# Wattage delivered using triple integral help

1. Oct 16, 2011

### Petyab

Useful equation.

Avg. Power
p(t)=(V^2(t))/R

My attempt at instantaneous power was

p(t,V,R)= ∫(0->1 for t ∫0->5 for V and ∫0->.1 for R V^2(t)/RdvdRdt

Integrating I go the triple integral of V^3t^2/6R^2

Substituting my values in gave a wattage of 1,250 watts/m^2 at t=1 second, v=5 volts, and R=1 ohms

Using Ohms Law V=IR or I=V/R I calculate now as I write this a current of
I=5/.1=50 amps

Is this correct?

2. Oct 16, 2011

### gsal

I do not know where you came up with the triple integral thing...you need to review this concept.

If anything, it should be the integral of V/R.

Is your R really varying? If not, it would simply be the integral of V

By the way, p(t) = (v(t))^2 / R is not average power, it is instantaneous power

The average power would by the integral of the instantaneous power over a period of time divided by that amount of time.

3. Oct 16, 2011

### Petyab

If R was varying would I be right for instantaneous power? The triple integral follows the rules for setting a triple integral in relation to the variable. The values given are set for R being constant though they should be the same in that case.