# Compute the charge and energy

1. Sep 27, 2013

### Zondrina

1. The problem statement, all variables and given/known data

The power absorbed by the box is $p(t) = 2.5e^{-4t}W$. Compute the charge and energy delivered to the box for $0 < t < 250ms$.

An image of the problem: http://gyazo.com/7cb5858f681c3c9db33e20b434c2b782

2. Relevant equations

$i = \frac{dq}{dt}$
$q = \int_{-∞}^{t} i dx$

Note that energy over time is the same as work over time.
$\frac{dE}{dt} = p = vi$
$E = \int_{t_1}^{t_2} p dt = \int_{t_1}^{t_2} iv dt$

$v = 50e^{-t}V$
$p = 2.5e^{-4t}W$
$0 < t < 250ms$

3. The attempt at a solution

The question asks to compute the charge and energy delivered for $0 < t < 250ms$.

I don't have any current to work with, but I do have power and voltage.

Wouldn't I just integrate $p(t)$ from 0 to 250? That would give me energy.

Integrating $p(t)/v(t)$ from 0 to 250 would give me charge wouldn't it?

Something doesn't feel right. The answers are different.

The answer for the energy is listed as: 395.1mJ
The answer for the charge is listed as: 8.8mC

Last edited: Sep 27, 2013
2. Sep 27, 2013

### rude man

You're right on both counts.

W = ∫p(t)dt. And you can't determine charge unless you're given either R of the box or V across it.

3. Sep 27, 2013

### Zondrina

4. Sep 27, 2013

### rude man

Change your upper limit of integration to the correct number!

PS what the h*ll is that n in the wolfram calculation??? Has to do with the fact that e is an irrational number, but what is n? Anyway, just pick the first term in their answer which agrees with your given answer.

5. Sep 27, 2013

### Zondrina

I see, so I have to change quantities into seconds with the way things are defined.

250ms = 0.250s

The integral now produces the correct answer of 0.395075(W*s) = 0.395075J ~~ 395.1mJ

For the part of the question about the charge. I can't get the charge from the information I've been given, can I.

6. Sep 27, 2013

### rude man

You always have to use consistent units. joules = watts x seconds. Always convert to the SI units (assuming your course uses it).

No, you cannot. You need either the voltage or the real part of impedance (aka resistance).

7. Sep 27, 2013

### Zondrina

Hmm, out of my own curiosity, hypothetically I attach a load to the box. The load would be absorbing power as per the convention.

Would the load be given in terms of $r(t) = (constant)e^{something}$?

Then I'm guessing using $v = ir$ in some way would help me find the charge.

8. Sep 27, 2013

### rude man

What is r(t)? Some kind of time-varying resistance???

Attaching an external load does not help since you still wouldn't know the current inside the box.

You need R in the box, that way P = I^2 R which you know, then you could solve for I and then Q = ∫I dt.

Or if you knew V, then V^2/R = P = VI & again solve for I and Q.

9. Sep 27, 2013

### Zondrina

Yes, I remember those, they're alternate equations for power dissipation.

Thank you for all your help.

P.S When I said attach a load, I meant put a resistive element into the schematic of the box.

Last edited: Sep 27, 2013
10. Oct 19, 2015

### EEfailiure

Isn't the voltage given as v(t)=50*e-t