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Homework Statement
http://imageshack.us/a/img832/1651/homeworkprobsg32.jpg
At t= 0, the switch is opened. Calculate the current i(t) for t>0
Homework Equations
V(t) = L* dI/dt
V = IR
I = V/R
voltage division, current division,
KCL, KVL
The Attempt at a Solution
Not so experienced with inductors but, at the beginning with the switch connecting:
voltage division:
V across right branch: (12Ω/14Ω)*36V
V = 30.85V
so this must be the total voltage across the branch with the 6Ω and 2H
Voltage across 2Ω = 36V - 30.85V
= 5.142V
I across 2Ω resistor = 5.142V/2Ω
= 2.571A, and
2.571A enters the parallel branches
so this is the total current combined through the parallel branches
I know now that the voltage across an inductor is 0 if the current does not change
so with that and current division:
I = (12/18)(2.571A)
= 1.714A through the inductor branch with the switch connected at the beginning
and then the current through the other branch is 0.857A at the beginning.
My problem I think is calculus related again since V(t) = L* dI/dt
is the only equation I know for an inductor, I try to set up a problem going like
v(t) = 2H*dI(dt) + 6Ω(1.714A)
but not sure
and couldn't the current be 0 if the switch is opened?
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