# Energy in inductor and capacitor

1. Aug 21, 2010

### smslca

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. Aug 21, 2010

### Anamitra

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

3. Aug 21, 2010

### smslca

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
4. Aug 21, 2010

### Anamitra

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