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Inductors sharing energy

  1. Jun 23, 2009 #1
    When an uncharged capacitor is connected to a charged capacitor there is a loss in energy due to radiation (the calculation is in the attachment "inductors_share_energy", after the inductor sharing energy calculation to this message).
    When a current carrying inductor(shorted on itself) is connected to another inductor will there be a similar loss ? My calculation (see attachment) shows that this does not happen.
    Are my assumptions for the inductor correct ?

    Attached Files:

    Last edited: Jun 23, 2009
  2. jcsd
  3. Jun 23, 2009 #2


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    Come on be serious. The very first step in "your calculation" was to assume that energy was conserved. How the heck can you be surprised that your so called calculations showed that energy was not lost when your starting assumption and subsequent calculation was based on energy being conserved?

    The real solution is to note that after t=0+ that the inductors share a common voltage and hence a common L di/dt (so a common di/dt given that both inductors are of equal value). From this it follows that the change in current must be equal for each inductor, so the decrease in current in the first inductor must equal the increase of current in the second inductor.

    Clearly then your solution of i(0+) = 1/sqrt(2) is not possible, as it corresponds to an increase in current of approx 0.71 Amps in the second inductor but a decrease of only approx 0.29 Amps in the first inductor. The real solution is i(0+) = 1/2 amp and as you can easily verify this corresponds to a loss of half the initial energy, exactly analogous to the two capacitor case.

    In the ideal case (no stray or parasitic capacitance) the voltage must be an impulse in nature, again this is analogous to the two capacitor case where the current is impulsive in nature under the ideal stated conditions.
    Last edited: Jun 23, 2009
  4. Jun 24, 2009 #3
    Thanks for the clarification.

    There is one point that still bothers me though, which is that the current conservation will be violated when the current in L1 is decreasing and the current in L2 is increasing.
    Second, I could figure out one equation: -L1 di1/dt = L2 di2/dt; the second one eludes me.
  5. Jun 24, 2009 #4


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    Yes that's why the current has to change instantaneously, requiring an impulsive (as in amplitude goes to infinity) voltage pulse. Of course it's unrealistic, every bit as unrealistic as the two capacitor problem. In fact these two problems are precise duals of each other and they both represent the unrealistic limiting cases as certain parameters go to zero. In any realistic model you'd have to include finite parasitic capacitance and/or parasitic shunt conductance and the infinities would disappear.
  6. Jun 24, 2009 #5
    I could get the expressions for the capacitor case from Physics by Resnick and Haliday and the solution was simple and elegant. But I could not find "corresponding" expressions that would fit the inductor sharing energy case in i. Resnick ii. Engg circuit analysis by William Hayt nor in iii. Schaum's electric circuits. Could you give me your recommendations.
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