L-C-R Circuit - finding current

  • Thread starter Thread starter ttiger2k7
  • Start date Start date
  • Tags Tags
    Circuit Current
Click For Summary
SUMMARY

The discussion focuses on solving an L-C-R circuit problem where the voltage (V) is 100.0 V, resistances (R1 = 40.0 Ohms, R2 = R3 = 36.0 Ohms), and inductance (L = 17.0 H) are given. The current (i1) immediately after closing the switch is determined using Kirchhoff's loop rule. The circuit analysis reveals that the inductor behaves like an open switch at the moment the switch is closed, resulting in no initial current through the inductor.

PREREQUISITES
  • Understanding of Kirchhoff's loop rule
  • Knowledge of inductors and their behavior in circuits
  • Familiarity with Ohm's Law
  • Basic circuit analysis techniques
NEXT STEPS
  • Study the application of Kirchhoff's loop rule in complex circuits
  • Learn about the time constant in L-R circuits
  • Explore the behavior of inductors in transient analysis
  • Investigate series and parallel resistor combinations in circuit analysis
USEFUL FOR

Electrical engineering students, circuit designers, and anyone interested in analyzing L-C-R circuits and understanding transient responses in electrical systems.

ttiger2k7
Messages
57
Reaction score
0
[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: [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]?
 
Last edited by a moderator:
Physics news on Phys.org
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!
 

Similar threads

  • · Replies 42 ·
2
Replies
42
Views
6K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
3
Views
2K
  • · Replies 49 ·
2
Replies
49
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K