RC circuit problem with time constants

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Homework Statement



http://img193.imageshack.us/img193/3696/aasdff.th.png [Broken]

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The Attempt at a Solution



(a) No current, so no time constant since C = 0 for it stores no charge

(b) Current immediately is driven through so [tex]\tau = (R_{1}+R_{2})C[/tex]

(c) I pulled the formula out of the book

[tex]I(t) = \frac{\varepsilon }{R}e^\frac{-t}{\tau}[/tex]

But I am not sure what they want after...
 
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Answers and Replies

  • #2
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Wait I just realized (c) is wrong because there are two resistors...
 
  • #3
ideasrule
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(a) No current, so no time constant since C = 0 for it stores no charge
No. The circuit is in equilibrium at t=0, but that doesn't mean it has no time constant because the time constant depends only on the configuration of the circuit. Also, the capacitor does have charge at t=0, since it had plenty of time to charge up before that time.

(b) Current immediately is driven through so [tex]\tau = (R_{1}+R_{2})C[/tex]
No, because the wire in the middle--containing the switch--provides a short circuit. Current from the capacitor completely bypasses the left part of the circuit.

(c) I pulled the formula out of the book

[tex]I(t) = \frac{\varepsilon }{R}e^\frac{-t}{\tau}[/tex]

But I am not sure what they want after...

Don't just pull formulas out of the book. Part c can easily be solved using Kirchoff's laws. Figure out what the voltage difference across the capacitor is, and try applying the voltage law to the loop at the right. Then consider what contribution the left part of the circuit makes to the current across S.
 
  • #4
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No. The circuit is in equilibrium at t=0, but that doesn't mean it has no time constant because the time constant depends only on the configuration of the circuit. Also, the capacitor does have charge at t=0, since it had plenty of time to charge up before that time.

How do we know it was charged before? Actually how do we know that someone charged it before I am tackling this problem?

No, because the wire in the middle--containing the switch--provides a short circuit. Current from the capacitor completely bypasses the left part of the circuit.

Not 100% what that means lol


Don't just pull formulas out of the book.


Noted!

Part c can easily be solved using Kirchoff's laws. Figure out what the voltage difference across the capacitor is, and try applying the voltage law to the loop at the right. Then consider what contribution the left part of the circuit makes to the current across S.

[tex]\varepsilon - I_1R_1 \pm \frac{q}{c}-I_1R_2=0[/tex]

I used a plus or minus because they didn't indicate the positive and negative terminals of the capacitor

[tex]\varepsilon - I_1(R_1 + R_2) \pm \frac{q}{c}=0[/tex]

[tex]\varepsilon - \frac{\mathrm{d}q }{\mathrm{d} t}(R_1 + R_2) \pm \frac{q}{c}=0[/tex]

[tex] - \frac{\mathrm{d}q }{\mathrm{d} t}(R_1 + R_2) \pm \frac{q}{c}= -\varepsilon[/tex]

[tex] \frac{\mathrm{d}q }{\mathrm{d} t}(R_1 + R_2) = \pm \frac{q}{c} + \varepsilon[/tex]

[tex] \frac{\mathrm{d}q }{\mathrm{d} t}(R_1 + R_2) = \pm \frac{q}{c} + \frac{\varepsilon c}{c}[/tex]

[tex] \frac{\mathrm{d}q }{\pm q+\varepsilon c} = \frac{dt}{c(R_1 + R_2)}[/tex]

Now I stopped here because I am not sure how I should let u = ±q and if du = ±dq or du = 干dq
 
  • #5
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Please...
 
  • #6
ideasrule
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How do we know it was charged before? Actually how do we know that someone charged it before I am tackling this problem?

The question says the switch was closed for a long time before t<0. That means the circuit has been charging for a long time.

Not 100% what that means lol

Try writing out Kirchoff's voltage law for the loop at the right. Now try writing out the law if the loop was completely separate from the rest of the circuit. (Hint: the two equations should be identical)

[tex]\varepsilon - I_1R_1 \pm \frac{q}{c}-I_1R_2=0[/tex]

I used a plus or minus because they didn't indicate the positive and negative terminals of the capacitor

When the circuit charges up, the top of the capacitor becomes positive and the bottom becomes negative, because the top is connected to the positive side of the battery.

