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RL-circuit with a switch, need help with the differential equation.

  1. May 13, 2012 #1
    Hello there!
    Now if I have a very very simple RL-circuit with a voltage source and we have a switch which is suddenly flipped in the off-position... does the SAME differential equation apply as if it's flipped ON?
    This I mean:
    [itex] \frac{\partial I }{\partial t}L + RI = U [/itex]

    I do understand how that is formed when the switch is flipped ON, but if it's flipped OFF doesn't that mean that there goes NO current through, and thus that equation is kinda pointless?
    Now my friend says that in his book it says that after it's flipped off the equation is:

    [itex] \frac{\partial I }{\partial t}L = RI [/itex]

    I don't get it where the [itex] U [/itex] suddenly vanishes and neither does she so I though I'd give it a shot here.
     
  2. jcsd
  3. May 13, 2012 #2
    DemoniWaari
    It will not be the same differential equation because it is not the same circuit.
    You have added something in series with your loop, the open switch.
    If we say it is an ideal switch: no current can flow through it, it can have voltage across it (call it V).
    Apply KVL around this loop:
    L(∂I/∂t) +RI = U + V

    Now since I = 0 (because of the switch):
    L (∂I/∂t) = U + V

    Now because there was current before the instantaneous switch opening (∂I/∂t = ∞):

    L(∞) = U + V

    And since U and L are finite:

    V = ∞

    In practice, with a large inductor, you will create a big spark when you open this switch. The energy that was in the inductor dissipates in all forms (heat, light, sound, chemical).
     
  4. May 14, 2012 #3
    Oh yes, I get now the new equation part! But what baffles me still is the fact that my friend's textbook says that
    [itex]\frac{\partial I }{\partial t}L = RI [/itex] after the switch is open. But if the current is zero, how can that be? Or is it a misprint?

    Edit:

    And in real world situation the change in current is never infinite, so... how does one try to solve the new dif.equation? Is the V a function of I or what?
     
    Last edited: May 14, 2012
  5. May 14, 2012 #4
    Look carefully at the schematic in your friends textbook.
    That equation is for a circuit consisting of just resistor and inductor with current in the inductor as an initial condition.

    I showed you how to solve the new differential equation. The math is showing us that we have a singularity when the switch opens. The mathematical singularity is due to simplifying assumptions we are making. We are assuming, for example, that the switch is ideal and no current can flow through it no matter what voltage is across it.
    You are right in that we will not actually see the voltage to to infinity. The voltage may go to thousands of volts then arc across the switch contacts. If we want to remove the singularity and analyze the circuit at switch opening more formally we need a lot more information about the switch, how fast it actually opens, how much voltage can it withstand before arcing etc.
     
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