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Unformatted text preview: ECE231 Circuits and Systems I Fall 2011 Session 4 Dr. Stewart Personick Office: ECEC Room 331 [email protected] Replacing a voltage source with a current source (and viceversa) What is inside the box? + c b V= 1.6V Rsource= 2 Real voltage source What is inside the box? + c b V= 1.6V i1 How much current, i1 , will flow if we place a short circuit (a wire) between the output terminals (c and d) of this box? Answer: i1= 1.6V/2 = 0.8A Real voltage source Rsource= 2 What is inside the box? c b i= 0.8A Rsource= 2 i1 This real current source is equivalent to the real voltage source on the previous slides in the sense that : from the outside of the box , given access only to terminals c and d, one cannot observe any difference between this box containing a real current source and the previous box containing a real voltage source Real current source What is inside the box? + c b V= 1.6V i1 Example (not a proof): i1 = 1.6V / (2 + R) Vcb = 1.6V x R / (2 + R) R Real voltage source Rsource= 2 What is inside the box? c b i= 0.8A Rsource= 2 i1 Example (not a proof): i1 = 0.8A x 2 / (2 + R) = 1.6V / (2 + R) Vcb = i1 x R = 1.6V x R / (2 + R) R Real current source Replacing a box containing sources and resistors with its Thevenin Equivalent Circuit What is inside the box? + c b V= 60V 10 20 What is the open circuit voltage? Answer: 60V x 20 / 30 = 40V What is the short circuit current? Answer: 60V / 10 = 6A What is the ratio of open circuit voltage to short circuit current? Answer: 40V / 6A = 6.667 (rounded to 4 significant figures) Box 1 What is inside the box? + c b V= 40V 6.667 From the outside of the box , given access only to terminals c and d, one cannot observe any difference between this box (Box 2) and the previous box (Box 1) The circuit inside this box is called the Thevenin Equivalent circuit for the circuitry inside Box 1....
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 Spring '08
 PIETRUCHA
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