- #1
- 58
- 0
[SOLVED] L-C-R Circuit - finding current
1. 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 [Broken]
2. Homework Equations
Kirchoff's loop rule
Voltage across Inductor: [tex]L\frac{di}{dt}[/tex]
3. 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]?
1. 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 [Broken]
2. Homework Equations
Kirchoff's loop rule
Voltage across Inductor: [tex]L\frac{di}{dt}[/tex]
3. 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]?
Last edited by a moderator: