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ECE 221 Electric Circuit Analysis I Chapter 14 Maximum Power Transmission

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

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ECE 221 Electric Circuit Analysis I Chapter 14 Maximum Power Transmission

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  1. ECE 221Electric Circuit Analysis IChapter 14Maximum Power Transmission (From [1] chapter 4.12, p. 120-122) Herbert G. Mayer, PSU Status 11/2/2015

  2. Syllabus Motivation Thévenin Equivalent First Derivative Formulae Identify Maximum p in RL 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 RL 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 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: The maximum power delivered in the load resistor RL And the efficiency: dissipated over delivered power, in % of the original delivered

  5. Thévenin Equivalent a a RTh iL iL + VTh - RL RL b b Green box: resistive network containing independent and dependent sources Equivalent CVS at terminals a, b with computable VTh and RTh

  6. First Derivative Formulae

  7. 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

  8. Identify Maximum p in RL

  9. 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

  10. 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

  11. 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

  12. 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)

  13. 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?

  14. Sample Maximum Power Transfer

  15. (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 = 360 * 150 / 180 = 300 V iSC = 360 / 30 = 12 A RTh = 300 / 12 = 25 Ω

  16. Sample Maximum Power Transfer a a 30 Ω 25 Ω iL iL + 360 V - + 300 V - RL RL 150 Ω b b Thévenin Equivalent of Sample CVS Sample CVS with 30 and 150 Ω Resistors

  17. (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

  18. (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 = 7 A P360 = 360 * 7 = 2520 W = 2.52 kW Percent = 900 / 2,520 = 0.357129 = 35.71 %

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