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DC Circuits: Inductors

  1. Apr 14, 2010 #1
    if the inductor is at dc steady state, the inductor would act like a short.

    in this case, why would the voltage across the inductor be zero?

    thanks in advance.
     
  2. jcsd
  3. Apr 14, 2010 #2

    Integral

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    According to Ohm's law E=IR if R=0 (a short) then E is also 0.
     
  4. Apr 14, 2010 #3
    A perfect electrical short means there is no electrical resistance. If there is no electrical resistance then there can be no voltage across the short. E=IR or Voltage=Amps times Resistance. As you can see as the resistance decreases so does the voltage.
     
  5. Apr 14, 2010 #4
    wow thanks, i cant believe i didnt look at ohm's law's simplest.

    i was looking for V=Ldi/dt, trying to figure that out but confused myself.

    thanks ruko.
     
  6. Apr 14, 2010 #5
    At t=[tex]\infty[/tex], the current through the inductor is maximum (for "charging" phase) or minimum (for "discharging" phase) and is no longer changing. Therefore, di/dt=0 amps/sec, so the voltage drop across the inductor is V= L(di/dt) = 0 volts.
     
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