L-C-R Circuit - finding current

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The discussion focuses on solving for the current i1 in an L-C-R circuit immediately after a switch is closed. The circuit parameters include a voltage of 100 V, resistances of 40 Ohms, and two resistors of 36 Ohms each in series, along with an inductor of 17 H. The application of Kirchhoff's loop rule is confirmed as appropriate for this analysis. Initially, the current through the inductor is zero, behaving like an open switch, which simplifies the calculation. The current will eventually increase over time, approaching a final steady-state value.
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[SOLVED] L-C-R Circuit - finding current

Homework Statement


In the figure below, V = 100.0 V, R1 = 40.0 Ohms, R2= R3 = 36.0 Ohms , and L = 17.0 H. No current flows until switch S is closed at t=0. Find the magnitude of the current i1 immediately after the switch is closed.

http://calculus.unl.edu/edu/classes/JF05/LRC.gif

Homework Equations


Kirchoff's loop rule

Voltage across Inductor: L\frac{di}{dt}

The Attempt at a Solution



Since R_{2} and R_{3} are in series I can add them:

36 Ohms + 36 Ohms = 72 Ohms

And applying the loop rule:

V - iR_{1} - L\frac{di}{dt} - iR_{2+3} = 0
100 V - i(40 Ohms) - ? - i(72 Ohms) = 0

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I have two questions regarding this problem:

Is applying the Loop rule the right way to go?

And if so, what exactly is L\frac{di}{dt}? Isn't it just \frac{\epsilon}{L}?
 
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It's much easier than you thought. "Immediately after the switch is closed" is the easiest kind of question to answer. It takes time for the current through an inductor to build up. (Or, if this were a capacitor problem instead of an inductor problem, I would be saying: it takes time for the voltage across a capacitor to build up).

So right after the switch is closed, with i3 beginning at zero, the inductor is like an "open switch".

But remember that this is true only for a point in time -- then i3 will increase asymptotically toward some final value.
 
thank you!
 
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