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Back Emf same as battery emf?

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data
    A coil of inductance 0.5 H is connected to a 18V battery. Calculate the rate of growth of current?

    2. Relevant equations
    E=-L(dI/dt)



    3. The attempt at a solution
    Actually I have the solution but the problem is that in the book they have assumed that the back emf that would develop due to increasing current would be the same as emf of the battery? i.e its written as dI/dt=E/L=18/0.5= 36 A/s
    How is it possible? Or am I thinking wrong
     
    Last edited: Feb 11, 2013
  2. jcsd
  3. Feb 11, 2013 #2

    gneill

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    Staff: Mentor

    The potential across the coil cannot be anything else but the same as that of the battery, assuming an ideal coil and battery. An ideal voltage source will produce ANY amount of current required to maintain its constant potential difference.
     
  4. Feb 11, 2013 #3
    But I don't understand why the back emf generated by 18V i.e same as the battery.
    We say that inductur has no resistance of itself. Then there is no potential drop when the current is steady
    And when the current is changing then EMF is induced which is opposes the cause that produces it . And in the formula E=-LdI/dt, the E is the back emf right? And not the battery potential?
     
  5. Feb 11, 2013 #4

    gneill

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    Staff: Mentor

    The current will definitely NOT be steady. There will be no steady state for this circuit since there's no resistance to limit the current.

    Write the KVL loop equation for the circuit. Solve the resulting differential equation for the current. You'll see that the current increases without bound over time, but that dI/dt is a contant. Thus the back-emf is a constant, and it happens to equal the battery voltage.
     
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