The switch in the circuit in has been closed for a long time. At the switch is opened. a) Find i0 for t>=0 (b) find v0 for t>=0
I don't understand the effect of the switch in the position it is in. It's acting as a short circuit, I think, so presumably the capacitor doesn't charge. But I don't understand how they have calculated the initial current. Also, for the second part of the solution, they seem to have inserted a value of 0.1 for p in the voltage near the inductor. I am not sure where they got that from either.
The solution is given below, but there are a couple of things I don't understand:
The Attempt at a Solution
I really wasn't sure why they did what they did. I tried to do a voltage divider to get the voltage at v(0) but then realised it was irrelevant anyway as it acted as a short when the switch was closed so the inductor wasn't charged anyway.
I now realise where they got 0.1 from - the initial current, as the capacitor is shorted and the inductor also acts as a short, so the initial current is 0.1 (from 80/800). But I still don't understand where they got their equation from for initial current, namely: io(0−) = 20/4000 = 5 mA
13.1 KB Views: 582
64.7 KB Views: 291