Explain how to calculate voltage (transients)

Discussion in 'Homework Help' started by Winten, Nov 14, 2010.

  1. Winten

    Thread Starter New Member

    Nov 12, 2010
    2
    0
    Lets say we have this circuit: [​IMG]
    Lets say that it it was still, and at t=1 sec the switch was turned down. Then from 1-5 switch up, and 5-6 switch down again. I calculated currents in this current time:

    So, at t=1, I = 5
    t=5, I=7.16
    t=6, I=6,54

    So now how can I calculate inductor voltage for these time moments.

    Can i simply use ohms law? How does the graph suppose to look like, does voltage always going down?

    And how it is known that voltage of the inductor starts of the 5V?
     
    Last edited: Nov 14, 2010
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Well at time T=1 the coil current is equal
    IL = 15V/2Ω = 7.5A and voltage on inductor is 0V
    And now if you open the switch the current in the inductor wants to decrease to 15V/3Ω=5A with time constant L/R = 0.333s.

    So when we open the switch the voltage in the inductor will change from 0V to UL = 15V - ( 3Ω * 7.5A) = -7.5V
    Becaues inductance wants 7.5A current to flow in the circuit.
    And after 5*L/R = 1.66s ( T = 2.66s) current in the inductor will be equal 5A and voltage across inductor equal 0V.
    And now at time T=5s we close the switch.
    Coil wants 5A continues to flow, so inductor voltage will be equal
    UL = 15V - (5A*2
    Ω) = 5V
    And current in inductor
    will keep increasing from 5A to 7.5 with time constant L/R = 0.5s
    And at time 6s when we again open the switch current in the inductor reach 0.86 * 7.5A = 6.48A
    So induce voltage will be equal UL = 15V - (3Ω*6.48A) -4.45V.
     
  3. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
    VL=L*dI/dt

    substitute I with the exponential equation for I in each case and you will get the voltage across the inductor.
     
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