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

  • Thread starter Thread starter Ognerok
  • Start date Start date
  • Tags Tags
    Power
AI Thread Summary
The discussion centers on the derivation of the power equation from the electrical energy equation U = QV. The user struggles with the differentiation process, initially treating charge Q as a constant, leading to confusion about the relationship between power and voltage. Clarifications are provided that power (P) is derived as P = dU/dt = V*dQ/dt, emphasizing that Q represents the rate of charge flow rather than a constant value. The distinction is made that the energy acquired by a charge relates to potential difference, not directly to power dissipation in resistance. Understanding these relationships is crucial for grasping electrical engineering concepts.
Ognerok
Messages
5
Reaction score
0

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.
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top