Question:

The circuit shown is at steady state before the switch opens at time t=0. Determine the voltage accross the inductor for t>0. The op-amp is assumed to be ideal.

Attempt:

I'm following the method the book uses for these types of questions which is to first convert the circuit connected to the capacitor to its Thévenin equivalent. So first I find the open circuit voltage and I get this to be 10V. Next I want the short circuit current so that I can get the Thévenin resistance. But if I replace the capacitor by a short circuit then I have a problem. The op-amp is ideal so no current flows along the inputs, which means that the short circuit current must flow around the outer loop. But then i'm going to get 3 voltage drops and no voltage rise

Thanks for any help!
Jon.

 

If switch is close at t < 0 (steady state before the switch opens)
So the capacitor voltage is equal 5V. And since we have the ideal opamp in the circuit Rout=0.

http://en.wikipedia.org/wiki/Operational_amplifier#Ideal_op-amps

So the only resistance it is seen by the capacitor is 20K resistor.
And at time T=0 (switch opens) the opamp voltage rise to 10V.