How Does Ohm's Law Apply to an Inductor in a DC Circuit?

[FancyChancey]

The resistance of a 4 H inductor is 50.0 Ω. The inductor is suddenly connected to a 5.00 V cell. What is the final steady current in the circuit?
I have no idea where to start with this. Can someone point me in the right direction please? Thank you.

 

[Feldoh]

Hi FancyChancey,

Start by drawing a diagram of the circuit you made when you connected the inductor to a voltage source.

 

[FancyChancey]

I drew the diagram but I still don't get it.

 

[Gnosis]

The resistance of a 4 H inductor is 50.0 Ω. The inductor is suddenly connected to a 5.00 V cell. What is the final steady current in the circuit?

I have no idea where to start with this. Can someone point me in the right direction please? Thank you.

Per your scenario, it is a simple Ohm's Law problem. 5 volts DC is being applied across the inductor, which has a 50 ohm resistance. The equation to use is:

I = V / R

The "final steady current" implies; after any initial short-lived current opposition due to the expanding magnetic field around the inductor has passed.

Therefore, a steady DC voltage produces a steady current in the inductor, which in turn produces a steady magnetic field around the inductor's wire. Since the steady magnetic field has no motion (it is neither expanding or collapsing), it can’t produce any opposing back emf effects to reduce current drawn from the 5 volt source therefore, steady current is derived by Ohm’s Law equation, I = V / R.