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Homework Help: Rate of change of current in an R-L Circuit

  1. May 18, 2018 #1
    1. The problem statement, all variables and given/known data
    upload_2018-5-18_12-18-4.png

    2. Relevant equations
    Current cannot change suddenly through inductor unless input is impulse voltage source.

    3. The attempt at a solution
    since input is DC source we have i(0+) = i(0-) so we have di/dt at t = 0+ is 0.
    Is it correct? Book has mentioned answer as B which is weird cause there is no Ls in circuit.
     
  2. jcsd
  3. May 18, 2018 #2

    cnh1995

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    It's probably a typo. They meant Is instead of Ls.

    No.
    How about using source transformation? It will make the circuit look more familiar.
     
  4. May 18, 2018 #3
    Wow.... I used laplace and then got the answer. You're right.
    upload_2018-5-18_16-41-37.png
    But shouldn't it be zero as per concept? Unless the input is an impulse voltage source the current across inductor cannot change so quickly. V = L di/dt. So unless V is an impulse, di/dt has to be zero.
     
  5. May 18, 2018 #4

    cnh1995

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    Well, if you use source transformation, you end up with a series RL circuit with Rtotal=Rs+R and inductance L, excited by a voltage source of emf IsRs.
    At t=0+, no current flows through the inductor, which means the entire source voltage appears across the inductor.
    So, IsRs=Ldi/dt, which gives di/dt=IsRs/L.
    No, here VL at t=0 is not zero, hence di/dt is also not zero. You can't compare the currents at t=0- and t=0+ to find di/dt at t=0+ because at t=0-, the inductor was not excited.
     
  6. May 18, 2018 #5
    Well you're method is way quicker than mine.
    Thanks a lot. Learnt new thing today.
     
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