Solve RC Circuit Discharge Homework: 6V, 60Ω, 0.02F

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Homework Help Overview

The discussion revolves around an RC circuit consisting of a 6 V source, a 60 ohm resistor, and a 0.02 F capacitor, with the capacitor initially uncharged. The problem asks for the current flowing through the capacitor after a long time once the voltage source is activated.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the time constant of the circuit and how it relates to the current over time. Questions arise regarding the interpretation of "a very long time" and its implications for the current in the circuit. Some participants explore the behavior of the exponential function in this context.

Discussion Status

The discussion is active, with participants offering insights into the behavior of the circuit over time. There is a focus on understanding the exponential decay of current and the steady state of the circuit. Multiple interpretations of the problem are being explored, particularly regarding the meaning of "long time."

Contextual Notes

Participants note the absence of a specific time frame for "a long time," leading to uncertainty about how to approach the problem. There is an emphasis on understanding the implications of the capacitor's behavior in the circuit as time progresses.

purduegirl
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Homework Statement



A circuit consists of 3 components connected between 2 terminals: A 6 V source, a 60 ohms resistor and a 0.02 F capacitor. The capacitor is initially uncharged. The voltage source is turned on. What is the current flowing through the above capacitor after a very long time?


Homework Equations



I(t) = E/R *e^(-t/time constant)
time constant = (R)(C)

The Attempt at a Solution


time constant = 1.2 s

Because the time is a long time, I thought that I wouldn't have to enter anything in for time. I then used the I= E/R*e^(-1/time constant) to solve for the current. I got I = (6 V/60 ohms)*e^(-1/1.2s) = 4.35E-1. I'm pretty sure I'm wrong here so any help to tell me what I am doing wrong would be greatly appreciated.
 
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Hi purduegirl,

purduegirl said:

Homework Statement



A circuit consists of 3 components connected between 2 terminals: A 6 V source, a 60 ohms resistor and a 0.02 F capacitor. The capacitor is initially uncharged. The voltage source is turned on. What is the current flowing through the above capacitor after a very long time?


Homework Equations



I(t) = E/R *e^(-t/time constant)
time constant = (R)(C)

The Attempt at a Solution


time constant = 1.2 s

Because the time is a long time, I thought that I wouldn't have to enter anything in for time. I then used the I= E/R*e^(-1/time constant) to solve for the current. I got I = (6 V/60 ohms)*e^(-1/1.2s) = 4.35E-1. I'm pretty sure I'm wrong here so any help to tell me what I am doing wrong would be greatly appreciated.

But you did enter something here for time--you entered 1, so you found the current at 1 second.

When they say the time is a very long time, what does that mean? Or alternatively, what is [itex]e^{-t/(\mbox{time constant})}[/itex] for very large t (if you plot it, for example)?
 
They didn't give a specified time for " a long time". That's why I'm not sure what they want here.
 
purduegirl said:
They didn't give a specified time for " a long time". That's why I'm not sure what they want here.

No, there's not a specified number; that's not what they want here.

But try answering my second question; the problem is understanding the behavior of the exponential function, which is important for understanding RC circuits. Make a plot of [itex]\exp(-t/1.2)[/itex]. What happens to the value of that as t gets larger and larger?

Once you know what happens to the exponential function, you'll know what happens to the current, since it's just a constant multiplied by the exponential function.
 
If this is a series circuit, remember that capacitors are technically breaks in the circuit. After a long time, the circuit is in steady state, and the current across branches involving capacitors can set to 0A. When referring to a long time, take the limit as t approaches infinite.
 
So from what I just read in my textbook, long after the switch is closed, the potential difference across the capacitor is nearly equal to the emf and the current is small. Looking at a graph of current as a function of time, I can see that the longer the circuit is flowing, the smaller the current gets.
 
Unforunatly, the only exponential examples my book gives are for finding the voltage across the capacitor (0.632*Voltage of Battery) and for finding the current when the time equals the time constant (0.368*(Voltage of Battery/Resisitivity). I tried the last equation to no avail.
 
purduegirl said:
So from what I just read in my textbook, long after the switch is closed, the potential difference across the capacitor is nearly equal to the emf and the current is small. Looking at a graph of current as a function of time, I can see that the longer the circuit is flowing, the smaller the current gets.

That's right; and as t gets larger and larger (goes to infinity), what would you say the current is? That is the answer they are looking for here.
 
purduegirl said:
So from what I just read in my textbook, long after the switch is closed, the potential difference across the capacitor is nearly equal to the emf and the current is small. Looking at a graph of current as a function of time, I can see that the longer the circuit is flowing, the smaller the current gets.

What could they mean by 'after a very long time'? There's no number to put in corresponding to a 'very long time' is there? I think you've just stated the answer, metaphorically.
 
  • #10
Thanks to all! My prof gave us homework on RC circuits, but never lectured over it. Thanks so much for the help!
 

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