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## Homework Statement

http://javid.us/phys22/week14q28.jpg

E=50.0 V, R=250 ohm, and C=0.50 microF.

The switch is closed for a long time and no voltage is measured across the capacitor. Just after the switch is opened, the voltage across the capacitor is measured to be 150 V. what is the inductance L?

Inductor and capacitor are ideal.

## Homework Equations

q=Qmax*cos(wt+phi)

i=dq/dt=-Imax*w*sin(wt+phi)

U=1/2*L*Imax^2=1/2*(1/C)*Vmax^2=1/2*L*i^2+1/2*(1/C)*v^2

## The Attempt at a Solution

I am having trouble understanding (or my teacher is having trouble understanding) how this circuit will behave.

The way I understand it, when the switch has been closed for a long time, the inductor offers no resistance or emf, therefore the capacitor is shorted, and cannot have any charge built up on it, and therefore cannot have a voltage difference across it.

If at this point we are told that the switch is thrown, then told that the voltage reaches a maximum of 150 V, then 1/2*L*Imax^2=1/2*(1/C)*Vmax^2 , and we get our answer.

However, this is not what the problem states (although I have a hunch it is what was meant) and therefore I am stuck trying to solve what may be unsolvable with the information given. My professor claims that it is doable, and that my logic is flawed, so allow me to share my logic with you and please feel free to point out any errors.

At steady state, the current through the circuit will flow exclusively through the inductor, will be steady, and will have a value of E/R. Since all the components in the circuit are ideal, at steady state, the only component draining power is the resistor, which will then have a voltage difference across it equal in magnitude and opposite in polarity to the battery. Thus I conclude that the voltage across the capacitor and inductor is zero.

Here is where I have the issue. How is it that "just after the switch is opened", there is a voltage across the capacitor of 150 V.

Please help.

I have also been discussing this on my professors forum, http://jinxer.com/topic.asp?TOPIC_ID=424 .

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