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Behaviour of a RC circuit charging

  • Thread starter Cade
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  • #1
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Homework Statement



A circuit is powered by a 10 V power supply and has a resistor of 500 ohms in series with a capacitor. After 4 seconds, the value of ln(1-(Vc/Vo)) is -2, where Vc is the voltage in the capacitor at a given time and Vo is 10 V.

1) Find the capacitance.
2) Find how long it takes for the potential difference across the capacitor to rise from 0 V to 4 V.

Homework Equations



Vc = Vo(1 - e^(-t/RC))
Tau = RC

The Attempt at a Solution


ln(1-(Vc/Vo))/t = -2/4 = 0.5

Let's try to figure out what this represents
Vc = Vo(1 - e^(-t/RC))
ln Vc = ln Vo(1 - e^(-t/RC))
ln Vc = ln Vo + ln(1 - e^(-t/RC))
ln Vc - ln Vo = ln(1 - e^(-t/RC))
Vc/Vo = 1 - e^(-t/RC)
e^(-t/RC) = 1 - Vc/Vo
-t/RC = ln(1 - Vc/Vo)
-1/RC = ln(1 - Vc/Vo)/t = -0.5

So I have 1/RC = 0.5, therefore the time constant Tau = RC = 2 seconds.

1) (500)(C) = 2 -> Capacitance = 0.004 F

2) If -t/RC = ln(1 - Vc/Vo) then
t = -RC*ln(1 - Vc/v0) = -2*ln(1-4/10) = 1.021 seconds.

Is this correct? I've been told that I have a mistake somewhere, but I don't know where.
 

Answers and Replies

  • #2
ehild
Homework Helper
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3. The Attempt at a Solution
ln(1-(Vc/Vo))/t = -2/4 = 0.5

Let's try to figure out what this represents
Vc = Vo(1 - e^(-t/RC))
ln Vc = ln (Vo(1 - e^(-t/RC)))
ln Vc = ln Vo + ln(1 - e^(-t/RC))
ln Vc - ln Vo = ln(1 - e^(-t/RC))


This whole thing was not needed...

Except for the missing parentheses, everything is correct.


ehild
 
  • #3
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Oops... sorry, I typed it up because I had written it, and I forgot the parenthesis :smile:

I only had the formula for voltage when discharging, so I thought I had to figure out what ln(1-Vc/Vo) meant before I could use it. Thanks for going through it, I will ask my instructor again.
 
  • #4
ehild
Homework Helper
15,427
1,823
You have a rounding error. T=1.02165 s which has to be rounded to t=1.022 s, if you give the result with 3 decimals.

ehild
 
  • #5
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Thanks, but it isn't an online homework problem, the instructor told me the working is incorrect. I'll ask him again when I next see him.
 
  • #6
318
1
1. Homework Statement

A circuit is powered by a 10 V power supply and has a resistor of 500 ohms in series with a capacitor. After 4 seconds, the value of ln(1-(Vc/Vo)) is -2, where Vc is the voltage in the capacitor at a given time and Vo is 10 V.

1) Find the capacitance.
2) Find how long it takes for the potential difference across the capacitor to rise from 0 V to 4 V.

2. Homework Equations

Vc = Vo(1 - e^(-t/RC))
Tau = RC

3. The Attempt at a Solution
ln(1-(Vc/Vo))/t = -2/4 = 0.5

Let's try to figure out what this represents
Vc = Vo(1 - e^(-t/RC))
ln Vc = ln Vo(1 - e^(-t/RC))
ln Vc = ln Vo + ln(1 - e^(-t/RC))
ln Vc - ln Vo = ln(1 - e^(-t/RC))
Vc/Vo = 1 - e^(-t/RC)
e^(-t/RC) = 1 - Vc/Vo
-t/RC = ln(1 - Vc/Vo)
-1/RC = ln(1 - Vc/Vo)/t = -0.5

So I have 1/RC = 0.5, therefore the time constant Tau = RC = 2 seconds.

1) (500)(C) = 2 -> Capacitance = 0.004 F

2) If -t/RC = ln(1 - Vc/Vo) then
t = -RC*ln(1 - Vc/v0) = -2*ln(1-4/10) = 1.021 seconds.

Is this correct? I've been told that I have a mistake somewhere, but I don't know where.
You seem to be junior member like me.
Have you read https://www.physicsforums.com/showthread.php?t=386951 [Broken].
When i read your answer it was to cumbersome that's why i give you this link. If you use tex tags then it became easy to understand what you have written.
 
Last edited by a moderator:
  • #7
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Thanks, I ought to learn LaTeX, but the equations I used didn't have many terms.
 

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