Basic Electric Circuits Session 10

Basic Electric Circuits – © 2020 Mouli Sankaran Email: mouli.sankaran@yahoo.com 2
Session 10: Focus
 Handy Circuit Analysis Techniques
 Linearity Definition
 Linear Circuit Elements (learnt so far)
◦ Independent and dependent Sources (voltage and
current)
◦ Resistor
 Linear Circuits
 Superposition
◦ Example
◦ Problem
 Home Work Problems

Handy
Circuit Analysis Techniques

Handy Circuit Analysis Techniques
 Linearity
 Superposition
 Source Transformations
 Thévenin’s Theorem
 Norton’s Theorem
 Thévenin and Norton Equivalent Networks
 Maximum Power Transfer
 Δ ↔Y Transformations for Resistive Networks
 Selecting a Particular Combination of Analysis
Techniques

Linearity Definition

Linearity Definition
 Linearity requires both additivity and
homogeneity (scaling) properties to be satisfied
 Additivity:
◦ It means that the output due to two or more inputs is
equivalent to adding individual outputs due to each applied
input
 Homogeneity (scaling):
◦ It means that the output is proportional to the input
◦ If the input is doubled the output would also be doubled

Linear Circuits and Superposition

Linearity and Superposition
 All the circuits that we analyze can be classified as
linear circuits
 The most important consequence of Linearity
◦ Is the principle of superposition
 This principle is very basic and will appear
repeatedly in the study of linear circuit analysis
 As a matter of fact, the non-applicability of
superposition to nonlinear circuits is
◦ The very reason that they are so difficult to analyze!

Linear Elements
 We define a linear element as a passive element that
has a linear voltage-current relationship
 It simply means that multiplication of current
through the element by a constant K results in the
multiplication of voltage across the element by the
same constant K
 At this time, only one passive element has been
defined (the resistor)
◦ And its voltage-current relationship
◦ v(t) = R i(t)
 As a matter of fact, if v(t) is plotted as a function of
i (t), the result is a straight line

V-I Relationship of a Resistor
 The slope of this line = R (Ohms) = V/I
Current through R
Voltage across R
R

Linear Dependent Sources
Linear dependent source
 A dependent current or voltage source whose output
current or voltage is
◦ Proportional only to the first power of a specified current or
voltage variable in the circuit
 The dependent voltage source
◦ vx = 0.6i - 14v1 is
◦ and vx = 0.6i1
2 and vx = 0.6i1v2 are
linear
non-linear

Linear Circuits
 We now define a linear circuit as a circuit composed
entirely of
◦ Independent sources,
◦ Linear dependent sources, and
◦ Linear elements

Superposition Principle

Superposition Principle
 Response from a linear circuit is proportional to the
source
◦ i.e., multiplication of all independent source voltages and
currents by a constant K
◦ Increases all the current and voltage responses by the same
factor K
◦ Including the dependent source voltage or current outputs

Forcing function and Response
 Forcing functions here are the two independent
current sources in the circuit
◦ ia and ib
 Responses are the nodal voltages
◦ v1 and v2
 Since the nodal voltages are driven by the current
sources
◦ Given the rest of the passive circuit elements are not changing

Superposition
 The nodal equations of the circuit are:
 Let us perform an experiment x, by changing the two forcing
functions to
◦ iax and ibx
 Then nodal equations of this circuit would be
vax
ibx
iax
vbx

Superposition … contd.
 Let us perform an experiment y, by changing the two forcing
functions to
◦ iay and iby
 Then nodal equations of this circuit would be
vay
iby
iay
vby

Superposition Explained
 These three sets of equations describe the same circuit with three
different sets of source currents.
 Let us add or “superpose’’ the last two sets of equations from
the experiments x and y
 If we select iax and iay such that their sum is ia and select ibx
and iby such that their sum is ib,
◦ Then the desired responses v1 and v2 may be found by adding v1x to v1y and
v2x to v2y

Superposition Theorem

Disabling Voltage and Current Sources
 Making a voltage source equal to zero
 Making a current source equal to zero

Superposition Theorem
 In any linear resistive network, the voltage across or
the current through any resistor or source may be
calculated by adding algebraically all the individual
voltages or currents caused by the separate
independent sources acting alone, with all other
independent voltage sources replaced by short circuits
and all other independent current sources replaced by
open circuits
Let us apply
Superposition
to this circuit

Applying Superposition
Calculate v1a and v2a
with only ia in the circuit
Calculate v1b and v2b
with only ib in the circuit
Adding both nodal voltages
would yield v1 and v2
v1a v2a
v1b v2b
Open circuit
Open circuit

Example: Problem 1
Note: This is Example 5.1 on page 126, Figure 5.3 in the Ref 1 book (by Hayt)
 Use superposition to find ix :
is is removed vs is removed
1.0 A

Example: Problem 2
Note: This is Example 5.3 on page 129, Figure 5.6 in the Ref 1 book (by Hayt)
 Use superposition to find ix : 1.4 A
(a)
(a)
v’’
(b)
(b)

Superposition Theorem:
Summary of Steps
1. Select one of the independent sources. Set all
other independent sources to zero.
◦ This means, voltage sources are replaced with short circuits
◦ Current sources are replaced with open circuits
◦ Leave dependent sources in the circuit undisturbed
2. Re-label voltages and currents using suitable
notation
◦ Example: v’, i’,v1a etc.
◦ Be sure to re-label controlling variables of dependent
sources also to avoid confusion and incorrect computation
3. Analyze the simplified circuit to find the desired
currents and/or voltages.

Superposition Theorem:
Summary of Steps … contd.
4. Repeat steps 1 through 3 until each independent
source has been considered.
5. Add the partial currents and/or voltages obtained
from the separate analyses.
◦ Pay careful attention to voltage signs and current directions
when summing.
6. Do not add power quantities.
◦ If power quantities are required, calculate only after partial
voltages and/or currents have been summed.

Home Work Problems

S10_HW_Problem1: Superposition
Note: This is Practice 5.1 on page 127, Figure 5.4 in the Ref 1 book (by Hayt)
 Use superposition to find ix : 0.66 A

S10_HW_Problem2: Superposition
Note: This is Practice 5.2 on page 129, Figure 5.7 in the Ref 1 book (by Hayt)
 Use superposition to find voltage across each current source:
v1 = 11.147 V v2 = -1.394 V

Session 10: Summary
 Handy Circuit Analysis Techniques
 Linearity Definition
 Linear Circuit Elements (learnt so far)
◦ Independent and dependent Sources (voltage and
current)
◦ Resistor
 Linear Circuits
 Superposition
◦ Example
◦ Problem
 Home Work Problem

References
Ref 1 Ref 2

Basic Electric Circuits Session 10

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