DC Circuits: Why Is Voltage Across Inductor 0 at Steady State?

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In a DC steady state, an inductor behaves like a short circuit, resulting in zero voltage across it. According to Ohm's law, with resistance at zero, the voltage must also be zero. As the current through the inductor reaches its maximum or minimum and stops changing, the rate of change of current (di/dt) becomes zero. This leads to a voltage drop of zero across the inductor, confirming that V = L(di/dt) equals zero volts. Understanding these principles clarifies the behavior of inductors in steady-state DC circuits.
dleccord
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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.
 
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According to Ohm's law E=IR if R=0 (a short) then E is also 0.
 
dleccord said:
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.

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.
 
wow thanks, i can't 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.
 
At t=\infty, 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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