The figure above has been set up for a long time prior to t=0 with the switch cl

Question

The figure above has been set up for a long time prior to t=0 with the switch closed.

A.) Find vc prior to t=0.

B.) Find the steady state value of vc after the switch has been opened for a long time. 

With the switch closed, it short circuits the initial resistor and with the circuit left open for a long time the inductor short circuits and the capacitor is an open circuit, right? Then Vc=12? I'm confused.

And with t>0, isn't Vc the same thing because the circuit doesn't break anywhere so no charge is degrading or anything? ANY help would be SOOO appreciated!

Explanation / Answer

A) For part (A), you are right. Vc=12 V B) For part (B), the question asks for steady state response, so I=V/R=12/(20+40)=0.2 A so Vc=RI=40 (0.2)=8 V

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