How is the equation for Power [U'] derived?

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Homework Statement



While going through a basic electrical engineering overview, I came across the equation U = QV, which is defined to be electrical energy. For Power, which is explained in the next section, P is basically the derivative of U = QV. I tried following the derivation myself, but for some reason, what I got was different from the book.


Homework Equations



U= QV

q = -1.6x10^-19 Coulombs ( basically a constant)

The derivative of this turned out to be d/dt = V*d[Q]/dt. My question is HOW.

The Attempt at a Solution




P = d/dt = Q * d[v]/dt; treated Q as a constant, but this doesn't line up. My Calculus must be rusty. Anyway, the full statement from the book:

P = U'= V*Q' = VI= v^2/R = I^2*R

-Thanks.
 
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Ognerok said:

Homework Statement



While going through a basic electrical engineering overview, I came across the equation U = QV, which is defined to be electrical energy. For Power, which is explained in the next section, P is basically the derivative of U = QV. I tried following the derivation myself, but for some reason, what I got was different from the book.



P = d/dt = Q * d[v]/dt; treated Q as a constant, but this doesn't line up. My Calculus must be rusty. Anyway, the full statement from the book:

P = U'= V*Q' = VI= v^2/R = I^2*R

-Thanks.

Power is the rate of work done against the opposition to the flow of charges in the conductor. While doing so the voltage across the conductor is constant.
So P = U' = V*dQ/dt.
 
rl.bhat said:
Power is the rate of work done against the opposition to the flow of charges in the conductor. While doing so the voltage across the conductor is constant.
So P = U' = V*dQ/dt.


That makes more sense. But...I thought Q itself was just a constant, you know, an electron's charge...it must be Q in the sense that Q = u/V? :)
 
Ognerok said:
That makes more sense. But...I thought Q itself was just a constant, you know, an electron's charge...it must be Q in the sense that Q = u/V? :)
It is different. It is the energy acquired by a charge when it is accelerated the a potential difference V. It is nothing to do with the power dissipated in a resistance.