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ECE 221
Electric Circuit Analysis I
Chapter 14
Maximum Power Transmission
(From [1] chapter 4.12, p. 120-122)
Herbert G. Mayer, PSU
Status 11/30/2014
For use at Changchun University of Technology CCUT
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Syllabus
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Motivation
Thévenin Equivalent
First Derivative Formulae
Identify Maximum
Derivative p’
Maximum Power
Sample
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Motivation
 When utility systems transmit electrical
energy across power lines, this must be done
with utmost efficiency, lest the universe gets
heated up 
 When electric signals of typically low voltage
get transmitted, ultimate goal is sending the
signal as strongly as possible, even it that
means losing a great % of power, i.e. even if
inefficiently!
 Here we discuss efficient power transmission
from a source, with unknown dependent and
independent sources and resistive networks
 How do we find the load RL at terminals a and
b such that the least % of power is wasted?
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Motivation
 Given such a network, we can always
determine the Thévenin equivalent CVS, with
VTh and RTh in series to the CVS
 For ease of computation, we then replace the
actual circuit by the Thévenin equivalent
network, and compute:
1. The maximum power delivered in the load
resistor RL
2. And the efficiency: dissipated over delivered
power, in % of the original delivered
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Thévenin Equivalent
a
RTh
iL
a
+
VTh
-
RL
b
iL
RL
b
Green box: resistive network
containing independent and
dependent sources
Equivalent CVS at terminals a, b
with computable VTh and RTh
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First Derivative Formulae
f(x)
f’(x), with u(x) and v(x) functions of x
f(x) = some variable
f’(x) = 0
f(x) = constant
f’(x) = 0
f(x) = x
f’(x) = 1
f(x) = u + v
f’(x) = u’ + v’
f(x) = u - v
f’(x) = u’ - v’
f(x) = u * v
f’(x) = u’*v + u*v’
f(x) = u / v
f’(x)= (u’*v -u*v’) / v2
f(x) = ln(u)
f’(x) = u’ / u
f(x) = u ^ v
f’(x) = u' * v * u^(v - 1) + ln(u) * v' * u ^ v
Example: f(x) = x ^ x f’(x) = 1 * x * x^(x - 1) + ln(x) * 1 * x^x
= x^x + ln(x) * x^x
= (ln(x) + 1) * x^x
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Identify Maximum
 Given a function y = f(x), with x occurring to
the nth power, there will be n-1 maxima
 To find the points of maximal values, compute
the derivative f’(x)
 Set the derivative f’(x) to zero, this is the
incline of the tangent
 Compute values of x, where in fact f’(x) = 0
 At those values for x, f(x) has highest −or
lowest− values
 Since the tangent will be horizontal
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Identify Maximum
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Identify Maximum
 Function (a) above is a 3rd power of x, there
will be 2 maxima
 There are 2 values for x, where f’(x) is 0, or
the tangent is horizontal
 Function (b) above is a 2nd power of x, there
will be 1 maximum
 Again, this is where the derivative f’(x) is 0, or
the tangent is horizontal
 We see that the power function p = u*i is a
second order function of the resistor RL,
hence we have 1 solution for the maximum
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Derivative p’ = f(RL)
p = i * v
p =
= i * i * RL
(Eq 1)
i2 * RL
p = (VTh/(RTh + RL))2 * RL
p =
RL * VTh2/(RTh + RL)2
Power p is max, when the second-order function p = f( RL
) has a tangent with 0 incline, i.e. when the derivative
dp/dRL is = 0, so we form first derivative!
Use product rule and quotient rule!
Note that VTh and RTh are not functions of RL so they are
constant, and are 0 when derived toward RL
Use product rule, quotient rule, and set : dp / dRL = 0
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Derivative p’ = f(RL)
dp/dRL= p’
P’ = (VTh2*(RTh+RL)2-VTh2*RL*2*(RTh+RL))/(RTh+RL)4
P’ = VTh2 *((RTh+RL)2-RL*2*(RTh+RL))/(RTh+RL)4
P’ = 0
0
= (RTh+RL)2 - RL* 2 * (RTh + RL)
0
= (RTh+RL) - RL* 2
0
= RTh + RL - RL* 2
RTh = RL
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Maximum Power
Maximum Power transmission occurs, when the load
resistance RL equals the Thévenin resistance RTh
How large is that Pmax?
With RL = RTh and Eq 1:
Pmax = VTh2 * RL / (2 * RL)2
Pmax = VTh2 / (4 * RL)
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Sample Maximum Power Transfer
(a) Given the circuit on the next page, find the
Thévenin equivalent with VTh and RTh
(b) What is the maximum power delivered to RL?
(c) What is the percentage of power delivered?
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Sample Maximum Power Transfer
30 Ω
+
360 V
-
a
150 Ω
iL
RL
b
Sample CVS with 30 and 150 Ω Resistors
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(a) Find Thévenin Equivalent
VTh is voltage at open plugs a and b
iTh is short-circuit current at plugs a and b
RTh = VTh / iTh
VTh
= 360 * 150 / 180= 300 V
iTh
= 360 / 30
=
12 A
RTh
= 300 / 12
=
25 Ω
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Sample Maximum Power Transfer
30 Ω
+
360 V
-
a
150 Ω
25 Ω
iL
a
+
300 V
-
RL
b
iL
RL
b
Sample CVS with 30 and 150 Ω Resistors
Thevenin Equivalent of Sample CVS
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(b) Maximum Power Delivered to RL
Maximum power is delivered with RL = RTh
pTh = i * v
Voltage at RL is half of VTh = 300 V, VL = 150 V
pTh
= 150 * 300/50 = 900 W
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(c) Percent of Power Delivered
RL in original circuit is = RTh = 25 Ω
RL parallel to 150 Ω in series with 30 Ω is 360/7
Percent = pTh / p360
p360
= v*i
i
= v / (360 / 7)= 360*7 / 360
= 7A
P360
= 360 * 7
= 2.52 kW
= 2520 W
Percent = 2,520 / 900 = 0.357129
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= 35.71 %