How do I determine the energy stored in an inductor after 2 seconds?

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Homework Statement



A resistor [tex]R[/tex] is connected in series with an inductor [tex]L[/tex]. The battery is connected at time [tex]t = 0[/tex]. How much of this energy after 2 seconds is stored in the magnetic field of the inductor?


Homework Equations



[tex]U_{L}=\frac{1}{2}Li^{2}[/tex]
[tex]i(t)=i_{0}(1-e^{\frac{-t}{\tau}})[/tex]

The Attempt at a Solution



I know that you're supposed to square [tex]i(t)[/tex] and then multiply by [tex]\frac{L}{2}[/tex]. However, when I looked at the solution they have it as:
[tex]U=\frac{L}{2}\int{i(t)^{2}dt[/tex]

why do you need to multiply by the integral of current squared instead of just the current squared? what is the final answer telling me if i multiply by the current squared versus the integral of the current squared?
 
It is because the energy stored changes over time due to how the current changes over time...its sort of strange though...one would expect that the energy stored was a state function and only depended on the current at that moment...but I guess I might have been off in that reasoning.
(The integral is essentially stating that the stored energy is not a state function, and that it does depend on the process...I didn't think it was like that but that's apparently what that answer tells)
 
Last edited:
homomorphism said:

Homework Statement



A resistor [tex]R[/tex] is connected in series with an inductor [tex]L[/tex]. The battery is connected at time [tex]t = 0[/tex]. How much of this energy after 2 seconds is stored in the magnetic field of the inductor?


Homework Equations



[tex]U_{L}=\frac{1}{2}Li^{2}[/tex]
[tex]i(t)=i_{0}(1-e^{\frac{-t}{\tau}})[/tex]

The Attempt at a Solution



I know that you're supposed to square [tex]i(t)[/tex] and then multiply by [tex]\frac{L}{2}[/tex]. However, when I looked at the solution they have it as:
[tex]U=\frac{L}{2}\int{i(t)^{2}dt[/tex]

The solution said this was the energy? It does not even have units of energy on the right hand side.

Was that the exact statement of the problem? (The wording seemed a bit strange to me.)
 

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