# RL Circuit Series

Jzhang27143
Suppose there is a circuit consisting of a battery, a resistor, and an inductor, and that there is initially no current. As current is increasing, the inductor creates an induced current in the opposite direction. I am confused as to which end of the inductor is at a higher voltage. Since the induced current is in the opposite direction as the natural current, shouldn't the natural current enter the inductor through the lower voltage end? I know this is not true because this would mean that there is an increase in voltage which would completely change the differential equation using Kirchoff. Why is the end of the inductor that the current first enters the higher voltage end?

Gold Member
Lenz' Law tells you that the induced voltage opposes the source. So the back emf (so called) will raise the voltage the L end of the resistor, to reduce the current that would flow if the L were just a piece of wire. If you have a very high Inductance, there could be no detectable current.
the Lenz law thing can be confusing at first but following the rule, 'verbally' can often tell you the result. It's not always at all obvious is you just try to follow intuition.

Homework Helper
Gold Member
..., the inductor creates an induced current in the opposite direction. ...
For what it's worth, I don't believe the inductor creates an induced current: I believe it creates an induced emf. I believe there is only one current in the circuit and that it depends on the sum of all the emfs and resistances.
The only emf across the ideal inductance would be the induced emf and it would not generate its own current, but simply help determine the circuit current.

Why is the end of the inductor that the current first enters the higher voltage end?
Because it is not a resistor (assuming we are talking ideal elements.)