1 ECE 221 Electric Circuit Analysis I Chapter 14 Maximum Power Transmission (From [1] chapter 4.12, p ) Herbert G. Mayer, PSU Status 11/2/2015

2 Syllabus Motivation Thévenin Equivalent First Derivative Formulae Identify Maximum p in R L Derivative p’ Maximum Power Sample

3 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 R L at terminals a and b such that the least % of power is wasted?

4 Motivation Given such a network, we can always determine the Thévenin equivalent CVS, with V Th and R Th in series to the CVS For ease of computation, we then replace the actual circuit by the Thévenin equivalent network, and compute:   The maximum power delivered in the load resistor R L   And the efficiency: dissipated over delivered power, in % of the original delivered

5 Thévenin Equivalent R Th + V Th - Equivalent CVS at terminals a, b with computable V Th and R Th RLRL iL a b RLRL Green box: resistive network containing independent and dependent sources a b

6 First Derivative Formulae f(x)f’(x), with u(x) and v(x) functions of x f(x) = some variablef’(x) = 0 f(x) = constantf’(x) = 0 f(x) = xf’(x) = 1 f(x) = u + vf’(x) = u’ + v’ f(x) = u - vf’(x) = u’ - v’ f(x) = u * vf’(x) = u’*v + u*v’ f(x) = u / vf’(x)= (u’*v -u*v’) / v 2 f(x) = ln(u)f’(x) = u’ / u f(x) = u ^ vf’(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

7 Identify Maximum p in R L Given a function y = f(x), with x occurring to the n th 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

8 Identify Maximum p in R L

9 Function (a) above is a 3 rd 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 2 nd 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 R L, hence we have 1 solution for the maximum

10 Derivative p’ = f( R L ) p=i * v =i * i * R L (Eq 1) p=i 2 * R L p=(V Th /(R Th + R L )) 2 * R L p = R L * V Th 2 /(R Th + R L ) 2 R TH here is constant, but R L is the variable load resistance, or which a max must be found Power p is max, when function p = g( R L ) has a tangent with 0 incline, i.e. when the derivative dp/dR L is = 0 So we compute the first derivative! Use product rule and quotient rule! V Th and R Th are not functions of R L so they are constants, and are 0 when derived toward R L Use product rule, quotient rule, and set : dp / dR L = 0

11 Derivative p’ = f(R L ) dp/dR L = p’ P’ = (V Th 2 *(R Th +R L ) 2 -V Th 2 *R L *2*(R Th +R L ))/(R Th +R L ) 4 P’ = V Th 2 *((R Th +R L ) 2 -R L *2*(R Th +R L ))/(R Th +R L ) 4 P’ = 0 0 = (R Th +R L ) 2 - R L * 2 * (R Th + R L ) 0 = (R Th +R L ) - R L * 2 0 = R Th + R L - R L * 2 R Th = R L

12 Maximum Power é Maximum Power transmission occurs, when the load resistance R L equals the Thévenin resistance R Th So the next question: How large is that P max ? With R L = R Th and Eq 1: P max = V Th 2 * R L / (2 * R L ) 2 P max = V Th 2 / (4 * R L )

13 A Sample Maximum Power Transfer é (a) Given the circuit on the next page, find the Thévenin equivalent with V Th and R Th (b) What is the maximum power delivered to R L ? (c) What is the percentage of power delivered?

14 Sample Maximum Power Transfer

15 (a) Find Thévenin Equivalent V Th is voltage at open plugs a and b i Th is short-circuit current at plugs a and b R Th = V Th / i SC V Th =360 * 150 / 180=300 V i SC =360 / 30= 12 A R Th =300 / 12= 25 Ω

16 Sample Maximum Power Transfer 25 Ω V - Thévenin Equivalent of Sample CVS RLRL iL a b 30 Ω V - Sample CVS with 30 and 150 Ω Resistors RLRL iL a b 150 Ω

17 (b) Maximum Power Delivered to R L Maximum power is delivered with R L = R Th p= i * v P max = i * v Voltage at R L is half of V Th = 300 V, V L = 150 V p max = 150 * 300/50=900 W --another way: P max = i * v= i * i * R L = (300/50) 2 * 25=900 W

18 (c) Percent of Power Delivered R L in original circuit is = R Th = 25 Ω R L parallel to 150 Ω in series with 30 Ω is 360/7 Percent = p Th / p 360 p 360 =v * i i=v / (360 / 7)=360*7 / 360=7 A P 360 =360 * 7=2520 W=2.52 kW Percent=900 / 2,520= =35.71 %