Solve RL Circuit Problem: Homework Statement & Equations

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SUMMARY

The forum discussion focuses on solving an RL circuit problem involving a switch that opens at t = 0. The initial current through the inductor is calculated using the equations V = iR and τ = L/R, leading to the expression i(t) = 5e^(-2t) A for t > 0. The participants clarify that the 5 Ohm resistor does not carry current due to both ends being at ground potential, allowing for simplification in the calculations. The use of current division across resistors is emphasized as a key technique in deriving the solution.

PREREQUISITES
  • Understanding of RL circuits and their behavior over time
  • Familiarity with Ohm's Law (V = iR)
  • Knowledge of time constants (τ = L/R) in electrical circuits
  • Experience with current division in parallel resistor networks
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  • Learn about source transformations in circuit analysis
  • Explore current division techniques in parallel circuits
  • Review the behavior of inductors in DC circuits
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Electrical engineering students, circuit designers, and anyone involved in analyzing RL circuits and their transient responses.

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Homework Statement


The attachment shows the circuit in question. Pardon the crude drawing. The top circuit is the circuit drawn in the text. The middle circuit is how the circuit is supposed to behave for t < 0 (switched closed). The bottom circuit is how the circuit is supposed to behave for t > 0 (switch open). The current source in the top circuit is constant.The switch is closed for a very long time and then at t = 0 it opens.
Derive an expression for i(t) for t > 0.

Homework Equations


V=iR
τ=L/R
i(t)=i(0)e^-(t/τ)

The Attempt at a Solution


Looking at the top circuit:
I handled the case for t<0 to find the Initial current through the inductor (is the initial current in the equation the initial current through the inductor or is it the current flowing through the resistor?) by closing the switch for a longtime and then did a source transformation transforming the current source with a resistor in parallel to a voltage source with a resistor in series.
vsc=(15A)(24Ω)=360VNow I'm at the middle circuit:
I understand that direct current shorts an inductor.
This in turn shorts out the 5ohm resistor ( Why? I know that current takes the path of least resistance but...)

So I can combine the 12Ω and 8Ω resistor and add it to the 24Ω resistor.
Req=(12||8)+24=36.5Ω

Therefore Ieq=360/36.5=9.86A

Then I used current division across the 12Ω and 8Ω resistors in parallel.
Iο=((8)/(12+8))(9.86A)=3.94A

After finding the initial current I open the switch and thus arrive at the bottom circuit:
From here I find RL and then find τ which should be my final answer.

The answer is 5e^(-2t) A. Could someone please help? Thank you.
 

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The issue is with how you're drawing the middle circuit.

The 5 Ohm resistor is still there. The thing is, there's no voltage difference across it, because the end of the other two resistors are both connected to ground (so the voltage on the left is 0 and the voltage on the right is 0). So you can conclude that no current passes through the 5 Ohm resistor. So now, you know you only have to find the current through the 12 Ohm resistor to find the current through the inductor (which is short circuited since the switch has been closed for a long time).

I found it easier to leave the current source in and use current division twice, but to each his own.
 
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Thank you axmls.
 

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