Dr. Wael Mohamed Fayek
Contents
Circuit elements,
Phases,
Techniques of circuit analysis,
Power calculations,
Circuit theorems,
Transformers,
Operational amplifiers,
Two‐port circuits,
RLC circuits,
CAD tools used in circuit projects.
Natural and step responses,
Series, parallel and resonant circuits,
Sinusoidal steady state analysis,
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Circuits Theory (POW1705)
Definitions of charge, current, & voltage.
Definitions of electric power, & energy.
Definitions of node, branch, & loop.
Circuit elements (Sources, R, L, & C).
Ohm’s law, & Kirchhoff’s Laws.
Resistances connections (series, parallel, delta, star, …).
Circuit analysis (mesh, & nodal analysis).
Circuit theorems (source transformation, & super‐position, Thevenin’s, Norton’s,
& maximum power transfer)
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Source Transformation‐ Simplest Circuit Forms
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Thevenin’s Theorem
It often occurs in practice that a
particular element in a circuit is variable
(the load) while other elements are fixed.
For example, a household outlet terminal
may be connected to different appliances
constituting a variable load.
Each time the load changes, the entire
circuit has to be analyzed all over again.
Thevenin’s theorem provides a technique
by which the fixed part of the circuit is
replaced by an equivalent circuit.
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Thevenin’s Theorem
The theorem states that a linear two‐terminal circuit can be replaced by an
equivalent circuit consisting of a voltage source VTh in series with a resistor RTh.
Where VTh is the open‐circuit voltage at the terminals and RTh is the input
resistance at the terminals when the independent sources are turned off.
The theorem was developed in 1883 by M. Leon Thevenin (1857–1926), a French
telegraph engineer.
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Finding Rth
Case 1: The network has no dependent sources:
Turn off all independent sources, Rth is the
input resistance of the network looking
between terminals a and b.
Case 2: The network has dependent sources:
Turn off all independent sources, dependent
sources are not to be turned off because they
are controlled by circuit variables.
Apply a voltage/current source at terminals a
and b and determine the resulting
current/voltage. Then Rth = vo/io
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Finding Rth
Sometimes it occurs that Rth takes a negative value:
In this case, the negative resistance (V=‐RthI) implies that the circuit is supplying
power.
This is possible in a circuit with dependent sources
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Load Current and Voltage
The current through the load IL and
the voltage across the load VL are
easily determined once the Thevenin
equivalent of the circuit at the load’s
terminals is obtained.
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Example 1:
Find the Thevenin equivalent circuit of the circuit shown in figure, to the left of
the terminals a, b.
Then find the current through RL = 6, 16, and 36Ω
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Solution:
I2 = ‐2A
32 = (4+12)I1 – 12 I2
I1 = 0.5 A
Vth = 12(I1‐I2) = 30V
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Solution:
To find Rth:
Voltage sources = 0 Short Circuit
Current sources = 0 Open Circuit
Rth = 1 + (4//12)
Rth = 1 + (4x12)/(4+12)
Rth = 4Ω
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Solution:
IL = Vth/(Rth+RL)
IL = 30/(4+RL)
RL = 6Ω
IL = 30/(4+6) = 3A
RL = 16Ω
IL = 30/(4+16) = 1.5A
RL = 36Ω
IL = 30/(4+36) = 0.75A
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Example 2:
Find the Thevenin equivalent of the
circuit shown figure.
Solution:
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Solution:
To find Vth:
Mesh equations:
I1 = 5A
0 = ‐4I1 + (4+2+6)I2 ‐2I3
2Vx = ‐2I2 + 2I3
Vx = 4(I1 – I2)
Solving these equations, I2 = 10/3A
Vth = Voc = 6I2 = 20V
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Solution:
To find Rth:
Remove all independent sources.
Current source = 0 Open circuit.
Add external source (vo) and find its
voltage and current.
Mesh equations:
2Vx = 2I1 – 2I2
0 = ‐2I1 + (4+2+6)I2 – 6I3
‐1 = ‐6I2 + (6+2)I3
Vx = ‐4I2, Io = ‐I3 = 1/6A
Rth = Vo/Io = 6Ω
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Solution:
The Thevenin equivalent circuit is
shown in figure.
