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Thevenin Probs, 7/24/01- P6.1 - @R.A. Decarlo & P. M. LinPROBLEM SOLUTIONS CHAPTER 6SOLUTION 6.1. (a)Vocis found by removing RLand doing voltage division.VVVOC28100600600300)700||600()700||600(63=+×+=RTHis found by setting the source to zero and by calculating the equivalent resistance seen looking backbetween the A and B terminal.Ω=+=200600||]100)600||300[(THR(b) UsingLLIRP⋅=the power for each resistance may be found by substituting the appropriate RLin the following equation.LLTHOCRRRVP⋅+=2For 50 Ω, 200 Ω, and 800 Ω, 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 RTH, open circuit the current source and short-circuit the voltage source. Theresulting resistance seen from terminal A-B is 1 kΩ. Using superposition, the contribution of the currentand voltage source at the open circuit output may be summed as 30 V (2 k/4 k) + 10 mA (2 k/4 k) (2 kΩ).VOCis then 25 V and ISC= VOC/RTHis 25 mA.(b)Following is a plot ofLLTHOCRRRVP⋅+=2for RLfrom 100 Ωto 4 k Ω.
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Thevenin Probs, 7/24/01- P6.2 - @R.A. Decarlo & P. M. LinSOLUTION 6.3. (a)Turning off the two sourcesRTH=(60+60)||40=30Ω,and using superpositionVOC=6V4040+60+60+0.1A6040+60+6040=3 V.ISCis the obtained as 100 mA.(b)UsingLLTHOCRRRVP⋅+=2a load of 90 Ωwill absorb 56.25 mW.(c)It absorbs 75 mW; hence the 30 Ωresistor absorbs more power.SOLUTION 6.4. As both resistor divider ration are the same (3/6), the voltage at A and B is the sameresulting in a VOCof 0 V.RTH=(3K||6K)+(9K||18K)=8 kΩThe relation VOC/ISCcannot be used in this situation.SOLUTION 6.5. Using superpositionVOC=2020×10320×103+5×103-105×10320×103+5×103+20⋅20sin(50t)=14+400sin(50t)V,and