ECE 221
Electric Circuit Analysis I
Chapter 14
Maximum Power Transmission
(From [1] chapter 4.12, p. 120-122)
Herbert G. Mayer, PSU
Status 11/2/2015
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Syllabus
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Motivation
Thévenin Equivalent
First Derivative Formulae
Identify Maximum p in RL
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 via RF, ultimate goal is sending the signal
as strongly as possible, even it that means losing a
great % of source power (not signal 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 some 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
f’(x) = 1 * x * x^(x - 1) + ln(x) * 1 * x^x
Example: f(x) = x ^ x
= x^x + ln(x) * x^x
= (ln(x) + 1) * x^x
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Identify Maximum p in RL
 Given a function y = f(x), with x occurring to
the nth power, there will be n-1 points of
maximum values, AKA the maxima
 To find all maxima, compute the derivative f’(x)
 Set the derivative f’(x) to zero, yielding the
point where the incline of the tangent is
horizontal
 Compute values of x, where 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 p in RL
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Identify Maximum p in RL
 Function (a) above is a 3rd power of x, there
will be 2 maxima
 Function (a) has 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
 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
= i * i * RL
(Eq 1)
p = i2 * RL
p = (VTh/(RTh + RL))2 * RL
p = RL * VTh2/(RTh + RL)2
RTH here is constant, but RL is the variable load
resistance, or which a max must be found
Power p is max, when function p = g( RL ) has a tangent
with 0 incline, i.e. when the derivative dp/dRL is = 0
So we compute the first derivative!
Use product rule and quotient rule!
VTh and RTh are not functions of RL so they are constants,
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
So the next question: 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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A 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
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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 / iSC
VTh
iSC
RTh
= 360 * 150 / 180= 300 V
= 360 / 30
=
= 300 / 12
=
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12 A
25 Ω
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
Thévenin Equivalent of Sample CVS
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(b) Maximum Power Delivered to RL
Maximum power is delivered with RL = RTh
p
=i*v
Pmax = i * v
Voltage at RL is half of VTh = 300 V, VL = 150 V
pmax = 150 * 300/50
= 900 W
--another way:
Pmax = i * v
= i * i * RL = (300/50)2 * 25 = 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 = 900 / 2,520 = 0.357129
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= 35.71 %