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ttiger2k7
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[SOLVED] L-C-R Circuit - finding current
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
Kirchoff's loop rule
Voltage across Inductor: [tex]L\frac{di}{dt}[/tex]
Since [tex]R_{2}[/tex] and [tex]R_{3}[/tex] are in series I can add them:
36 Ohms + 36 Ohms = 72 Ohms
And applying the loop rule:
[tex]V - iR_{1} - L\frac{di}{dt} - iR_{2+3} = 0[/tex]
[tex]100 V - i(40 Ohms) - ? - i(72 Ohms) = 0[/tex]
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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 [tex]L\frac{di}{dt}[/tex]? Isn't it just [tex]\frac{\epsilon}{L}[/tex]?
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: [tex]L\frac{di}{dt}[/tex]
The Attempt at a Solution
Since [tex]R_{2}[/tex] and [tex]R_{3}[/tex] are in series I can add them:
36 Ohms + 36 Ohms = 72 Ohms
And applying the loop rule:
[tex]V - iR_{1} - L\frac{di}{dt} - iR_{2+3} = 0[/tex]
[tex]100 V - i(40 Ohms) - ? - i(72 Ohms) = 0[/tex]
---
I have two questions regarding this problem:
Is applying the Loop rule the right way to go?
And if so, what exactly is [tex]L\frac{di}{dt}[/tex]? Isn't it just [tex]\frac{\epsilon}{L}[/tex]?
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