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Unformatted text preview: We then find: S1 = -400 rad/s S2 = -1600 rad/s Note that the question states that the switch has been open for a long time. This implies that the inductor is acting like a short circuit. The question also states that no energy is stored in the capacitor. Once the switch closes, the capacitor is still bypassed using the short assumption of the inductor. No charge accumulates in the capacitor, and the 30mA will circulate through the inductor and the current source. The circuit reaches steady state when the switch is closed. (Of course, this wouldnt be the case in a real life circuit, but in this ideal situation, no current reaches the capacitor)....
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- Spring '05
- Electrical Engineering