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Self Induction

  1. Feb 7, 2009 #1
    This question is about the self induction:

    The resistance of a coil is 15 ohm, its self induction is 0.6 Henry, it is connected to a source of direct current which gives 120 volts. Calculate the rate of growth of the current at the moment when the current reaches 80 % of its maximum value.

    My Answer:

    First I calculated the maximum current:
    I= V/R = 120/5 =8 Ampere

    Therefore: the 80 % of the maximum current =6.4 A

    Then the electromotive force will be:
    V= I * R = 6.4 * 15 = 96 volt

    And since: the induced e.m.f in the coil = -L * (dI/dt)

    Therefore: 96 = 0.6 * (dI/dt)

    Therefore: dI/dt = 96/0.6 = 160 A/s

    I found the answer in my book as follows:

    Since e.m.f = -L*(dI/dt)

    Therefore: (120-96) = 0.6 * (dI/dt)

    Therefore: dI/dt = 24/0.6 = 40 A/s

    So, why did he substitute the e.m.f in the equation with the difference in volts (120-96)? Why didn't he substitute with the (96 volts)?

    Thanks in advance,
  2. jcsd
  3. Feb 7, 2009 #2
    Consider the emf produced by the coil and the battery as separate and think about Lenz' Law.
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