Analyzing Current in RC Circuits: A Scientific Approach

[timnswede]

Homework Statement

In the circuit below the switch is initially open and both capacitors initially uncharged. All resistors have the same value R.
a.) Find the current through R2 just after the switch is closed.
b.) Find the current through R2 a long time after the switch has been closed.
c.) Describe qualitatively how you expect the current in R3 to behave after switch is closed.

0

Homework Equations

V=IR
Kirchhoff's Rules

The Attempt at a Solution

I have only attempted part a so far because I don't think I even did that part right. Well right after the switch is closed the capacitors are an open circuit right? So the current only goes through R1 and R2, so I if current goes up from the positive terminal of the battery and my loop is clockwise I got V-2R=0, so R=V/2.

 

[gneill]

I have only attempted part a so far because I don't think I even did that part right. Well right after the switch is closed the capacitors are an open circuit right?

Oops. Quite the reverse. The capacitors look like short circuits. The charge on a capacitor cannot change instantaneously so immediately after the switch is closed they will still have 0 V as a potential difference.

 

[timnswede]

OK, since there is no potential difference on either of the capacitors that would mean that there is no potential difference across R2 either then since it is in parallel with C1? And I just realized I did Kirchhoff's rules wrong, but then for part b) it would be V-2RI=0, so I=V/2R, since there is no current through the capacitors a long time after the switch has been closed.

 

[gneill]

Yes, that looks good.

 

[timnswede]

Yes, that looks good.

Actually after thinking about it a bit more, I think I confused myself more. Since there is no potential difference on the capacitor then electrons won't go there, but the battery still has a potential difference, so wouldn't the electrons still flow through R2, R1, the switch and then to the positive plate of the battery?
Also for part c, electrons won't go to R3 right when the switch is closed, or after a very long time, but they will go to R3 when the capacitors are charging right?

 

[gneill]

Actually after thinking about it a bit more, I think I confused myself more. Since there is no potential difference on the capacitor then electrons won't go there, but the battery still has a potential difference, so wouldn't the electrons still flow through R2, R1, the switch and then to the positive plate of the battery?

You should get used to working with conventional current (positive charge carriers) rather than electron flow. All the literature you will use will assume conventional current and you don't need the added "translation" headache when things get complicated.

Current will definitely flow to the capacitors. Immediately after the switch closes they behave just like wires (no resistance). They will begin to charge up and a potential difference will develop over time. C1 being in parallel with R2 will initially grab all the current.

Also for part c, electrons won't go to R3 right when the switch is closed, or after a very long time, but they will go to R3 when the capacitors are charging right?

Right.

 

[timnswede]

You should get used to working with conventional current (positive charge carriers) rather than electron flow. All the literature you will use will assume conventional current and you don't need the added "translation" headache when things get complicated.

Current will definitely flow to the capacitors. Immediately after the switch closes they behave just like wires (no resistance). They will begin to charge up and a potential difference will develop over time. C1 being in parallel with R2 will initially grab all the current.

Right.

Ohh OK, thank you for the explanation. It makes sense to me now.