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RL circuit equation

  1. Sep 15, 2012 #1
    Suppose a circuit with resistor R and inductor L with no source. I am trying to find kirchoffs equation for this circuit - I am getting iR -Ldi/dt = 0 as my equation which is apparently wrong. I just cannot understand how do I make equations for such circuits.
     
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  3. Sep 15, 2012 #2

    CWatters

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    Should that be iR + Ldi/dt = 0
     
  4. Sep 16, 2012 #3
    Yes it should be but I dont get it why. Voltage across resistor is iR and then voltage across inductor decreases by Ldi/dt so iR-Ldi/dt=0
     
  5. Sep 16, 2012 #4

    vanhees71

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    Draw your circuit and indicate an (arbitrary) direction, in which you want to count the current positive. Then use the right-hand rule to attach the surface-normal vector oriented positive relative to that direction of the current. Finally use Faraday's Law,
    [tex]\partial_t \vec{B}=-\vec{\nabla} \times \vec{E},[/tex]
    and integrate (line integral) along the circuit in direction of the positve current. Then the left-hand side translates into [itex]L \frac{\mathrm{d} i}{\mathrm{d}t}[/itex] for compact circuits, and the right-hand side you can transform into an integral along the surface, translating into [itex]-R i[/itex], where we have made use of Ohm's Law, [itex]\vec{E}=\vec{j}/\sigma[/itex]. From this you get the desired equation,
    [tex]L \frac{\mathrm{d} i}{\mathrm{d} t}=-R i.[/tex]
     
  6. Sep 16, 2012 #5
    Sorry I am unaware of some of the things you said. Can you simplify please?
     
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