Decaying current power calculation

AI Thread Summary
To calculate power loss over 0.3 seconds with a resistance of 0.0015 ohms and a current function I = e^(-15.24t) - e^(-39984.75t), it's essential to focus on energy loss rather than charge. The instantaneous power can be expressed as p(t) = (i(t))^2 R, and the energy loss is found by integrating this power function over the time interval. The correct approach is to integrate from 0 to 0.3 seconds to obtain the total energy loss during that period. This method ensures that the varying power throughout the interval is accurately accounted for. Understanding these steps clarifies the calculation process for energy loss in electrical circuits.
Mzzed
Messages
67
Reaction score
5
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.
 
Engineering news on Phys.org
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.
 
  • Like
Likes 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?
 
Just integral from 0 to 0.3
 
  • Like
Likes Mzzed
Thread 'Weird near-field phenomenon I get in my EM simulation'
I recently made a basic simulation of wire antennas and I am not sure if the near field in my simulation is modeled correctly. One of the things that worry me is the fact that sometimes I see in my simulation "movements" in the near field that seems to be faster than the speed of wave propagation I defined (the speed of light in the simulation). Specifically I see "nodes" of low amplitude in the E field that are quickly "emitted" from the antenna and then slow down as they approach the far...
Hello dear reader, a brief introduction: Some 4 years ago someone started developing health related issues, apparently due to exposure to RF & ELF related frequencies and/or fields (Magnetic). This is currently becoming known as EHS. (Electromagnetic hypersensitivity is a claimed sensitivity to electromagnetic fields, to which adverse symptoms are attributed.) She experiences a deep burning sensation throughout her entire body, leaving her in pain and exhausted after a pulse has occurred...
Back
Top