Behaviour Of A RC Circuit Charging - Physics Forums

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Behaviour of a RC circuit charging
  • Thread starter Thread starter Cade
  • Start date Start date Oct 24, 2011
  • Tags Tags Charging Circuit Rc Rc circuit
Click For Summary

Homework Help Overview

The discussion revolves around the behavior of an RC circuit charging with a 10 V power supply, a 500-ohm resistor, and a capacitor. The original poster presents a problem involving the voltage across the capacitor after a specific time and seeks to find the capacitance and the time taken for the voltage to rise to a certain level.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive relationships involving the voltage across the capacitor using the natural logarithm and exponential functions. They express uncertainty about their calculations and seek clarification on potential mistakes.

Discussion Status

Participants are engaged in examining the calculations presented by the original poster. Some have pointed out specific errors, such as a rounding issue, while others have suggested that the original poster may have misinterpreted the use of logarithmic expressions. There is an ongoing exploration of the assumptions made in the problem setup.

Contextual Notes

Participants note that the problem is not part of an online homework assignment, and the original poster plans to consult their instructor for further clarification on the issues raised in the discussion.

Cade Messages 90 Reaction score 0

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. Physics news on Phys.org
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ehild Science Advisor Homework Helper Messages 15,536 Reaction score 1,917
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 Cade Messages 90 Reaction score 0 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. ehild Science Advisor Homework Helper Messages 15,536 Reaction score 1,917 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 Cade Messages 90 Reaction score 0 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. vkash Messages 316 Reaction score 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 . 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: May 5, 2017 Cade Messages 90 Reaction score 0 Thanks, I ought to learn LaTeX, but the equations I used didn't have many terms.

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