DeCarloCh6Sol - Thevenin Probs, 7/24/01 - P6.1 - @R.A....

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Thevenin Probs, 7/24/01 - P6.1 - PROBLEM SOLUTIONS CHAPTER 6 SOLUTION 6.1 . (a) V oc is found by removing R L and doing voltage division. V V V OC 28 100 600 600 300 ) 700 || 600 ( ) 700 || 600 ( 63 = + × + = R TH is found by setting the source to zero and by calculating the equivalent resistance seen looking back between the A and B terminal. = + = 200 600 || ] 100 ) 600 || 300 [( TH R (b) Using L L I R P = the power for each resistance may be found by substituting the appropriate R L in the following equation. L L TH OC R R R V P + = 2 For 50 , 200 , and 800 , the power obtained is 627.2 mW, 980.0 mW, and 627.2 mW respectively. The use of Thevenin equivalent does reduce the effort in obtaining the answer. SOLUTION 6.2 . (a) To find R TH , open circuit the current source and short-circuit the voltage source. The resulting resistance seen from terminal A-B is 1 k . Using superposition, the contribution of the current and voltage source at the open circuit output may be summed as 30 V (2 k/4 k) + 10 mA (2 k/4 k) (2 k ). V OC is then 25 V and I SC = V OC / R TH is 25 mA. (b) Following is a plot of L L TH OC R R R V P + = 2 for R L from 100 to 4 k .
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Thevenin Probs, 7/24/01 - P6.2 - SOLUTION 6.3 . (a) Turning off the two sources R TH = (60 + 60)||40 = 30 , and using superposition V OC = 6 V 40 40 + 60 + 60 + 0.1 A 60 40 + 60 + 60 40 = 3 V. I SC is the obtained as 100 mA. (b) Using L L TH OC R R R V P + = 2 a load of 90 will absorb 56.25 mW. (c) It absorbs 75 mW; hence the 30 resistor absorbs more power. SOLUTION 6.4 . As both resistor divider ration are the same (3/6), the voltage at A and B is the same resulting in a V OC of 0 V. R TH = (3 K ||6 K ) + (9 K ||18 K ) = 8 k The relation V OC / I SC cannot be used in this situation. SOLUTION 6.5 . Using superposition V OC = 20 20 × 10 3 20 × 10 3 + 5 × 10 3 - 10 5 × 10 3 20 × 10 3 + 5 × 10 3 + 20 20sin(50 t ) = 14 + 400sin(50 t ) V, and
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Thevenin Probs, 7/24/01 - P6.3 - R TH = (20 k ||5 k ) + 20 k = 24 k . I SC can then be found using the V OC / R TH relationship as 0.58+20sin(50t) mA. SOLUTION 6.6. Once again using superposition V K mA K K K K K mA K K K K V V OC 204 2 54 6 5 8 4 4 54 4 8 6 6 72 = + + + + + + = , and = + + + = K K K K K K R TH 10 2 4 ] 6 || ) 8 4 [( SOLUTION 6.7 . Using source transformation, (a) is obtained from the original circuit. Then combining in series the resistors and voltage sources, and retransforming them (b) is obtained. Finally adding the two currents and transforming back the circuit to its Thevenin form (c) is obtained.
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This note was uploaded on 03/08/2009 for the course ECE 201 taught by Professor All during the Spring '08 term at Purdue University-West Lafayette.

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DeCarloCh6Sol - Thevenin Probs, 7/24/01 - P6.1 - @R.A....

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