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Energy in inductor and capacitor

  1. Aug 21, 2010 #1
    Energy in inductor is given as (Li2])/2
    but energy is power absorbed in t secs is E=integral(from 0 to t)Lidi here i is current.
    since this integral stretches from 0 to t After doing the integral the current i must be turned into the time variable i.e E=(Lt2)/2;
    Then why are we writing it as i2 what is this i represent. Is this the current i and this Energy i the same or different.

    Same problem for capacitance in terms of voltage
     
    Last edited: Aug 21, 2010
  2. jcsd
  3. Aug 21, 2010 #2
    Power,P=Ei where E stands for the emf due to the inductive effect
    Energy spent in time [0 to t]= integral[0 to t]Eidt

    |E|= L di/dt

    Energy expended= integral[0 to t]Li di/dt * dt
    = integral[0 to i]Li di [please do note the change of variable here]
    = 1/2 Li^2
     
  4. Aug 21, 2010 #3
    since they may not be equal how could time change into current.
    Let us suppose we want to find energy for 2 secs with current of x amps , and inductance 1 henry , then
    IS E=0.5x joules using i as variable in boundary value correct or E=2 joules using t as variable in boundary value is a correct one.
     
    Last edited: Aug 21, 2010
  5. Aug 21, 2010 #4
    Things would become clear if you know how current is changing with time i=i(t)

    if you know the above function you can again find f(t) =di/dt
    Then you find:

    integral[0 to 2]L*i(t)* f(t)dt ------------- (1)

    OR
    You use the relation i=i(t) to calculate the current at time=2 seconds
    that is you find i(2),noting i(0)=0

    Then
    Energy expended=
    = integral[0 to i(2)]Li di
    = 1/2 Li(2)^2 ---------------- (2)

    (1) and (2) should give you the same result for any function i=i(t)provided they are differentiable and provided i(0)=0
     
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