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I kindly ask for assistance in derivation of the equation for instantaneous power in an electric circuit, P(t) = V(t) I(t). I want to derive it as rigorously as possible. Here's what I got:
We start with [itex]P = {\bf F} \cdot {\bf v}[/itex], where [itex]{\bf v} = \frac{d\bf r}{dt}
[/itex]
We know that the force exerted on a test charge q is given by [itex]{\bf F} = {\bf E} q[/itex], and for voltage we know [itex]dV = - {\bf E} \cdot dx[/itex].
Inserting F in equation for power, we get [itex]P = {\bf E}q \cdot \frac{d\bf x}{dt} = {\bf E}\cdot dx \frac{q}{dt} = - dV \frac{q}{dt} .[/itex]
How would I go from this, to the desired result, without taking a "quantum leap"?
Is there a better way to actually derive this mathematically impeccably?
We start with [itex]P = {\bf F} \cdot {\bf v}[/itex], where [itex]{\bf v} = \frac{d\bf r}{dt}
[/itex]
We know that the force exerted on a test charge q is given by [itex]{\bf F} = {\bf E} q[/itex], and for voltage we know [itex]dV = - {\bf E} \cdot dx[/itex].
Inserting F in equation for power, we get [itex]P = {\bf E}q \cdot \frac{d\bf x}{dt} = {\bf E}\cdot dx \frac{q}{dt} = - dV \frac{q}{dt} .[/itex]
How would I go from this, to the desired result, without taking a "quantum leap"?
Is there a better way to actually derive this mathematically impeccably?