However, you don't need to do any of this calculus. At t=0, you know that the capacitor's voltage is epsilon, the battery's emf. Since the switch is closed, you can treat the left and right loops as completely separate. If you do this, what do you get for the current flowing through the two loops? Add them together, taking direction into account, and you'll get the answer.
 
  • #7
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The question says the switch was closed for a long time before t<0. That means the circuit has been charging for a long time.

It says it has been opened for a long time

Try writing out Kirchoff's voltage law for the loop at the right. Now try writing out the law if the loop was completely separate from the rest of the circuit. (Hint: the two equations should be identical)

For the loop on the right, I denote I2 and the one on the left I1

http://img687.imageshack.us/img687/4417/98175083.th.png [Broken]

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[tex]-I_2 R_2 - q/C = 0[/tex]

[tex] \varepsilon - I_1 R_1 = 0[/tex]

I have qualms about the one on the right because I have a negative number equal to a positive number...



When the circuit charges up, the top of the capacitor becomes positive and the bottom becomes negative, because the top is connected to the positive side of the battery.

However, you don't need to do any of this calculus. At t=0, you know that the capacitor's voltage is epsilon, the battery's emf. Since the switch is closed, you can treat the left and right loops as completely separate. If you do this, what do you get for the current flowing through the two loops? Add them together, taking direction into account, and you'll get the answer.

Since my answer for the previous one is wrong, I'll continued from my calculus and I got

[tex]q(t) = e^{\frac{-t}{C(R_1 + R_2)} - \varepsilon c[/tex]

[tex]q'(t) = -\frac{1}{C(R_1 + R_2)}e^\frac{-t}{C(R_1 + R_2)}[/tex]
 
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  • #8
gneill
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When the switch is closed, the loop on the right has only one resistance in it. It's a simple capacitor discharging through a resistor, which as you should know, follows an exponential discharge curve. What is the initial current in that loop when the switch is closed at t=0?
 
  • #9
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Wait so I don't have to worry about the left circuit?

I just want to know, if for some reason that battery was not there and the capacitor was initially charged and placed in a circuit (series for simplicity) with a resistor and it is being discharged as the voltage drops (that is when I connect the capacitor the circuit is open with the resistor and then I close it), how do I write out the equation?

I mean how do I know the positive and negative terminal without the presence's of the battery to tell me like this question?
 
  • #10
gneill
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Wait so I don't have to worry about the left circuit?

Not for the discharge of the capacitor. It only comes into play when you want ot work out the total current flowing through the switch. You can treat both circuits independently and then add the current contributions.

I just want to know, if for some reason that battery was not there and the capacitor was initially charged and placed in a circuit (series for simplicity) with a resistor and it is being discharged as the voltage drops (that is when I connect the capacitor the circuit is open with the resistor and then I close it), how do I write out the equation?

I mean how do I know the positive and negative terminal without the presence's of the battery to tell me like this question?

In that case the problem statement will either tell you the initial polarity, or they will ask a question in such a way that it doesn't matter what the polarity is, such as: "What is the magnitude of the current flowing after 1.32 milliseconds".

For a discharging capacitor the current, charge, and capacitor voltage always go as exp(t/τ), and it's usually not to difficult to determine the τ and the initial value.
 
  • #11
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Okay so here it is

a) [tex]\tau = C(R_1 + R_2)[/tex]

b) [tex]\tau = C(R_2)[/tex]

c)

[tex]-IR_{2} - \frac{q}{C} = 0 [/tex]

[tex]-\frac{dq}{dt}R_2 = \frac{q}{C}[/tex]

[tex]\frac{dq}{q}R_2 = \frac{-dt}{CR_2}[/tex]

[tex]\int_{C\varepsilon }^{q}\frac{dq}{q}R_2 = \int_{0}^{t}\frac{-dt}{CR_2}[/tex]

[tex]ln\left | \frac{q}{C\varepsilon }\right | = \frac{-t}{CR_2}[/tex]

[tex]\left | \frac{q}{C\varepsilon }\right | = e^\frac{-t}{CR_2}[/tex]

Now let [tex]\tau = CR_2[/tex]

[tex]q(t) = C\varepsilon e^\frac{-t}{\tau}[/tex]

[tex]q'(t) = C\varepsilon\frac{-1}{\tau} e^\frac{-t}{\tau}[/tex]

Am I done? Is that what they wanted?
 
