2007quiz3sol

2007quiz3sol - I 1 = 5 6-45 o . + _ 1 Z V oc th I 1...

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Solutions for Quiz #3 Problem 1: Because we have an AC voltage source and a DC current source, they are not at the same frequency and we have to solve this problem by superposition. Consider Frst the contribution from the DC current source. To zero out the voltage source, we have to replace it by a wire (short circuit). At DC, the inductor behaves like a short circuit as well. Therefore the left branch of the circuit is just a wire: + - 0.1 i(t) 2A 5 - + V (t) o1 i(t) and so the contribution from the DC current source is V o 1 = 0 V . Consider now the contribution from the voltage source. We zero out the current source by making it an open circuit, and we can also relabel the inductor with its impedance value for this voltage source: + - 0.1 i(t) 5 + - Z=10j - + V (t) 10 o2 i(t) 0 Using KVL: 10 = (5 . 1 + 10 j ) I I = 0 . 89 6 - 63 o V o 2 = 5 I + 0 . 1 I = 4 . 54 6 - 63 o V o ( t ) = V o 2 ( t ) = 4 . 54 cos (1000 t - 63 o ) 1
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Problem 2: Network A can be represented by its Thevenin equivalent circuit, which looks like this: + _ Z V oc th When we hook up the 1Ω resistor, the current that fows is
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Unformatted text preview: I 1 = 5 6-45 o . + _ 1 Z V oc th I 1 Likewise when we hook up the Thevenin equivalent to the 1F capacitor, the current that fows is I 2 = 10 6-45 o . Two KVL equations can now be written as ollows, or these two circuits: V oc = (1 + Z th ) 5 6-45 o V oc = ( Z th-j ) 10 6-45 o Now we have two equations with two unknowns, V oc and Z th . We can equate the right hand sides o the two equations to eliminate V oc and nd Z th . (1 + Z th ) 5 6-45 o = ( Z th-j ) 10 6-45 o This can be manipulated around to Z th = 2 j + 1 and then we can plug back in to either o those original equations to obtain V oc = 10 2 6 o Problem 3: Part (a) o this problem is Bobrow problem 4.22. Part (b) o this problem is Bobrow problem 4.32. Both o these are very similar to problems you were assigned. The solutions or these two Bobrow problems are in a separate le. 2...
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This note was uploaded on 05/07/2008 for the course ECE 53 taught by Professor Cheng during the Spring '08 term at UCSD.

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2007quiz3sol - I 1 = 5 6-45 o . + _ 1 Z V oc th I 1...

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