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RiceKernel
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Homework Statement
Homework Equations
The voltage across an inductor is V(T) = L i'
Basic current division and voltage division .
Step function where u(T) : when t is smaller than 0 : u(t) = 0
t >0 : u(t) = 1
The Attempt at a Solution
The voltage source has the equation 40 - 40u(t).
When t <0 , this means that the value of the VS is 40V.
What happens to the current iL through the inductor...? Does the inductor acts as a short-circuit or open-circuit?
Assuming an Inductor acts as a short-circuit under these conditions, iL= 40V/40ohms =1 Amps
This is my i(0).
Therefore Rth = 20 ohms
t = L/R = 10 / 20 = 1/2
So iL(t) is = 1A(e^(-2t))
When t > 0 , Vin becomes 0 V so the whole circuit becomes an RL circuit which can be simplified using equivalent resistance equation and the final circuit becomes a 20 ohms resistor in series with the inductor.
The energy dissipated by an Inductor is E = (L*i^2) /2 and since we know the initial energy stored in the inductor is (10H( 1A)^2) /2 = 5 Joules,
It must be dissipated by the resistor. So the total energy dissipated by the circuit is 5 J?
The main problem here is with me understand how an inductor behave in this kind of situation (DC I presume). How do we find the initial current and current at infinity? Kind of confused...