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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