  • #12
gneill
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Not quite done. Note that τ = R2*C. So your C*ε/τ becomes ε/R2, the initial current that will come from the capacitor when it is fully charged at time t = 0. You didn't have to resort to calculus to determine this, it should be obvious that this will be the initial current given the initial voltage on the capacitor is ε. Then the exponential decay is "tacked on" to obtain

[tex] I2(t) = \frac{\epsilon}{R2}e^{-t/\tau}[/tex]

I2 is the current through the switch, with respect to time, due to the discharging capacitor.

Now just add to that the current due to the battery pushing current through its resistor, R1.

[tex] I(t) = \frac{\epsilon}{R1} + \frac{\epsilon}{R2}e^{-t/\tau}[/tex]
 
  • #13
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Not quite done. Note that τ = R2*C. So your C*ε/τ becomes ε/R2, the initial current that will come from the capacitor when it is fully charged at time t = 0. You didn't have to resort to calculus to determine this, it should be obvious that this will be the initial current given the initial voltage on the capacitor is ε. Then the exponential decay is "tacked on" to obtain

[tex] I2(t) = \frac{\epsilon}{R2}e^{-t/\tau}[/tex]

What do you mean I don't need calculus...? THERE IS AN EASSIER WAY??!!
 
  • #14
gneill
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What do you mean I don't need calculus...? THERE IS AN EASSIER WAY??!!

Yes. Once you've analyzed the RC or RL circuit once via calculus, you should recognize that all such circuits will have solutions of the same form (because their differential equations all have the same form). The only difference is the value of the time constant and the initial and final conditions. The solutions will always look like A*exp(t/tau) or A*(1 - exp(t/tau)) for discharging or charging respectively. Here A is the magnitude of the circuit quantity you're after: q, or V, or I.

In this case the capacitor C is discharging from an initial voltage of ε down to zero through a resistance R2. So the voltage on the capacitor will have the expression:

[tex] V(t) = \epsilon e^{-t/\tau} [/tex]

and the current will have the expression

[tex] I(t) = \frac{\epsilon}{R2} e^{-t/\tau} [/tex]

The charge will be

[tex] q(t) = \epsilon C e^{-t/\tau} [/tex]

For this problem, all you really had to do was determine the initial voltage on the capacitor thus the initial current, and the applicable time constant. Then the result can be written by inspection. This will save you a lot of time in exams!

In some circuits the charging or discharging will occur over some fixed range of voltages; for example the circuit may be set up so that the capacitor starts at 10V and discharges down to 3V as determined by some external components. Then your ΔV will be the difference between the starting and ending values, in this case 7V.
 
  • #15
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Can I just ask again for part (b), why did we only concern ourselves with the right loop? When the switch is closed, the battery is still hooked up to the circuit, so why do we disregard it?
 
  • #16
gneill
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Can I just ask again for part (b), why did we only concern ourselves with the right loop? When the switch is closed, the battery is still hooked up to the circuit, so why do we disregard it?

It's not disregarded, it's treated separately. When the switch is closed, the battery and its resistor play no part in the operation of the capacitor circuit -- the common component between the loops is a zero ohm "resistor" (the switch).
 
  • #17
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Doesn't the battery still drive charges through the circuit? Even if it splits into the two junction entering the right loop?
 
  • #18
gneill
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Doesn't the battery still drive charges through the circuit? Even if it splits into the two junction entering the right loop?

No, how can it? No potential is developed across zero resistance.
 
  • #19
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No, how can it? No potential is developed across zero resistance.

What does that mean? "developed across zero resistance"
 
  • #20
gneill
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What does that mean? "developed across zero resistance"

Put a current I through resistance R. What potential V is developed across the resistance?
 
  • #21
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V = IR, you mean R2?
 
  • #22
gneill
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No, any resistance! In the case of this circuit, the resistance of the switch is taken to be zero ohms. So no potential can be developed across it no matter how large the current passing through it. With no potential difference, there is nothing to drive current into the next loop. No current from the battery loop can pass into the capacitor loop. All of it takes the zero resistance path back to the battery negative terminal.
 
  • #23
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Wait, how does having zero resistance prevent current from coming through? Shouldn't it having infinite resistance be the shield that blocks the current?
 
  • #24
gneill
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Look at the circuit diagram. The switch connects R1 directly to the negative terminal of the battery. There is no potential from the battery that is part of the other loop, so it cannot drive current there.

Why don't you write the KVL equations for the following circuit and tell me what the value of i2 is.
 

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  • #25
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Oh so the currents are different
 

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