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Find the Time Constant of an RC Circuit

  • Thread starter tomrja
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  • #1
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Homework Statement


Two capacitors (c1 = 11 and c2 = 15 μF) are charged in series by a 11 V battery. Find the time constant of the charging circuit. (r = 4 Ω).


Homework Equations


V=E(1-e^(-t/T)) Where T=Tau=RC, E is the Voltage of the power source, and V is the voltage of the capacitor.



The Attempt at a Solution


I am fairly certain that Tau is the time constant in this case, so I solved for Tau, but I don't know what to plug in for the voltage of the capacitor (V). Am I even headed in the right direction with this?
 

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Answers and Replies

  • #2
SammyS
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You have two capacitors in series with a resistor. What is the equivalent capacitance of the two capacitors?

V represents the sum of the voltages across C1 & C2. Instead of using V=E(1‒e(‒ t/τ)), use I = (I0)e(‒ t/τ).
 
  • #3
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I got an equivalent capacitance of 6.3462e-6F. I guess I am not understanding where the currents in the equation come into play. Would I use Q=CV? Where would that get me? Also, I found an equation for the circuit to be V-(R1*I)-(Q/C)=0. What is the difference between I and Io? Isn't there only on current flowing through the circuit? Thanks for helping me out... Im a little lost.
 
  • #4
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I is the current at time t
Io is the max current
 
  • #5
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I is the current at time t
Io is the max current
Ahh okay. But how am I supposed to know what I and Io are?
 
  • #6
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I will be changing with time ... Io is max current ... when do you think that current will be maximum

but i think that time constant should be just RC (c is net capacitance) ... does it give correct answer?
 
  • #7
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and eqn of I given by sammy is wrong and also a bit unusable ...

you know how to define q in capacitor at any time ... just differentiate it with dt
 
  • #8
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and eqn of I given by sammy is wrong and also a bit unusable ...

you know how to define q in capacitor at any time ... just differentiate it with dt
haha... ok. I'm sorry, I'm still lost... I'll show you what I have done and maybe you can guide me from there.

I found a equivalent capacitance and that was 6.3462e-6F. I know that Q=CV=(6.3462e-6F)(11)=6.98077e-5 which is the max charge on the capacitor. I know that I=V/R=(11V)/(4ohm)=2.75A. I have an equation for an RC circuit that is charging: V=E(1-e^(-t/T)) Where T=Tau=RC, E is the Voltage of the power source, and V is the voltage of the capacitor. Is the voltage of the capacitor different from the voltage of the power source, because If they were the same then I would get 0=e^(-t/T) and you cant take the ln of 0. Am I doing something right?
 
  • #9
18
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I got it... I went way more into this problem than I should have. It just ended up being T=RC.

Thanks for you help

Take Care
 
  • #10
1,137
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I got it... I went way more into this problem than I should have. It just ended up being T=RC.

Thanks for you help

Take Care
HA HA HA ...

I was right ...
 

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