# Back EMF of an inductor in a DC circuit?

Will an inductor in a DC circuit act as a voltage source and oppose the applied current creating the magnetic field? Or is it because of the steady state direct current there is no change in I there is no back EMF opposing the power supply?

simulate this circuit – Schematic created using CircuitLab

I'm confused, going from 0A to xA wouldn't that be a change? Creating a growing magnetic field...?

Assuming the above circuit has been sitting for a while since the power supply was turned on and has reached steady state, then the current in the circuit will be the normal expected I=V/R. At steady state, inductors are basically just resistors. There will be no voltage drop or rise across L1 (assuming it is ideal and has no resistance) because the current is constant so di/dt=0.

Inductors only act differently than resistors when the current though them is changing. If you start with PS at zero volts and ramp up to a final voltage, then di/dt will be non-zero during the ramp up period and the inductor will have a negative voltage across it resisting the current flow until it reaches steady state.

Make sense?

Check out the resources listed at the end of this answer...

How is the current induced in this circuit?

...for lots of great references I think you'll enjoy!

Inductors need a time varying current to flow through them to produce an time varying field across it and thus produce an self induced EMF opposing the original current flowing through it.The source provided by you shows the usage of constant non varying current(DC),so there won't be any induced emf around it initially,but as the circuit reaches steady state the inductor behaves as if it is an short circuit because usually steady state implies a lot of time after application of input or theoretically infinite time after any recent changes to circuit, thus time tends to infinite implies voltage tends to zero as per formula above for back emf. this is the transient behavior of inductor to DC source. as you are asking, yes, it's because of the constant DC the back emf across inductor is null. but, once you remove the source and connect the circuit without any source in it, inductor acts as the source for sometime and supplies the same current flowing through it previously when acting like a short circuit thus behaves like a source, so the current direction remains still,but the emf polarity reverses to indicate that it is no more passive element as it is supplying current now. this is because of the fact that inductor can't change current flowing through it instantaneously.thus the difference of AC source from that of DC is the back emf and resultant impedance while steady state current is flowing. thank you .i hope this is useful to u.

You are correct. When the power supply is switched on the current through the inductor will be increasing (di/dt > 0). This will produce a transient voltage across the inductor that resists the applied voltage (a "back emf", following the passive sign convention for i and vL). Once the current reaches its maximum, di/dt = 0 and the voltage across the inductor is also zero. The circuit has reaches a steady state, or DC ("Doesn't Change").

However, another change will occur when the power supply is suddenly switched off. di/dt < 0, so the inductor assumes a negative voltage that tries to keep the current pumping. Both of these inductive voltage spikes can damage other parts of the circuit, particularly integrated circuits and transistors used for switching. This is why inductive loads (e.g. motors, solenoids, relays, etc.) are provided with flyback diodes which allow the current to keep flowing and dissipate the stored energy of the inductor's magnetic field:

simulate this circuit – Schematic created using CircuitLab