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

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Discussion Overview

The discussion revolves around the behavior of a capacitor in a circuit when a switch is opened after being closed for a long time. Participants explore the implications of current flow in the capacitor despite the plates not being directly connected, examining concepts related to circuit analysis, capacitor polarity, and the role of inductors.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • Some participants explain that when the switch is closed, the inductor behaves as a short circuit and the capacitor as an open circuit, allowing current to flow only through the resistor.
  • There is confusion regarding the polarity of the capacitor in the solutions, with some participants questioning why it is marked opposite to the voltage source.
  • One participant suggests that the polarity of the capacitor may need to be considered when the switch opens, as it could affect the current flow.
  • Another participant raises concerns about the potential damage to a polar capacitor if the switch is opened without a catch diode, implying that the inductor's energy release could be significant.
  • Some participants discuss the interpretation of current direction and voltage across the capacitor, suggesting that conventions may dictate how these are represented in the circuit diagrams.
  • A participant mentions that the existence of polar capacitors is not typically covered in introductory courses, questioning the relevance of such details in the context of the problem.
  • One participant provides a basic explanation of how current appears to flow through a capacitor, emphasizing the role of electric fields and electron movement between the plates.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of capacitor polarity and current flow, indicating that multiple competing perspectives remain unresolved. There is no consensus on the implications of the capacitor's behavior when the switch is opened.

Contextual Notes

Participants note that the discussion involves assumptions about circuit behavior, the role of inductors, and the characteristics of polar capacitors, which may not be fully addressed in the problem statement.

NewtonianAlch
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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.
 
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NewtonianAlch said:

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?
 
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NewtonianAlch said:
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"?
 
NascentOxygen said:
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... :bugeye:
 
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?
 
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berkeman said:
Yikes, good point! Without a catch diode, that big polar cap is going to be toast if that switch gets opened... :bugeye:

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?
 
This is it:

http://img855.imageshack.us/img855/8157/40697987.jpg
 
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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, vc=-12 V.
 
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.
 
  • #10
Were they to show the left vertical bus grounded, then there would be no room for speculation. It would follow that Vc was defined. :smile:
 
  • #11
berkeman said:
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? :rolleyes:

BTW a supercapacitor would not blow up if reverse-biased. It would just build up its ESR.
 
  • #12
NewtonianAlch said:

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?
 
Last edited by a moderator:
  • #13
sumanth248 said:
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.
 

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