Ω; I2 = (V1 – V2) / 3Ω; I3 = V2 / 4Ω; I4 = V2 / 6Ω
The nodal equations become
(V1 / 5Ω) + ((V1 – V2) / 3Ω) = 2A
(V2 / 4Ω) + (V2 / 6Ω) = ((V1 – V2) / 3Ω) + 3A
Simplifies as
(1/5 + 1/3)V1 - (1/3)V2 = 2A
-(1/3)V1 + (1/4 + 1/6 + 1/3)V2 = 3A

Lecture 5 Page 5 of 6
EC Lecture 5 2020s2 - Circuit Analysis.docx (Revised: 20/10/20)
TAFE SA Associate Degree in Biomedical Engineering
Associate Degree in Electronic Engineering
Associate Degree in Electrical Engineering
ELECTRONIC CIRCUITS
Step 7: Using simultaneous equations or determinants solve to give V1 = 8.65V and V2 = 7.85A.

5 AC analysis
If a circuit contains AC sources with resistors, capacitors and inductor the calculation for the resultant AC
voltage/current is the same as DC. Except now phasors will be used to represent all impedance, voltages and
currents.

Example: Convert the AC voltage source to AC current source.

ZT = 3Ω + j4Ω = 5Ω ∠53.13°

I = 10V ∠0° / 5Ω ∠53.13° = 2A ∠-53.13°

Lecture 5 Page 6 of 6
EC Lecture 5 2020s2 - Circuit Analysis.docx (Revised: 20/10/20)