Decaying current power calculation

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Discussion Overview

The discussion focuses on calculating power loss due to resistance in an electrical circuit over a specified time period, specifically addressing the integration of current to find energy loss. The conversation involves theoretical aspects of power calculations and integration techniques.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant proposes using integration of current over a time period to find power loss, questioning whether this approach is valid.
  • Another participant clarifies that integrating current yields charge, not power, and suggests that the participant should instead find energy loss by integrating power as a function of time.
  • A later reply confirms the integration limits, indicating that the integration should be performed from 0 to 0.3 seconds.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial approach to calculating power loss, but there is agreement on the need to integrate power rather than current to find energy loss.

Contextual Notes

There is an implicit assumption that the resistance value and current function are correctly defined, but no further details on the derivation or implications of the current function are provided.

Mzzed
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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.
 
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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.
 
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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?
 
Just integral from 0 to 0.3
 
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