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Unformatted text preview: I 1 = 5 645 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 = 10 645 o . Two KVL equations can now be written as ±ollows, ±or these two circuits: V oc = (1 + Z th ) × 5 645 o V oc = ( Z thj ) × 10 645 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 ²nd Z th . (1 + Z th ) × 5 645 o = ( Z thj ) × 10 645 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 ²le. 2...
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 Spring '08
 Cheng
 Alternating Current, Frequency, Volt, Thévenin's theorem, Voltage source, DC current source

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