ECE 231 -5

ECE 231 -5 - ECE-231 Section 101 Circuits and Systems I...

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ECE-231 Section 101 Circuits and Systems I Fall ’10 Session 5 Professor Stewart Personick Office: ECEC Room 321 Stewart.Personick@NJIT.edu
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Given a real source, how much power will it “deliver” to a “load” whose resistance is R?
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Start with a black box, represented by its Thevenin equivalent circuit + - a b V th = 16V R th = 8
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Place a “load” resistor across the output terminals + - a b V th = 16V R R th = 8 Power flows out of the voltage source. Some of this power flows into the series resistor, R th , inside the black box. The remainder of the power flows into the load resistor, R.
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Place a “load” resistor across the output terminals + - a b V th = 16V R R th = 8 The mesh current is: i = 16V/ (8 + R) The power flowing out of the voltage source = 16V x i = 16V x 16V/ (8 + R) The power flowing into R th = i 2 x R th = [16V/ (8 + R)] 2 x 8 The power flowing into the load = i 2 x R = [16V/ (8 + R)] 2 x R i
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Place a “load” resistor across the output terminals + - a b V th = 16V R R th = 8 The power flowing into the load = i 2 x R = [16V/ (8 + R)] 2 x R = P load (Watts) Examples: If R = 80 Ohms, then P load = 2.645 Watts; If R = 0.8 Ohms, then P load = 2.645 Watts; If R = 8 Ohms, then P load = 8 Watts i
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Power Delivered to the Load v. R 5W R 50 10 1W
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Place a “load” resistor across the output terminals + - a b V th R R th The power flowing into the load = i 2 x R = [V th / (R th + R)] 2 x R = P load (Watts) To find the value of R that maximizes P load (Watts): use calculus . Take the derivative of P load (Watts) with respect to R. Then find the value of R which makes d[ P load (Watts)]/dR = 0. i
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The power flowing into the load = i 2 x R = [V th / (R th + R)] 2 x R = P load (Watts) d {[V th / (R th + R)] 2 x R} /dR = [V th / (R th + R)] 2 - [2(V th ) 2 / (R th + R) 3 ] x R The derivative (above) = 0 when: 2 x R / (R th + R) = 1 Thus the value of R that maximizes the power delivered by the black box to R is: R = R th This value of load resistance is usually called a matched load . Note: when R = R th (i.e. when R is a matched load), the power flowing into the internal series resistor is equal to the power delivered to the load. Thus, with a matched load, half of the total power flowing out of the voltage source is dissipated within the black box. Therefore, the efficiency with which the black box delivers power to a matched load is: 50%
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Place a “load” resistor across the output terminals + - a b V th = 16V R = 8 R th = 8 The power flowing into the load = i 2 x R = [16V/ (8 + R)] 2 x R = P load (Watts) If R = 8 Ohms, then P load = 8 Watts i = 1A 16 Watts 8 Watts 8 Watts
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The concept of Superposition (Applied to the analysis of linear circuits)
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HW#1 Problem #4 + - 12 Volts 1 Ampere 10 Ohms Resistor Calculate the current I 1 What is the voltage, v ab , between node a and node b? Is power flowing into the voltage source or out of the voltage source?
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This note was uploaded on 03/26/2012 for the course ECE 231 taught by Professor Pietrucha during the Spring '08 term at NJIT.

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ECE 231 -5 - ECE-231 Section 101 Circuits and Systems I...

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