Current through an inductor after a switch closes

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SUMMARY

The discussion focuses on calculating the current through an inductor immediately after closing a switch in a circuit. The relevant equation is I(t) = E/R (1 - e^(-R/L*t)), where E is the voltage, R is the resistance, and L is the inductance. At the instant the switch closes (t=0), the inductor behaves as an open circuit, resulting in an initial current of 0 amps and a voltage across the inductor of 100 volts. This behavior is crucial for understanding inductor dynamics in electrical circuits.

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David Donald
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Homework Statement


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Let's say I have a circuit such as the one above, and let's say the circuit has been
open for a long time.

How would I find the current through the inductor, the instant after closing the switch?

and the voltage potential across the 40 mh inductor?

Homework Equations


I(t) = E/R ( 1 - e^(-R/L*t) )

The Attempt at a Solution


I'm a bit lost would the current just be the voltage/resistor = 100/120 = .83333 amps??

how would I find the current right after the switch closes?
and then how would I use that to find the voltage across the 30 mH inductor?

Could someone give me an explanation on how to find the current through the inductor and voltage potential, I'm quite lost and confused : /
 
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David Donald said:
would the current just be the voltage/resistor = 100/120 = .83333 amps??
No. Put t=0 in this equation.
David Donald said:
I(t) = E/R ( 1 - e^(-R/L*t) )
 
So the current would just then be 0 ? I(t) = E/R ( 1 - 1) (0) = (0)
 
David Donald said:
So the current would just then be 0 ?
Yes. Inductor acts as an open switch at this instant. What will be the voltage across this open switch?
 
cnh1995 said:
Yes. Inductor acts as an open switch at this instant. What will be the voltage across this open switch?

It's 100 volts, I'm not going to bother writing all the work out here since I just derived it on paper.
My confusion just stemmed from the behavior of the inductor itself...
 

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