DeCarloCh6Sol - Thevenin Probs [email protected] Decarlo P M Lin...

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Thevenin Probs, 7/24/01 - P6.1 - @R.A. Decarlo & P. M. Lin PROBLEM SOLUTIONS CHAPTER 6 S OLUTION 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. S OLUTION 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 - @R.A. Decarlo & P. M. Lin S OLUTION 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. S OLUTION 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. S OLUTION 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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