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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: L\frac{di}{dt}
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
---
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}?
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
---
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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