What is the formula for electrical power in a DC circuit?

[ximath]

I am studying DC circuits and trying to derive formula for power; that is in fact
[tex]P = I.Vab[/tex]

where Vab is the potential difference between two terminals of a circuit element.

From my previous studies, I know that.

[tex] P = dW / dt [/tex]

I assume that for a small interval, dt, a single charge q has a small displacement , dx.

Then;

[tex]dW = E q dx[/tex]

where E is magnitude of the electric field.

Hence, from the formula above,

[tex] P = E q dx / dt [/tex]

This is all I could come up with. I want to go on with this idea to prove

[tex] P = I Vab [/tex]


I know I need to substitute dq somewhere (to get I ) , somehow.. Simply replacing q with dq does not seem to work; it leads to an incorrect formula. (I find P = I dV ; I guess, if I do that.)

 

[MATLABdude]

...I assume that for a small interval, dt, a single charge q has a small displacement , dx.

Then;

[tex]dW = E q dx[/tex]

where E is magnitude of the electric field.

It would be more advantageous to assume that you have a differential charge dq moving a distance x in the electric field E. (And does dq/dt ring a bell?) Also I believe this approach is more correct, since you have numerous charges (instead of a single one) and you're attempting to find their aggregate behaviour when you use Ohm's law.

This derivation is also a very macroscopic one; the standard first-principles approach is given at Wikipedia:
http://en.wikipedia.org/wiki/Ohm's_law

 

[astrokreng]

Your approach is on the right track, but there are a few things that need to be clarified. First, let's define some variables:

P = power (in watts)
I = current (in amperes)
Vab = potential difference (in volts)
t = time (in seconds)
q = charge (in coulombs)
E = electric field (in volts per meter)
dx = displacement (in meters)

Now, let's start with the definition of power:

P = dW / dt

This means that power is the rate of change of work (W) over time (t). In a DC circuit, the work done is equal to the product of the charge (q) and the potential difference (Vab):

W = q * Vab

Substituting this into the power formula, we get:

P = d(q * Vab) / dt

Now, we can use the chain rule to expand the derivative:

P = (dq/dt) * Vab + q * (dVab/dt)

Remember, current (I) is defined as the rate of change of charge over time, so we can replace (dq/dt) with I:

P = I * Vab + q * (dVab/dt)

But, in a DC circuit, the potential difference (Vab) is constant, so (dVab/dt) is equal to 0. Therefore, the second term drops out and we are left with:

P = I * Vab

And there you have it! The formula for electrical power in a DC circuit is P = I * Vab. Keep in mind, this formula only applies to DC circuits where the potential difference is constant. In AC circuits, the formula is more complex and involves factors like resistance and reactance.