Behaviour of a RC circuit charging

In summary, 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.
  • #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.
 
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  • #2
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
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
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
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
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
Thanks, I ought to learn LaTeX, but the equations I used didn't have many terms.
 

1. What is a RC circuit charging?

A RC circuit charging refers to the process of charging a capacitor in a circuit that consists of a resistor and a capacitor. When a capacitor is connected to a voltage source through a resistor, it will gradually charge up to the same voltage as the source.

2. How does a RC circuit charging work?

In a RC circuit charging, the resistor limits the flow of current, causing the capacitor to charge slowly. As the capacitor charges, the voltage across it increases until it reaches the same voltage as the source. At this point, the capacitor stops charging and the circuit reaches steady state.

3. What is the time constant in a RC circuit charging?

The time constant in a RC circuit charging is the product of the resistance and capacitance in the circuit. It represents the time it takes for the capacitor to charge to 63.2% of the source voltage. A larger time constant means a slower charging process.

4. What factors affect the behaviour of a RC circuit charging?

The behaviour of a RC circuit charging is affected by several factors, including the resistance and capacitance values, the source voltage, and the initial charge on the capacitor. A higher resistance or capacitance will result in a longer charging time, while a higher source voltage will result in a faster charging time.

5. How is the voltage across the capacitor related to time in a RC circuit charging?

In a RC circuit charging, the voltage across the capacitor increases gradually over time until it reaches the same voltage as the source. The rate of increase is determined by the time constant, with the voltage increasing faster at the beginning and slowing down as it approaches the source voltage.

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