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 Vs in the simplified circuit Thus the Thevenin voltage is given by Vs 1 6R 12R 2 Vs 4 30 V Voc To find the Thevenin resistance we short circuit Vs 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 Rth 3R 900 b Using the Thevenin equivalent circuit to simplify our calculations 2 30 RL 900 RL 0 1875 W 0 24 W 0 244898 W RL 300 600 1200 PRL 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 effort needed to obtain the answers 1 c Using Maximum Power Transfer Theorem for maximum power transfer RL 900 and resultant power delivered to the load is given by Pmax 30 2 4 900 0 25 W d If Vs is doubled power becomes four times as Pmax Vs2 Thus power delivered to the load becomes Pload 0 25 4 1 W Problem 17 We proceed to find the Thevenin voltage Voc and Thevenin resistance Rth Norton current is then given by Isc Voc Rth and Norton resistance is same as Thevenin resistance Since this problem involves a dependent source we cannot apply the conventional method We write KVL equations to find the required values vAB 300iA vx 0 4 vx 200 iA vx 18000 vx 0 3 18000 iA 29 320 689655iA vx vAB vAB Rth iA Voc Voc 0 Rth 320 69 Isc 0 Rnorton 320 69 Problem 21 a Again we write KVL equations to find the Thevenin and Norton equivalent circuits vAB 40iA 200 iA kvx vx 50 0 Vs vx 40iA vAB 0 2 vAB 1 200 k 0 02 iA 240 8000 k 0 02 200Vs k 0 02 vAB 60iA 18 vAB iA Rth Voc Voc 18 V Rth 60 Isc 0 3 A b Clearly using the previous set of equations Voc is zero for k 0 02 and for this value of k Rth 240 Problem 37 a vAB iA Rth Voc 54 0 01Rth Voc 66 0 04Rth Voc Rth 400 Voc 50 V Isc 0 125 A b RL 400 Pmax Voc2 4RL 2500 4 400 1 5625 W 3
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