Current Through a Capacitor after switch closed

In summary: VR2=Vo, then Vo/(R2+R3)=Ic. Thanks for the help!In summary, current flows through the resistor and capacitor when the switch is closed, but when the switch is opened the current from the capacitor flows in the opposite direction as the current from R1 and R2.
  • #1
eatsleep
42
0
http://imgur.com/PQ4XCEo



2.V=IR



3. I know the capacitor has a voltage of 5.5 because it is charged to the supply voltage, I don't understand what happens to current when the switch is opened.

 

Attachments

  • asdf.png
    asdf.png
    9.7 KB · Views: 1,525
Physics news on Phys.org
  • #2
eatsleep said:
http://imgur.com/PQ4XCEo
2.V=IR
3. I know the capacitor has a voltage of 5.5 because it is charged to the supply voltage, I don't understand what happens to current when the switch is opened.

Homework Statement

Check your assumptions regarding the fully charged capacitor voltage.

Suppose you removed the capacitor from the circuit. What would be the voltage presented at the open terminals where it was connected?

When the switch opens, can you identify a closed path for current to flow (a circuit) that includes the capacitor as a source of voltage?
 
  • #3
When the capacitor is removed the open terminals have 0 voltage because there is no current flowing through the terminals.

Would the loop when the switch is open only include the R2 and R3 resistors and the capacitor? Those resistors are in series so the current through the capacitor is 5.5/(R2+R3)

Is this close?
 
  • #4
eatsleep said:
When the capacitor is removed the open terminals have 0 voltage because there is no current flowing through the terminals.

Would the loop when the switch is open only include the R2 and R3 resistors and the capacitor? Those resistors are in series so the current through the capacitor is 5.5/(R2+R3)

Is this close?

Close. You've identified the correct (sub)circuit. However, the voltage at the open capacitor terminals will not be zero. Consider, for example, an isolated battery. No current flows, yet there is still a potential difference at its open terminals... so potential between points does not require that a current flows.

In the given circuit when the switch is closed, some current will still flow through certain components even when the capacitor has reached its peak voltage and its current is zero. What current still flows in the circuit?
 
  • #5
gneill said:
Close. You've identified the correct (sub)circuit. However, the voltage at the open capacitor terminals will not be zero. Consider, for example, an isolated battery. No current flows, yet there is still a potential difference at its open terminals... so potential between points does not require that a current flows.

In the given circuit when the switch is closed, some current will still flow through certain components even when the capacitor has reached its peak voltage and its current is zero. What current still flows in the circuit?

When the switch is closed the current would be flowing through R1 and R2, but when the switch opens the current from the capacitor flows in the opposite direction as the current from R1 and R2 so the total is Ic-IR1R2?
 
  • #6
eatsleep said:
When the switch is closed the current would be flowing through R1 and R2, but when the switch opens the current from the capacitor flows in the opposite direction as the current from R1 and R2 so the total is Ic-IR1R2?

For the moment just consider the currents that flow while the switch is closed. When the capacitor is "full", Ic is zero. So what will be the potential difference V2?

attachment.php?attachmentid=55786&stc=1&d=1360966350.gif
 

Attachments

  • Fig1.gif
    Fig1.gif
    3.4 KB · Views: 1,985
  • #7
To find the potential difference, would that require that I just use voltage division to find the voltage across R2?
 
  • #8
eatsleep said:
To find the potential difference, would that require that I just use voltage division to find the voltage across R2?

Yup. There's only the one current flowing at steady state!
 
  • #9
gneill said:
Yup. There's only the one current flowing at steady state!

So when the capacitor is charged the only current flowing is through R2. Now when the switch is opened the capacitor starts to discharge. Can i use the equation (-Vo/R)*(-t/RC), substituting t=0 for immediately after the switch closes and use (R2+R3) as R?
 
  • #10
eatsleep said:
So when the capacitor is charged the only current flowing is through R2. Now when the switch is opened the capacitor starts to discharge. Can i use the equation (-Vo/R)e(-t/RC), substituting t=0 for immediately after the switch closes and use (R2+R3) as R?

(Don't forget the exponential) Yes. At the moment the switch opens, the full potential Vo drives the current, so the ##e^{(...)}## term is 1.

What value did you find for Vo?
 
  • #11
gneill said:
(Don't forget the exponential) Yes. At the moment the switch opens, the full potential Vo drives the current, so the ##e^{(...)}## term is 1.

What value did you find for Vo?

Vo would be equal to Vs?
 
  • #12
eatsleep said:
Vo would be equal to Vs?

Nope. Did you not look at the voltage divider (R1 and R2) that supplies the voltage to the capacitor branch while the capacitor is charging?
 
  • #13
gneill said:
Nope. Did you not look at the voltage divider (R1 and R2) that supplies the voltage to the capacitor branch while the capacitor is charging?

oh yea, so VR2=Vo, then Vo/(R2+R3)=Ic. Thanks for the all the help
 

FAQ: Current Through a Capacitor after switch closed

What is a capacitor?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductive plates separated by an insulating material, known as a dielectric.

How does a capacitor work?

When a voltage is applied to a capacitor, it causes the electrons on one plate to move towards the other plate, creating an electric field between them. The capacitor then stores this electrical energy until it is discharged.

What is "Current Through a Capacitor after switch closed"?

"Current Through a Capacitor after switch closed" refers to the flow of electric current through a capacitor after a switch is closed and a voltage is applied. This current is initially high, but decreases over time as the capacitor becomes charged.

Why does the current through a capacitor decrease over time?

The current through a capacitor decreases over time because the capacitor becomes fully charged, meaning that the electric field between the plates can no longer support the flow of current. As a result, the current through the capacitor decreases until it reaches zero.

What factors affect the current through a capacitor after switch closed?

The current through a capacitor after switch closed can be affected by the capacitance of the capacitor, the voltage applied, and the resistance of the circuit. A higher capacitance or voltage will result in a higher initial current, while a higher resistance will cause the current to decrease more slowly.

Similar threads

Replies
4
Views
1K
Replies
22
Views
4K
Replies
7
Views
3K
Replies
12
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
Replies
7
Views
5K

Back
Top