Why Does Current Flow in a Capacitor if the Plates Aren't Connected?

[NewtonianAlch]

Homework Statement

http://img803.imageshack.us/img803/382/74211498.jpg

Basically when the switch has been closed for a long time, the inductor acts as a short-circuit and the capacitor as an open-circuit so current only flows through the resistor and then back into the negative terminal of the voltage source.

In the solutions the polarity of the capacitor is marked as opposite to how it's marked for the voltage source, I'm not too sure why.

 

[berkeman]

Homework Statement

http://img803.imageshack.us/img803/382/74211498.jpg

Basically when the switch has been closed for a long time, the inductor acts as a short-circuit and the capacitor as an open-circuit so current only flows through the resistor and then back into the negative terminal of the voltage source.

In the solutions the polarity of the capacitor is marked as opposite to how it's marked for the voltage source, I'm not too sure why.

If it's a polar capacitor (which it would be for so large of a value), then its +/- markings must match the voltage source. Can you post something that shows the incorrect polarity?

 

[NascentOxygen]

Basically when the switch has been closed for a long time, the inductor acts as a short-circuit and the capacitor as an open-circuit so current only flows through the resistor and then back into the negative terminal of the voltage source.

True.

In the solutions the polarity of the capacitor is marked as opposite to how it's marked for the voltage source, I'm not too sure why.

In the solutions? Maybe you are looking at their answer to "Mark the capacitor polarity when the switch opens at t=0"?

 

[berkeman]

In the solutions? Maybe you are looking at their answer to "Mark the capacitor polarity when the switch opens at t=0"?

Yikes, good point! Without a catch diode, that big polar cap is going to be toast if that switch gets opened...

 

[NewtonianAlch]

The actual question was:

The switch in the circuit has been closed for a long time, but is opened at t = 0. Determine i(t) for t > 0.

Then it says for t < 0, the equivalent circuit is as follows, with a re-drawn circuit with a shorted inductor, and an open-circuited capacitor with the polarity reversed on the capacitor.

The only reason I can think of is because when a capacitor starts discharging the current flow is negative?

 

[NewtonianAlch]

Yikes, good point! Without a catch diode, that big polar cap is going to be toast if that switch gets opened...

I don't understand, is that because the value of the inductor is also rather high thus releasing a large amount of energy when that switch is opened? Which way would the current actually flow anyway if there was no diode?

 

[NewtonianAlch]

This is it:

http://img855.imageshack.us/img855/8157/40697987.jpg

 

[vela]

My guess would be because of the direction they assumed the current i(t) is flowing. By convention, you want the positive side to be the end where the current flows into the capacitor. At t=0, v_c=-12 V.

 

[gneill]

Vela's probably hit the nail on the head: the diagram (post #7) is specifying a polarity by which you are to interpret the voltage across the capacitor, not the potential at a particular instant. Why they so specify it is not clear, since it appears that they're looking for the current...

When the switch opens the current that's flowing through the inductor will have no choice but to find its way through the capacitor path. At t=0 the circuit will "look like" a typical LRC oscillator caught partway through a cycle. Component values should tell you whether it's underdamped, overdamped, or critically damped. That should give a pretty good idea of the general shape of the response to expect.

 

[NascentOxygen]

Were they to show the left vertical bus grounded, then there would be no room for speculation. It would follow that Vc was defined.

 

[rude man]

If it's a polar capacitor (which it would be for so large of a value), then its +/- markings must match the voltage source. Can you post something that shows the incorrect polarity?

I don't think the intent of the question was to indicate polarity markings on the capacitor; the existence of polar caps is not usually part of an introductory course ... if you think 1/4F is a big cap (not really, with supercaps now available) what about a 1/4 H inductor?

BTW a supercapacitor would not blow up if reverse-biased. It would just build up its ESR.

 

[sumanth248]

Homework Statement

http://img803.imageshack.us/img803/382/74211498.jpg

Basically when the switch has been closed for a long time, the inductor acts as a short-circuit and the capacitor as an open-circuit so current only flows through the resistor and then back into the negative terminal of the voltage source.

In the solutions the polarity of the capacitor is marked as opposite to how it's marked for the voltage source, I'm not too sure why.

i want to ask a basic question about capacitor.even though capacitor plates are not connected directly then how current starts to flow in circuit?

 

[NascentOxygen]

i want to ask a basic question about capacitor.even though capacitor plates are not connected directly then how current starts to flow in circuit?

Hi sumanth248, welcome to Physics Forums.

A capacitor consists of two parallel plates very close to each other. If electrons are pushed onto one plate, their electric field repels an equal number from the other plate. So if N electrons go into one terminal of a capacitor, N electrons are seen to emerge from the other terminal—thus it appears that current has gone right through it.