What is the Time Constant for a Capacitor Circuit with Changing Capacitance?

  • Thread starter jcpwn2004
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In summary: So you can't really answer that.Ok so I did V=IR 50V=.2R so R = 250 ohm. Then t=RC so the new RC = .25. So I(t)=.2e^-4t. then I'm lost...Well let's be careful about what we're doing here, we need to establish a few constant using Ohm's Law, and an additional constant, the time constant. We want I when the voltage drop across the resistor is 50V, and the initial current, when the voltage drop across the resistor is 100V, and tau. We want t to always be less than
  • #1
jcpwn2004
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Homework Statement



http://img231.imageshack.us/img231/7518/physicsdw3.jpg [Broken]

Homework Equations



Q=Qfinal(1-e^(-t/rc) I=Io(e^-t/rc) V=q/c T=RC

The Attempt at a Solution



A.) I think that'd it would be brighter since there is more current through the circuit but have no clue...

B.) Qfinal is .1 and Io = .2 I thought that T=RC though and I get .5, but the answer in the book RC = to -2...

D.) I thought I'd work backward for this problem. I found Ceq equal to 1200 uF. Then I did Q = V/C so Q = .0328. I know from the answer ke that Q(t) = .1(1-e^-2t) so then I get .0328 = .1(1=e^-2t) so e^-2t = .672 and then =2t = ln.672 but then i get t = .199 and the answer key says .416...
 
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  • #2
Let's start with part a. There are a few things you should know about the behavior of capacitors. In a short time, that is, for a time much less than the time constant [tex] \tau = RC [/tex], the capacitor behaves like a closed switch. For a long time, or a time much great than the time constant, it will behave like an open switch. Knowing this, what can you say about the brightness of the bulb immediately after the switch is closed? A long time after it is closed?
 
  • #3
Immediately after not much would happen and a long time after the bulb would get brighter?
 
  • #4
If the capacitor behaves like a closed switch in the short time (immediately after the switch is closed)..think about it..
 
  • #5
I really have no idea, if it acted ilke a closed switch that would mean it would get brighter immediately? But then wouldn't that mean that after a long time it would start to dim out?
 
  • #6
You've got it. After a long time the charge begins to build up on the capacitor, which slows the current until the current is zero. Think about Ohm's Law. No current, no voltage drop across the light bulb, right?

For part b, I don't have a calculator in front of me, but you have the correct equations for the current and charge as functions of time. The charge is gathering at an exponential rate on the capacitor, and the current is decreasing as mentioned above. Does that part make sense?
 
  • #7
Well i sovled for Qfinal and Io but I can't figure out how to solve for RC. I use [tex] \tau = RC [/tex] so T = 500x1000uF but that's .5 and the answer in the book has it at 2.
 
  • #8
jcpwn2004 said:
Well i sovled for Qfinal and Io but I can't figure out how to solve for RC. I use [tex] \tau = RC [/tex] so T = 500x1000uF but that's .5 and the answer in the book has it at 2.

It has 2 seconds for the time constant? If it does, I would disagree, I get 0.5 seconds for tau as well.
 
  • #9
hotcommodity said:
It has 2 seconds for the time constant? If it does, I would disagree, I get 0.5 seconds for tau as well.

yah the answer in the book is I(t)=.2e^-2t Q(t)=.1(1-e^-2t)
 
  • #10
OHH haha. That's not the time constant, they just simplified. They have:

[tex] e^{\frac{1*t}{.5}} = e^{2t} [/tex]

Keep in mind that the time constant tau is in the denominator.
 
  • #11
hotcommodity said:
OHH haha. That's not the time constant, they just simplified. They have:

[tex] e^{\frac{1*t}{.5}} = e^{2t} [/tex]

Keep in mind that the time constant tau is in the denominator.

oh duh lol cool, thanks for the help, so what am I doing wrong for part d?
 
  • #12
Hmm, I'm getting .346 seconds. I'll work through it one more time..just a sec.
 
  • #13
hotcommodity said:
Hmm, I'm getting .346 seconds. I'll work through it one more time..just a sec.

well the answer is .416 so we both suck lol
 
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  • #14
Ok I found the answer. You won't need the charge equation, you want the current equation. You're going to have a new time constant, do you see where I'm going with this?
 
  • #15
hotcommodity said:
Ok I found the answer. You won't need the charge equation, you want the current equation. You're going to have a new time constant, do you see where I'm going with this?

Ok so I did V=IR 50V=.2R so R = 250 ohm. Then t=RC so the new RC = .25. So I(t)=.2e^-4t. then I'm lost...
 
  • #16
Well let's be careful about what we're doing here, we need to establish a few constant using Ohm's Law, and an additional constant, the time constant. We want I when the voltage drop across the resistor is 50V, and the initial current, when the voltage drop across the resistor is 100V, and tau.

[tex] \tau = (.0012F)(500 \Omega) = blank [/tex]

I at the time Vdrop is 50V: I = V/R = fill in blank

Find I when Vdrop is 100V using Ohm's Law above.

Now you should be able to set up your equation, and solve for t.
 
  • #17
I'm not sure why you are using I(t) to calculate the time it takes for the voltage to drop to 50V.

You can use:

[tex]V(t)=V_0e^-^t^/^R^C[/tex]

You know the new capacitance, you know the voltage drop across the capacitors (which is the same as that of the resistor), so just calculate t.
 
  • #18
Snazzy said:
I'm not sure why you are using I(t) to calculate the time it takes for the voltage to drop to 50V.

You can use:

[tex]V(t)=V_0e^-^t^/^R^C[/tex]

You know the new capacitance, you know the voltage drop across the capacitors (which is the same as that of the resistor), so just calculate t.

It's the same thing except you're not using Ohm's Law to find constants in the equation.
 

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy in the form of an electric field. It consists of two conductors separated by an insulating material, known as a dielectric.

2. How does a capacitor work?

A capacitor works by storing charges on its two conductors, creating an electric field between them. When a voltage is applied to the capacitor, it charges up and stores electrical energy. This stored energy can then be released when needed.

3. What is the purpose of a capacitor?

Capacitors have various purposes in electronic circuits, including filtering out noise, stabilizing power supplies, and storing energy for quick release. They are also used in timing circuits and as coupling devices between different parts of a circuit.

4. How do I choose the right capacitor for my circuit?

The right capacitor for your circuit depends on factors such as capacitance, voltage rating, and frequency. You should also consider the type of dielectric material used, as different materials have different properties. Consult a capacitor datasheet or seek advice from an electronics expert to choose the best capacitor for your circuit.

5. What are some common problems with capacitors?

Some common problems with capacitors include leakage, short circuits, and breakdown. Leakage occurs when the dielectric material loses its insulating properties, leading to a decrease in capacitance. Short circuits can occur due to physical damage or manufacturing defects. Breakdown happens when the voltage across the capacitor exceeds its rating, causing it to fail. Proper handling and storing of capacitors can help prevent these problems.

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