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Unformatted text preview: Therefore . Finally, P8.36 Before the switch opens, . After the switch opens the part of the circuit connected to the inductor can be replaced by it's Norton equivalent circuit to get: Therefore . Next, Finally , P8.310 First, use source transformations to obtain the equivalent circuit for t < 0: for t > 0: P8.46 , and P8.68 For t < 0, the circuit is: After t = 0, replace the part of the circuit connected to the capacitor by its Thevenin equivalent circuit to get: P8.611 For t > 0 the circuit is at steady state so the inductor acts like a short circuit: Apply KVL to the supermesh corresponding to the dependent source to get Apply KVL to get so and (a) For t > 0, find the Norton equivalent circuit for the part of the circuit that is connected to the inductor. Apply KCL at the top node of the dependent source to see that . Then Apply KVL to the supermesh corresponding to the dependent source to get Apply KCL to get Then Then , so and now...
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 Winter '10
 xiejiang

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