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Norton’s Theorem
The theorem states that a linear two‐terminal circuit can be replaced by an
equivalent circuit consisting of a current source IN in parallel with a resistor RN,
where IN is the short‐circuit current through the terminals and RN is the input
resistance at the terminals when the independent sources are turned off.
In 1926, E. L. Norton, an American engineer at Bell Telephone Laboratories,
proposed his theorem. About 43 years after Thevenin published his theorem.
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Determination of IN, & RN
RN is found in the same Rth is found.
In fact, using source transformation, the Thevenin
and Norton resistances are equal.
RN = RTH,
IN = V TH/RTH,
RTH = V TH/IN
Source transformation is often called Thevenin‐
Norton transformation.
To find the Norton current, IN, determine the short
circuit current flowing from terminal a to b.
Dependent and independent sources are treated
the same way as in Thevenin’s theorem.
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Example 3:
Find the Norton equivalent circuit of the circuit shown in figure at terminals a‐b.
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Solution:
To find IN:
The Mesh equations are:
I1 = 2A
12 = ‐4I1 + (4+8+8)I2
I2 = IN = Isc = 1A
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Solution:
To find RN:
Norton equivalent circuit:
RN = 5 // (8+4+8)
RN = 5x20/(5+20) = 4
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Solution:
To check the answers find Vth:
I3 = 2A
12 = ‐4I3 + (4+8+5+8)I4
I4 = 0.8A
Vth = Voc = 5I4 = 4V
IN = Vth/Rth = 4/4 = 1A ###
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Example 4:
Find the Norton equivalent circuit of the circuit
shown in figure at terminals a‐b.
Solution:
Find Vth:
Ix = 10/4 = 2.5A
Voc = 5(2Ix) + 10
Vth = Voc = 35V
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Solution:
Find IN:
Ix = 10/4 = 2.5A
Isc = IN = 2Ix + 10/5
IN = 7A
Find RN:
RN = Vth/IN = 35/7 = 5
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Example 5:
Using Thevenin’s theorem, find io.
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Solution:
To find Vth:
Io = 0
I1 = 4A
5Io = ‐(3+1)I1 + (2+3+1)I2
I2 = 16/6 = 8/3A
Vth +20 – 5Io = (4+1)I1 – I2
Vth + 20 = 5x4 – 8/3
Vth = ‐8/3V
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Solution:
To Find IN:
Io = I1 – I3
I1 = 4A
5Io = ‐3I1 + (1+3+2)I2 – 1I3
20 – 5Io = ‐I2 + (4+1)I3
I3 = 8A
IN = Io = ‐4A
RN = Rth = Vth/IN = (‐8/3)/(‐4) = 2/3
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Solution:
Io = Vth/(Rth+5)
Io = ‐8/3(5+2/3) = ‐8/17A
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Maximum Power Transfer
There are applications in areas such as
communications where it is desirable to
maximize the power delivered to a load.
Assume that the load resistance is adjustable.
If the entire circuit is replaced by its Thevenin
equivalent except for the load, the power
delivered to the load is:
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Maximum Power Transfer
For a given circuit, Vth and Rth are fixed.
By varying the load resistance the power
delivered to the load varies as sketched
in Figure.
Maximum power is transferred to the
load when the load resistance equals the
Thevenin resistance as seen from the
load (RL = RTh).
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Maximum Power Transfer
To prove this:
2
0
2
0
0
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Maximum Power Transfer
Since:
For maximum power transfer RL = Rth:
2
4
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4
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Example 6:
Find the value of RL for maximum power transfer.
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Solution:
Rth = 2 +3 +6//12 = 9
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Solution:
To find Vth:
I2 = ‐2A
12 = (6+12)I1 – 12I2
I1 = ‐2/3 A
Vth = 12 ‐ 6I1 – 3I2
Vth = 22 V
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Solution:
For maximum power transfer:
RL = Rth = 9
The maximum power is:
Pmax = (22)2/(4x9) = 13.44 W
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Quiz:
Determine the value of RL that will
draw the maximum power from the
rest of the circuit shown in figure.
Calculate the maximum power.
Answer:
4.222 , 2.901 W.
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