Why Does the Voltmeter Read 20V When the Switch is Turned Off in an LC Circuit?

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The discussion centers on the behavior of an LC circuit when the switch is turned off at time t=0, resulting in a voltmeter reading of 20V. It is established that at t=0, the inductor behaves like infinite resistance, preventing current from passing through the first resistor, thus maintaining the voltage across the voltmeter at 20V. The analysis utilizes Kirchhoff's laws and differential equations to explain the transient behavior of inductors and capacitors, emphasizing that inductors resist changes in current while capacitors resist changes in voltage.

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Hello,

This is just a conceptual question I have for the image below.
http://img22.imageshack.us/img22/7489/44840348.jpg

If the switch is turned off at t=0, why is the voltage across the voltmeter=20V? I understand that current in all inductors is zero. But wouldn't at t=0, current would need to pass through the first resistor, and therefore a voltage has dropped making voltmeter reading less than 20V?

Thank you in advance.
 
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First consider a simple LC curcuit. Apply kisrchoff's and solve the differential equation for current. You will get that An inductor behaves like infinite resistance at t=0 and a pure conductor at t=infinity. Form this i think that you would be able to solve the problem

And just for extra information: A capictor behaves like a pure conductor at t=0 and as an infite resistance at t=infinity
 

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