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rms5643
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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.
Homework Equations
V=IR, KVL, Mesh Analysis, Voltage Division, Solution form of first order equations
The Attempt at a Solution
- 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+
- 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
- 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
[*]-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.