How Do You Calculate Charge and Energy from Power and Voltage Over Time?

[STEMucator]

Homework Statement

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

Homework Equations

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

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

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

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

 

[rude man]

Homework Statement

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

Homework Equations

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

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

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

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

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.

 

[STEMucator]

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.

If i integrate the power over the time interval, I get $0.625$:

http://www.wolframalpha.com/input/?i=integrate+2.5e^%28-4t%29+from+t%3D0+to+t%3D250

The answer is 395.1 mJ?

 

[rude man]

If i integrate the power over the time interval, I get $0.625$:

http://www.wolframalpha.com/input/?i=integrate+2.5e^%28-4t%29+from+t%3D0+to+t%3D250

The answer is 395.1 mJ?

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.

 

[STEMucator]

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.

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.

 

[rude man]

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

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

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

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

 

[STEMucator]

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).

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.

 

[rude man]

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.

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.

 

[STEMucator]

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.

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.

 

[EEfailiure]

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