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Induced Emf in an Inductor in an LRC circuit

  1. Apr 18, 2014 #1
    1. The problem statement, all variables and given/known data
    In the circuit shown in the figure, the switch S is closed at time t=0 with no initial charge on the capacitor. What is the induced emf if the inductor at time t=0?


    2. Relevant equations
    ε=-LdI/dt


    3. The attempt at a solution
    Initially the inductor acts like a broken wire in that no current can go through it so the potential difference across the 100 Ω resistor is 0. The capacitor initially has 0 resistance, so all the current will flow through there and the voltage across the 50Ω resistor in parallel with the capacitor (V4) will also be zero. So a time t=0 it acts like a simple circuit with resistance 50 Ω and εbattery 40 V. I am unsure if there would also be a potential difference across the inductor due to an induced emf. I don't really know how to find the current as a function of time for this combination circuit so I'm not sure if dI/dt would then be zero, making the induced emf zero. Any help is appreciated!
     

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  3. Apr 19, 2014 #2
    dI/dt does not equal zero in the inductor. Although at time t=0 there is no current in the inducter, at some future time there will be a current in the inductor.
     
    Last edited: Apr 20, 2014
  4. Apr 19, 2014 #3

    SammyS

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    To expand on this, what is the rate of change of the voltage across the capacitor immediately after the switch is closed?
     
  5. Apr 20, 2014 #4

    rude man

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    The rate of change of voltage on, and current thru, the capacitor is irrelevant.
    What is the initial voltage across the capacitor? Ergo, across the L-R arm? Ergo, across L?
     
  6. Apr 20, 2014 #5

    SammyS

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    Yes.

    rude man is correct.

    I must have had some brain flatulence .
     
  7. Apr 21, 2014 #6

    rude man

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    That's funny, Sammy!
    Anyway, welcome to the club ...
     
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