# DC Circuits: Inductors

1. Apr 14, 2010

### dleccord

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?

2. Apr 14, 2010

### Integral

Staff Emeritus
According to Ohm's law E=IR if R=0 (a short) then E is also 0.

3. Apr 14, 2010

### ruko

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.

4. Apr 14, 2010

### dleccord

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.

5. Apr 14, 2010

### GRB 080319B

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.