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LR circuit, energy dissipated?

  • Thread starter breez
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  • #1
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http://img527.imageshack.us/img527/1787/lccircuitrk6.th.gif [Broken]

The switch S is closed for a long time and then released again. How much energy is dissipated through the resistors after the switch is released?

It should be (LI^2)/2 shouldn't it? Where I is the current right before the switch is released. I even integrated the R(i(t))^2 where i(t) is the current decay function of time for an RC circuit. The answer I got was 81 J but apparently the answer is 36 J? I'm deeply confused. Can someone show me how to correctly solve this if my approach is wrong?
 
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Answers and Replies

  • #2
Defennder
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First find the current through the 6 ohm resistor in series with the inductor. Then when the switch is open, you can disregard the left hand side of the picture. The potential across the inductor is [tex]L\frac{dI}{dt}[/tex]. Use that along with the value of the resistors 6 and 12 ohms to construct an equation with KVL for the case when the switch is open.

Then solve that first order DE by separation of variables. Plug in the value of [tex]I_{0}[/tex] which is the initial current through the 6 ohm resistor in series with the inductor when the switch is just opened. You'll now have a decaying exponential function of t. That is the current through the circuit for t>0, assuming the switch is open at t=0. You can find the energy dissipated in the circuit by [tex]P = \frac{dE}{dt}[/tex] where P is power dissipated, E is energy dissipated. Using the expression for energy dissipated in the resistors, perform an integration from t=0 to t->infinity.

You'll get 36 J.
 
  • #3
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Oh I see, I got 36 J. You can skip the integration using equivalent resistance to find the current in the middle branch, then do U = LI^2/2. This gives 36 J. Since the final current is 0, the energy must all be dissipated.
 
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