# Decaying current power calculation

Tags:
1. Jul 3, 2017

### Mzzed

I would like to find the power loss over 0.3 seconds due to a resistance of 0.0015 ohms with a current following the function:

I = e-15.24t-e-39984.75t where t is time.

I would have imagined summing all the current over that 0.3 second period via integration would then allow me to find the power lost in that 0.3 second time period using this sum in place of the current (I) in the usual power formula P=I2R. my problem is I think the integral of current is charge? and then the working I have just shown wouldn't make sense I don't think?

Any help would be appreciated.

2. Jul 3, 2017

### Merlin3189

You're right that ∫ I dt would give you the charge and RQ2 does not give power, but that is not what you want to do.
You don't want to find charge, so don't integrate current.
You said you want to find "power loss", but the power loss over that period would vary during the period, just as the current does, so all you could get would be power as a function of time. What you actually mean is, you want to find energy loss over that time period.
Then it is straightforward. Energy is the integral of power. Make your expression for the instantaneous power and integrate that.
p(t)= ( i(t))2 R so E = R ∫ ( i(t) )2 dt
and dimensionally this works out fine, ∫ amp2 dt = amp coulomb , multiply by R volts/amp and get volt coulomb or joule for eg.

3. Jul 3, 2017

### Mzzed

Ah that makes sense, thankyou! So with the integration, do I integrate from 0 to 0.3 or would I just integrate and then sub in 0.3 for t?

4. Jul 3, 2017

### Merlin3189

Just integral from 0 to 0.3