Charging then discharging Capacitor.

[SherlockOhms]

Homework Statement

A capacitor is charged up to it's maximum charge capacity. Once it's reached it's maximum charge capacity, a switched is open which isolates the charged capacitor in a circuit of it's own, containing itself and a resistor (The emf used to charge the capacitor is no longer involved). Draw a rough graph outlining the current in the isolated loop's behaviour with respect to time.

Homework Equations

The Attempt at a Solution

Well, as the capacitor discharges, the current will increase to a maximum value and then eventually die off after the capacitor has fully discharged due to there being no emf to drive the current around the circuit. So, the graph would look something like a hill. Sort of going up (I increasing) and eventually going down ( I decreasing). Does this sound correct?

 

[gneill]

Homework Statement

A capacitor is charged up to it's maximum charge capacity. Once it's reached it's maximum charge capacity, a switched is open which isolates the charged capacitor in a circuit of it's own, containing itself and a resistor (The emf used to charge the capacitor is no longer involved). Draw a rough graph outlining the current in the isolated loop's behaviour with respect to time.

Homework Equations

The Attempt at a Solution

Well, as the capacitor discharges, the current will increase to a maximum value and then eventually die off after the capacitor has fully discharged due to there being no emf to drive the current around the circuit. So, the graph would look something like a hill. Sort of going up (I increasing) and eventually going down ( I decreasing). Does this sound correct?

Why would the current increase to a maximum? What would be the current before the switch opens? (Say it was closed for a very long time). Can it ever be higher than that?

 

[SherlockOhms]

I though that as the capacitor charges it eventually stops the current from flowing? So, I thought that once it begins discharging it would initially be 0.

 

[gneill]

I though that as the capacitor charges it eventually stops the current from flowing? So, I thought that once it begins discharging it would initially be 0.

That depends on where the switch and resistor are. Is the resistor across the capacitor at all times? What then is the voltage across the resistor when the capacitor is charged?

 

[SherlockOhms]

I'll draw out a picture of the circuit and then post the picture here in awhile.

 

[SherlockOhms]

If the capacitor was fully charged, wouldn't the current be pretty low and therefore the voltage drop across the resistor relatively low too?

 

[gneill]

If the capacitor was fully charged, wouldn't the current be pretty low and therefore the voltage drop across the resistor relatively low too?

It depends upon the details of the circuit. Can you provide a diagram?

 

[SherlockOhms]

There. So, S1 closes at the start (S2 is open) leaving the capacitor to charge fully. Once it's fully charged S2 is closed and S1 is opened (Them emf source is now excluded for the mini circuit to the right).

 

[gneill]

Okay. Presumably the capacitor has units of Farads (F), not Henrys (H).

What is the initial potential across the capacitor when S2 closes? What then is the potential across the 10 Ω resistor?

 

[SherlockOhms]

Yup. My bad. I was looking at an LR circuit just before drawing that! The potential would Q/C. Q being the charge to which was initially charged to. Yes?

 

[gneill]

What potential (volts) is on the capacitor?

 

[SherlockOhms]

As in the drop as you move across it? I would've thought it was Q/C. Seeing as Q = |V|C.

 

[SherlockOhms]

Well, that's the potential difference. Not necessarily a drop.

 

[gneill]

As in the drop as you move across it? I would've thought it was Q/C. Seeing as Q = |V|C.

Yes, well what's V? Give a number.

 

[SherlockOhms]

After S1 is opened and S2 is closed, it'll initially just be 2V. The emf supplied. Right?

 

[gneill]

After S1 is opened and S2 is closed, it'll initially just be 2V. The emf supplied. Right?

Correct. That's the potential across the capacitor at time t=0⁺. So what's the potential across the 10 Ω resistor at that time?

 

[SherlockOhms]

0V? That's if I'm sticking by my initial guess that I at time 0 is 0. This seems kinda odd though.

 

[SherlockOhms]

That can't be right actually can it? Then there'd be a net potential difference around a closed loop which doesn't make sense at all.

 

[gneill]

That can't be right actually can it? Then there'd be a net potential difference around a closed loop which doesn't make sense at all.

Right. So what must the potential across the resistor be?

 

[SherlockOhms]

2V. The current has to be 0.2 A initially. Does this current like instantly start flowing once the switches are opened and closed respectively? I mean, the current was 0 for t<0 and then at t=0 it just jumps to 0.2 A?

 

[gneill]

2V. The current has to be 0.2 A initially. Does this current like instantly start flowing once the switches are opened and closed respectively? I mean, the current was 0 for t<0 and then at t=0 it just jumps to 0.2 A?

Yes, it just jumps. Resistors have no trouble accommodating instantaneous current changes.

 

[SherlockOhms]

Cool. Thanks for the explanation, man.