# Homework Help: Energy in Capacitor

1. Jun 30, 2008

### purduegirl

1. The problem statement, all variables and given/known data
An uncharged 21 mF capacitor is completely charged by a 5.5 V voltage source in series with a 60 W resistor. Find the energy stored in the capacitor.

Find the total energy dissipated by the resistor during the charging process.

At what rate is energy being dissipated in the resistor after a time equal to the time constant of the circuit? (Time = 0 when the charging starts).

2. Relevant equations
W=1/2CV
p= V^2/R
RC = (Resistor)(Capacitance)

3. The attempt at a solution

For part one, I took the fact that storage in a capacitor is equal to W = 1/2CV. When I put in my numbers for it, I got 57.75 J. However, when I entered in my answer it is incorrect.

For part two, I used the second equation. When I plugged in my numbers I got 0.504. Again, this is not correct.

For part three, I took the RC equation and got 1.26E-3 s.

2. Jun 30, 2008

### Defennder

Just to check, you sure it's a 60 W and not 60 ohms resistor? Meaning to say that the power dissipated in the resistor is always 60 W?

3. Jun 30, 2008

### purduegirl

It's Ohms. When I copied and pasted the question over it turned ohms in W.

4. Jul 1, 2008

### Defennder

I don't know if there's any other way, but it looks as though you have to solve a differential equation to get the answer. The voltage through the resistor is 60 ohms, and the corresponding potential drop across the resistor is 60 (dq/dt) where I=dq/dt. Potential across the capacitor at any one time is q/C. So adding up these 2 potential drops should give you the voltage across the source:

60(dq/dt) + q/C = 5.5

Solve this DE, and input the initial value q(0)=0 since the capacitor is initially uncharged.

Once you do the above you can then solve every part of the question.

5. Jul 1, 2008

### Redbelly98

Staff Emeritus
You may want to double-check that formula.

6. Jul 1, 2008

### purduegirl

I double checked the formula and found that I wasn't squaring the Velocity. I am still not getting the last two parts so any help on those would be appreciated.

7. Jul 1, 2008

### cepheid

Staff Emeritus
Voltage. Not velocity.

This first part of the problem should be as simple as plugging in some values. Let us know if you still can't seem to get the correct answer.

8. Jul 1, 2008

### cepheid

Staff Emeritus
For the last two parts, Defennder's suggestion does seem to be the best choice. You are correct that the power dissipated in the resistor is given by P = V^2/R, where V is the voltage across the resistor and R is the resistance. *HOWEVER*, V is NOT constant. The voltage across the resistor is a function of time. It changes as the capacitor charges. In a series RC circuit such as this one, that function is determined by a first-order differential equation. Purduegirl, do you know differential equations?

If there's some sneaky way to do part two without calculus, I think we'd all have to think about it some more.

Defennder, I have a question. I'm assuming that your strategy was to use the ODE to solve for v(t), and then to use that to find the function p(t) for the resistor. Then p(t) would be integrated to find the total energy. Here's my question. Over WHAT interval would you integrate? My understanding is that the capacitor voltage doesn't reach the source voltage, it merely approaches it asymptotically. So would you just integrate from zero to infinity and say that that's a good approximation for any reasonably large finite time interval, since the capacitor voltage changes so negligibly after a certain amount of time? Or would you use a rule of thumb? What is the rule of thumb? Three time constants? Five time constants? I certainly can't remember.

Last edited: Jul 1, 2008
9. Jul 1, 2008

### purduegirl

I meant voltage. Was typing in a hurry!

10. Jul 1, 2008

### Defennder

In this case, what I did was to find the expression for current and note that it is an exponentially decaying function of time, meaning to say the capacitor is fully charged when t=infinity. So the energy dissipated in the resistor is just I^2R, and the integration for p(t) with respect to t is done from 0 to infinity. Because the exponential term has a negative power, that means that the first term in the definite integral when evaluating limits dies to 0. In the end this just works out to be a finite positive number. I personally don't use any rule of thumb in such questions, so I just take what the equations tell me at face value, in this case it occurs at t=infinity, so I just take limits at that point.

But your question reminds me of the time when I did a intro solid-state physics course and I came across a question where I was asked to find the penetration distance of the electron into a potential barrier. Now of course this confused me, since I thought if the electron were to penetrate into the barrier it would penetrate all the way and emerge on the other side or not penetrate it at all. I went to see my prof about this question and it turns out that the question should be interpreted to mean "Find the penetration distance of the electron into the potential barrier where the wavefunction of the electron has been found to decay by factor of e". This of course made the problem solvable (since he said that this problem was similar to the way the Beer-Lambert law for light intensity penetration was derived), but at the same time it made me realise that a lot of times it's often unclear in university physics the conventional interpretation of a question should be given before a solution is attempted. Interpret it one way, it's unsolvable and but do it the other way, all of a sudden it appears rather trivial.