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

In summary, for the given circuit consisting of a 6V source, a 60 ohms resistor, and a 0.02 F capacitor connected between two terminals, with the capacitor initially uncharged and the voltage source turned on, the current flowing through the capacitor after a very long time will approach 0A. This is due to the fact that, as time goes on, the potential difference across the capacitor will become nearly equal to the emf and the current will decrease. This behavior can be observed on a graph of current as a function of time, where the longer the circuit is flowing, the smaller the current gets. Therefore, the current at a very long time can be considered to be 0A.
  • #1
purduegirl
74
0

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.
 
Physics news on Phys.org
  • #2
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)?
 
  • #3
They didn't give a specified time for " a long time". That's why I'm not sure what they want here.
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 
  • #8
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.
 
  • #9
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!
 

1. How do I calculate the time constant for this RC circuit?

The time constant for an RC circuit is equal to the product of the resistance (R) and capacitance (C). In this case, the time constant would be 60Ω * 0.02F = 1.2 seconds.

2. What is the voltage across the capacitor after 1 second?

To calculate the voltage across the capacitor after a certain amount of time, you can use the formula V = V0 * e^(-t/RC), where V0 is the initial voltage, t is the time, R is the resistance, and C is the capacitance. Plugging in the given values, we get V = 6V * e^(-1/1.2) = 3.18V.

3. How long does it take for the capacitor to discharge to 10% of its initial voltage?

The time it takes for the capacitor to discharge to a certain percentage of its initial voltage can be calculated using the formula t = -RC * ln(Vt/V0), where Vt is the target voltage and V0 is the initial voltage. In this case, t = -(60Ω * 0.02F) * ln(0.1/6) = 4.8 seconds.

4. Can I use this circuit to power a 12V light bulb?

No, this circuit is not designed to power a 12V light bulb. The voltage across the capacitor will never exceed 6V, so it cannot power a 12V light bulb.

5. What happens to the voltage across the capacitor as time goes on?

As time goes on, the voltage across the capacitor will decrease exponentially. This means that it will decrease rapidly at first, and then gradually decrease until it reaches 0V. The rate of decrease is determined by the time constant, which is the product of the resistance and capacitance.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
962
  • Introductory Physics Homework Help
Replies
4
Views
266
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
6K
  • Introductory Physics Homework Help
Replies
9
Views
3K
Back
Top