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Electrical Circuits_Lecture4
Circuit Theorems
Lecture 4 Objectives:

To study the Superposition theorem

To study the Source Transformation
techniques

Voltage to current transformation

Current to voltage transformation
INTRODUCTION
A large
complex circuits
Simplify
circuit analysis
Circuit Theorems
‧Thevenin’s theorem
‧Circuit linearity
‧source transformation
‧ Norton theorem
‧ Superposition
‧ max. power transfer
Linearity Property

A linear circuit is one whose output is linearly
related (or directly proportional) ito its input.
I0
Suppose vs = 10 V gives i = 2 A. According to
the linearity principle, vs = 5 V will give i = 1 A.
V0
v
Superposition
The superposition principle states that the voltage
across (or current through) an element in a linear
circuit is the algebraic sum of the voltages across (or
currents
through)
that
element
independent source acting alone.
due
to
each
Steps to apply superposition principle
1.
Turn off all independent sources except one source. Find
the output (voltage or current) due to that active source
using nodal or mesh analysis.


2.
3.

Turn off voltages sources = short voltage sources; make it
equal to zero voltage
Turn off current sources = open current sources; make it
equal to zero current
Repeat step 1 for each of the other independent sources.
Find the total contribution by adding algebraically all the
contributions due to the independent sources.
Dependent sources are left intact.
Example 1: use Superposition principle to find Io
2mA Source Contribution
2kW
2mA
1kW
I’0
I’0 = -4/3 mA
2kW
4mA Source Contribution
2kW
4mA
1kW
I’’0
I’’0 = 0
2kW
12V Source Contribution
12V
2kW
– +
1kW
I’’’0
I’’’0 = -4 mA
2kW
Final Result
I’0 = -4/3 mA
I’’0 = 0
I’’’0 = -4 mA
I0 = I’0+ I’’0+ I’’’0 = -16/3 mA
Example 2: find v using superposition
one independent source at a time,
dependent source remains
KCL: i = i1 + i2
Ohm's law: i = v1 / 1 = v1
KVL: 5 = i (1 + 1) + i2(2)
KVL: 5 = i(1 + 1) + i1(2) + 2v1
10 = i(4) + (i1+i2)(2) + 2v1
10 = v1(4) + v1(2) + 2v1
v1 = 10/8 V
Consider the other independent source
KCL: i = i1 + i2
KVL: i(1 + 1) + i2(2) + 5 = 0
i2(2) + 5 = i1(2) + 2v2
Ohm's law: i(1) = v2
v2(2) + i2(2) +5 = 0 => i2 = (5+2v2)/2
i2(2) + 5 = i1(2) + 2v2
-2v2 = (i - i2)(2) + 2v2
-2v2 = [v2 + (5+2v2)/2](2) + 2v2
-4v2 = 2v2 + 5 +2v2
-8v2 = 5 => v2 = - 5/8 V
from superposition: v = -5/8 + 10/8
v = 5/8 V
Source Transformation

A 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
Source Transformation
vs
vs  is R or is 
R
Source Transformation
Vs  Rs I s
Vs
Is 
Rs
Source Transformation



Equivalent sources can be used to simplify
the analysis of some circuits.
A voltage source in series with a resistor is
transformed into a current source in parallel
with a resistor.
A current source in parallel with a resistor is
transformed into a voltage source in series
with a resistor.
Example 3: Use source transformation to find vo
in the circuit in the following figure.
we use current division to get
2
i
(2)  0.4A
28
and
vo  8i  8(0.4)  3.2V
Example5: Find vx in the circuit using source
transformation
 3  5i  vx  18  0
Applying
KVL around the loop igives
 3  5i  vx  18  0
(1)
Appling KVL to the loop containing only the 3V voltage source, the
resistor, and vx yields
(2)
 3  1i  vx  0  vx  3  i
Substituting this into Eq.(1), we obtain
15  5i  3  0  i  4.5A
Alternatively
 vx  4i  vx  18  0  i  4.5A
thus
vx  3  i  7.5V