HW13%20solution - ECE 201 Spring 2010 Homework 13 Solutions...

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ECE 201 Spring 2010 Homework 13 Solutions Problem 9 (a) 2R and 6R in series gives 8R. 8R and 8R in parallel gives 4R. Thus 12R is in series with V s in the simplifed circuit. Thus the Thevenin voltage is given by V oc = V s 12 R × 1 2 × 6 R = V s 4 = 30 V To fnd the Thevenin resistance, we short circuit V s . Thus 8R and 8R are in parallel, which gives 4R, which in turn is in series with 2R. This gives 6R and 6R in parallel. Thus R th =3 R = 900 Ω. (b) Using the Thevenin equivalent circuit to simpliFy our calculations, P R L = ± 30 900 + R L ² 2 R L =0 . 1875 W, 0 . 24 W, 0 . 244898 W ( R L = 300 Ω , 600 Ω , 1200 Ω) The Thevenin equivalent circuit analysis allows us to analyze the circuit without the load once and then plug in the various load resistor values to compute the relevant quantities. However, using earlier techniques would require 3 separate analyses to compute these values. Hence the use oF a Thevenin equivalent reduces the e±ort needed to obtain the answers.
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This note was uploaded on 01/22/2012 for the course IE 370 taught by Professor Chunghorng,r during the Spring '08 term at Purdue University-West Lafayette.

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HW13%20solution - ECE 201 Spring 2010 Homework 13 Solutions...

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