Need help analyzing an RC circuit

  • #1

Homework Statement


The circuit contains an ideal battery, two resistors and a capacitor (C= 250 μF).The switch is closed at time t = 0, and the voltage across the capacitor is recorded as a function of time as shown in the graph.

3IbTwPY.jpg


pcgoSMt.png


Homework Equations


time constant = RC


The Attempt at a Solution


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There are a few questions that accompany this diagram, but some of them I am confused about. I am asked for the voltage of the battery, which I put as 9.0 V, and the time constant, which I got as 2.5 s (because that's when the capacitor was charged to 63%). However, I am also asked a few more questions.

I have to find the resistance of resistor R1, and of resistor R2, and the voltage across resistor R1 at t = 2.0 seconds, and the voltage across resistor R2 at t = 2.0 seconds, and the total current produced by the battery at t = 0.The only one I feel like I had an idea as to what to do was the one where I have to find the total current, which I found using the equation time constant = R*C to solve for R, then plugging it into the equation I = V/R to find I. Was this the correct way to approach problem? And will I use the calculated I and Kirchhoff's Rules to find the resistance across the resistors? Because I'm having trouble solving for R1 and R2, so I don't know if I'm even supposed to be using it.
 

Answers and Replies

  • #2
BvU
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Hi there,

Your relevant equation is correct, but you want to be a bit more complete: you also use an equation for the current through the capacitor, so mention it.

From the graph and the description it is clear what happens from t = 0 to t = 15.
At t = 2 you have the voltage across the capacitor and you have the voltage the battery delivers. So what can you say about the voltage across R2 ?

When you calculate the total current, is that the total current the battery delivers, or only the total current in the R1 C branch ?

At t = 15 something changes too! What happens then ? And can you use that to say something about R2 ?
 
  • #3
Hi there,

Your relevant equation is correct, but you want to be a bit more complete: you also use an equation for the current through the capacitor, so mention it.

From the graph and the description it is clear what happens from t = 0 to t = 15.
At t = 2 you have the voltage across the capacitor and you have the voltage the battery delivers. So what can you say about the voltage across R2 ?

When you calculate the total current, is that the total current the battery delivers, or only the total current in the R1 C branch ?

At t = 15 something changes too! What happens then ? And can you use that to say something about R2 ?
Will the voltage across the capacitor at t = 2 be the same as the voltage across R1?
So I am thinking that the current I calculated was just the current for the R1 C branch. So this means if I know R1 and R2 and the current for the R1 C branch, I should be able to use KIrchhoff's Rules to find the total current? I also see that at t = 15, the capacitor is fully charged and starts discharging (?), but I have no clue what that has to do with R2.
 
  • #4
BvU
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Will the voltage across the capacitor at t = 2 be the same as the voltage across R1?
No.
So I am thinking that the current I calculated was just the current for the R1 C branch
Correct.
I also see that at t = 15, the capacitor is fully charged and starts discharging
Correct. (it's "full enough": ##e^{-t/RC} \approx 0## ).
no clue what that has to do with R2
If the switch is opened again at t = 15 (that's what I was fishing for), then what happens ?
 
  • #5
No.
Correct.
Correct. (it's "full enough": ##e^{-t/RC} \approx 0## ).
If the switch is opened again at t = 15 (that's what I was fishing for), then what happens ?
If it's opened, there is no more current flowing through the circuit..?
 
  • #6
BvU
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No. there is a way for current to flow
 
  • #7
No. there is a way for current to flow
I'm honestly so confused. Does the fact that the capacitor stored charge somehow mean that there is still current left that can flow?
 
  • #8
NascentOxygen
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I'm honestly so confused. Does the fact that the capacitor stored charge somehow mean that there is still current left that can flow?
After 15 secs, the capacitor voltage is constantly changing, so some of its charge must be going somewhere. What is the path that current from the capacitor might be taking?
 
  • #9
After 15 secs, the capacitor voltage is constantly changing, so some of its charge must be going somewhere. What is the path that current from the capacitor might be taking?
Do you miss that it's discharging? and wouldn't the current from the capacitor be split up between going to through R and to the battery?
 
Last edited:
  • #10
NascentOxygen
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Do you miss that it's discharging? and wouldn't the current from the capacitor be split up between going to through R and to the battery?
I said "changing", and this covers discharging as well as charging. The switch to the battery is open-circuit, so no current can flow to/from the battery after 15 secs.
 
  • #11
I said "changing", and this covers discharging as well as charging. The switch to the battery is open-circuit, so no current can flow to/from the battery after 15 secs.
So it has to go through R2? Is there reason that it can't go to the battery because since the circuit is open, there is nothing attracting the current towards the battery anymore?
 
  • #12
NascentOxygen
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So it has to go through R2? Is there reason that it can't go to the battery because since the circuit is open, there is nothing attracting the current towards the battery anymore?
"open" means there is no path, the copper path is interrupted by an air gap inside the switch.

I used the description "open-circuit" to distinguish this condition from its opposite, a "short-circuit". An ideal switch is either an open-circuit or a short-circuit in a conductive path.
 
  • #13
BvU
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I'd like to know if there is something in the problem statement about this t = 15 seconds. I have hypothesized that the switch is opened again, but until now we haven't had confirmation.
 
Last edited:
  • #14
NascentOxygen
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I have to find the total current, which I found using the equation time constant = R*C to solve for R, then plugging it into the equation I = V/R to find I. Was this the correct way to approach problem?
There are two time constants visible in the graph. There's that while charging the capacitor, which you found. There is another while the capacitor is discharging. You have not yet measured that second time constant, ##T_2##.
 
  • #15
NascentOxygen
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I'd ike to know if there is something in the problem statement about this t = 15 seconds. I have hypothesized that the switch is opened again, but until now we haven't had confirmation.
Is the v-t graphic not showing up in post #1 for you?
 
  • #16
BvU
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Yes and I know what it means. But I wonder if our poster has an idea, or if it's in a part of the complete problem statement that he/she didn't show us.
 

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