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Circuit Analysis Involving Inductance

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data

    inductancew.png



    3. The attempt at a solution
    Just was hoping someone here would be able to help me with the circuit above. I wanted to ensure that I analyzed it properly.

    I applied Ohm's Law to determine the initial current through the inductor. i(initial)=-16V/2.2kΩ=-7.27 mA. Negative sign indicating up through the inductor.

    Next, I applied Thevenin's Theorm to reduce the circuit to the left of the inductor. The right sight behaves as an open circuit due to the open switch.

    I get Rth = 8.2+2.2 k-Ohm=10.4 kΩ

    Additionally, I determine the Thevenin equivalent voltage is:
    Eth=(4mA)(8.2)=32.8V

    I can then determine the equation that describes the current, i, in the system.

    i = i(final)+[i(initial)-i(final)]e^(-t/τ)
    where i(final) is determined by assuming steady state conditions.

    i(final)=32.8V/10.4kΩ=3.1538mA
    thus,
    i = 3.14mA+[-7.27-3.15]e^(-t/τ) and τ= 500/10.4=48.08 s
    and
    i=3.15-(10.4)e^(-t/48.08)

    I found the voltage to be a bit more tricky and my result seems very high.... here is what i did:

    I noted the inital polarity of the inductor to be positive polarity at the ground, and negative at the top. I then applied KVL around the loop to determine the voltage.

    Vm = -32.8V - (7.27mA)(10.4kΩ) = -108.4V.

    Thus, V=-108e^(-t/48.08)

    Thanks a bunch in advance!
     
  2. jcsd
  3. Oct 5, 2011 #2

    gneill

    User Avatar

    Staff: Mentor

    Your analysis looks fine to me. I might quibble a bit about your choice of polarity for the inductor voltage though. It just seems natural to me to choose to measure the inductor voltage with a ground reference if one of its legs is grounded. But that's just me :smile:

    The instant when the switches are operated the current through the inductor will continue to flow from bottom to top. The voltage across the inductor will be such as to continue to drive this current back through Rth and Vth. In other words, it'll be about +108V with respect to ground.
     
  4. Oct 6, 2011 #3
    Alright. I ended up switching the polarity. The reason I had the polarity like that is because I figured that was the condition it would be left in from the initial switch 1.

    I switched the polarity and I have,
    v=+(108V)e^(-t/48.08s)

    now if I were to graph this, it would merely have a y-intercept of 108V, and decline to zero as t--->5τ
     
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