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Homework Help: First-Order Transient Circuits

  1. Apr 4, 2014 #1
    1. The problem statement, all variables and given/known data
    Use the step-by-step method to find vo(t) for t > 0 in the circuit in the figure below.

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

    3. 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,
        [*]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:

    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.​
  2. jcsd
  3. Apr 5, 2014 #2


    User Avatar

    Staff: Mentor

    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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