# Resistor Capacitor Circuit

1. Oct 3, 2012

### smithisize

1. The problem statement, all variables and given/known data

I'm really looking for help with A.) because I think that if I could figure out A.) the rest should be ok. So, here are all parts of the question, but I'm really only concerned with A at the moment!

The diagram below depicts an RC-circuit where C = 5.40 F, R0 = 14.5Ω , R1 = 24.5 Ω, R2 = 27.5 Ω, and V = 7.50 Volts. The capacitor is initially uncharged.

This is how I drew my currents etc:

The circles are for Kirchoff Loops I tried to construct, and they're going clockwise. Along with a loop around the outside of the circuit.

A.) What is the current through R1 immediately after the switch S is closed?
B.) What is the current through R2 immediately after the switch S is closed?
C.) What is the current through the capacitor immediately after the switch S is closed?
D.) What is the current through R1 after the switch S has been closed for a very long time? Assume that the battery does not go dead.
E.)
What is the current through R2 after the switch S has been closed for a very long time? Assume that the battery does not go dead.

2. Relevant equations

V=IR
C=(Q/V)

3. The attempt at a solution
According to Kirchoff's Laws: I1= I2 + I3
So, since V=IR and R1 and R2 are || : I1 = (Vparallel/R1)+(Vparallel/R2) --> I1= V*(1/R1 + 1/R2)
Therefore: I1/(1/R2 + 1/R1) = Vparallel. Where I1 = V/R0.
Then, I2 = Vparallel/R1

But, that's not correct unfortunately. Where am I going wrong? Thanks!

Side note, here are the eqns I got from kirchoff's laws (pretty sure I messed something up):

V - I1R0 - I2R1 = 0 Left loop
-I4R4 + I2R1 = 0 Right loop at t=0 (or basically zero) Also, I4=I3.
V-I1R0-I4R2 = 0 Entire loop again at t=0.

2. Oct 3, 2012

### Staff: Mentor

Hi smithisize; Welcome to Physics Forums.

I1 ≠ V/R0. The potential across resistor R0 is not V; Potential V is dropped across the sum of R0 and R1||R2, so all of V cannot be across just Ro.

You could use your KVL system of equations to determine the individual currents.

Or calculate the total resistance first and use it to find the current I1. With I1 in hand you can then use your other relationships to proceed.

3. Oct 3, 2012

### smithisize

Thanks, I appreciate it.

So, I think I should probably become more familiar with using the KVL sys of eqns. Not my strong point.

Given the eqns I posted above, I substituted I2+I3 for I1 then solved for I3 (using the 1st eqn) and got I3 = (V - I2R0)/(R0+R2)
Then I plugged that value into my third equation and solved for I2 and got I2 = (R2*V)/(R0+R1*R2+R1*R0)
Evaluated: .19765 A

Does that look correct??

EDIT: Well, I was able to solve for I1, and in turn solve for I2. I2 ended up being .144 A. So, which of my eqns is wrong? Where is my KVL going awry?

Last edited: Oct 3, 2012
4. Oct 3, 2012

### Staff: Mentor

No, something's gone wrong in your manipulations. Note that the denominator of your last expression has a resistance term added to resistance2 terms --- can't add mixed units!
0.144A is a good result.

5. Oct 4, 2012

### CWatters

While you can use KVL & KCL (and it might be good practice to do so) I probably would have treated this as a variant of a potential divider circuit to work out the voltage on R1 and from that the currents

For example, initially the capacitor is discharged so..

VR1 = V * (R1//R2) / {(R1//R2) + R0}

IR1 = VR1 / R1

6. Oct 5, 2012

### smithisize

Well thanks to both of you for your replies, really helped me out!