Solve LC Circuit Problem: Examples & Resources

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SUMMARY

The discussion focuses on solving an LC circuit problem involving two inductors and the application of Kirchhoff's laws and Laplace transforms. The user initially struggles to find the currents in the circuit after a switch is opened. Key steps include finding steady-state conditions, simplifying the circuit, and using the equation i(t) = i(0)e^(-(R/L)t) to express current as a function of time. The final results indicate a current of i(t) = 60/23e^(-15t/11) through the inductor and -60/23e^(-15t/11) through the resistor.

PREREQUISITES
  • Understanding of Kirchhoff's laws
  • Familiarity with differential equations
  • Knowledge of Laplace transforms
  • Basic concepts of RL circuits
NEXT STEPS
  • Study the application of Laplace transforms in circuit analysis
  • Learn about the behavior of RL circuits during transient states
  • Explore current division techniques in series circuits
  • Review examples of solving differential equations in electrical circuits
USEFUL FOR

Electrical engineering students, circuit designers, and anyone looking to deepen their understanding of RL circuit behavior and analysis techniques.

Schfra
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Homework Statement



An image of the problem is attached. I have to find 2 currents in an LC circuit.

Homework Equations



V = L(di/dt)

The Attempt at a Solution


[/B]
I have no clue how to do this type of problem. Are there any examples problems similar to this online? Or can someone give me the necessary information to solve the problem. I’m not looking for anyone to solve the problem for me, I’m just looking for somewhere to start.
 

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You've only two recourses, Kirchhoff's laws and differential equations, or Laplace transforms to do the "differential stuff" algebraically. Personally I'd choose the Laplace route.
 
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gneill said:
You've only two recourses, Kirchhoff's laws and differential equations, or Laplace transforms to do the "differential stuff" algebraically. Personally I'd choose the Laplace route.
What’s the process for solving a problem like this though? This is the first RL circuit problem I’ve been assigned to do so I’m pretty lost.

I assume I can replace the two inductors on the right with one inductor with a value of 2/3H.

After that, I have no idea what to do. The closest thing I’ve seen to this problem online was a video of someone solving for the current as a function of time before a switch was closed. In this problem, they seem to want current after the switch is opened.
 
Schfra said:
I assume I can replace the two inductors on the right with one inductor with a value of 2/3H.
Sure. Might as well incorporate the 3 H inductor as well.
Schfra said:
After that, I have no idea what to do. The closest thing I’ve seen to this problem online was a video of someone solving for the current as a function of time before a switch was closed. In this problem, they seem to want current after the switch is opened.
Find the steady state conditions for the circuit before the switch is opened. When the switch opens you've got an RL series circuit with an initial current.
 
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gneill said:
Sure. Might as well incorporate the 3 H inductor as well.

Find the steady state conditions for the circuit before the switch is opened. When the switch opens you've got an RL series circuit with an initial current.
For steady state, I believe I have to treat inductors as short circuits.

I’ve simplified the circuit and then used current division and got a current of 55/23A through the inductor. If I did that right, I think this should be the initial current. What would I then do with that to get current as a function of time?
 

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Re-check your current calculation.

What does the part of the circuit of interest look like when the switch opens? Can it be simplified?
 
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gneill said:
Re-check your current calculation.

What does the part of the circuit of interest look like when the switch opens? Can it be simplified?
I re-did the calculation and got 60/23 or about 2.61 for the current through the inductor.

Would the new circuit just be an inductor of 11/3H in series with a resistor of 5 Ohms?
 
Schfra said:
I re-did the calculation and got 60/23 or about 2.61 for the current through the inductor.

Would the new circuit just be an inductor of 11/3H in series with a resistor of 5 Ohms?
Yes and yes. What are the initial conditions (the instant following switch opening)? What have you learned about the behavior of an RL circuit?
 
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gneill said:
Yes and yes. What are the initial conditions (the instant following switch opening)? What have you learned about the behavior of an RL circuit?
Looking through my notes I found the equation i(t) = i(0)e^(-(R/L)t), which gives me an answer of i(t) = 60/23e^(-15t/11) for the current through the inductor and -60/23e^(-15t/11) through the resistor. I’m not sure if that’s what you meant by initial conditions, or if my answers are correct.
 
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Yes, and your results look good. Be sure to indicate the units (amps) when you submit your results.
 
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  • #11
gneill said:
Yes, and your results look good. Be sure to indicate the units (amps) when you submit your results.
Alright, thanks for the help.
 

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