Engineering Is This the Correct Approach for Solving First-Order Transient Circuits?

AI Thread Summary
The discussion focuses on solving a first-order transient circuit using the step-by-step method. The initial current through the inductor is determined to be 6A, leading to a voltage drop of 3V across a 2Ω resistor immediately after the switch is thrown. After a long time, the voltage drop across the same resistor is calculated to be -6V. The Thevenin equivalent circuit is analyzed, yielding a time constant of 0.1666 seconds. The final equation modeling the circuit behavior is confirmed, with a suggestion to express the time constant in a simplified form.
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Homework Statement


Use the step-by-step method to find vo(t) for t > 0 in the circuit in the figure below.
1G3Myiq.png


Homework Equations


V=IR, KVL, Mesh Analysis, Voltage Division, Solution form of first order equations

The Attempt at a Solution


  1. Finding the current through the inductor before the switch is thrown:
    • Since the circuit has reached steady state, the inductor can be replaced with a short circuit, circumventing the 4Ω resistor.
    • Mesh Analysis: With Y being the current in the left loop and X being the current in the right loop, both clockwise
    • -12+(Y-X)*2-12=0,
      12+(X-Y)*2+X*2=0
      [*]X=6, Y=18
      [*]Current through inductor is 6A at t=0-

    [*]Finding the voltage drop across the 2Ω resistor after the switch has been thrown:
    • At the moment the switch is thrown, the inductor can be replaced by a 6A current source. The voltage source on the left is thrown out of the circuit.
    • Loop Analysis clockwise:
    • 12+X*2+(X-6)*4+X*2=0
    • X=1.5
    • 1.5A*2Ω=3V
    • Voltage drop across the resistor in question is 3V at t=0+
    [*]Finding the voltage drop across the 2Ω resistor after the switch has been thrown for a very long time:
    • The inductor acts a short circuit, thus circumventing the 4Ω resistor at the top
    • Voltage division:
    • -12*2/4=-6V
    • Voltage drop across the resistor in question is -6V at t=infinity
    [*]Finding the Thevenin equivalent circuit & calculating the time constant τ
    • Since the 12V source becomes an open circuit, we only have the 3 resistors
    • 2Ω is in series with the other 2Ω, and the 4Ω equivalent resistor is in parallel with the other 4Ω resistor at the top.
    • The Thevenin resistance is 2Ω
    • Then, τ=(1/3H)/2Ω => 0.1666
    [*]Therefore, the equation modeling this circuit is:
    [*]-6+[3--6]*e^(-t/0.166)
    [*]-6+9*e^(-t/0.166)


Does this look correct? I've redone my steps and I haven't found an error except in the calculation in my time constant, as you can see the change from 0.08333 to 0.1666. Though, I have one attempt left at this and would like to make sure the previous steps are correct.​
 
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Your calculations look good to me.

Note that your time constant works out to 0.1666 which is to say, 1/6. So you could write ##\left( -\frac{t}{\tau} \right)## as (-6t).
 

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