ee215-examfs-w04

# ee215-examfs-w04 - EE215 Fundamentals of Electrical...

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EE215: Fundamentals of Electrical Engineering Final Exam March 15, 2004 Time: 2:30 – 4:20pm, 110 minutes Class: _ EE215 _____________ Section: ____________________ Name: ____________________ Student ID: ____________________ Please show all steps in solving the problems. Clearly mark your final answers and don’t forget the units. Total number of points: 200. Problem 1 (40 points) (a) By source transformation, find the Norton equivalent with respect to the terminals a and b for the circuit above. (20 points) Answer: 30V source transforms to 6A and 5 is in parallel with 20 . 5||20 = 4. 6A source is transformed to 24V and 4 is in series with 8 , which combine to 12 . 24V source is then transformed again to 2A which combines with 8A having a 6A source pointing down. Another transformation gives us a 72V source, which allows us to combine 12 and 6 to 18 . The last transformation gives us a 4A source with 18 and 6 in parallel. 18||6 = 4.5. This is the Norton equivalent. + 30V 8 6 20 6 b a 5 8A 6A 8 6 20 6 b a 5 8A

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2A 6 6 b a 8A 12 24V 8 6 4 6 b a 8A + 72V 6 6 b a + 12 4A 6 b a 18 4.5 4A
(b) Find the Thévenin equivalent circuit. (5 points) Answer: (c) An easier method to find R th is to deactivate the independent sources and simply calculate the equivalent resistance. Redraw the original circuit with the sources deactivated and verify the value of R th . (10 points) Answer: Voltage source is replaced by short, and current source is replaced by open circuit. R th = (6+(5||20)+8)||6 = 18||6 = 4.5 . (d) Now imagine that both independent sources of the original circuit are replaced by current

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ee215-examfs-w04 - EE215 Fundamentals of Electrical...

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