Calculating Potential Difference in Self-Induction

[Haru Yo Koi]

Homework Statement

A coil has 3.65 \Omega resistance and 410 mH inductance.
If the current is 3.94 A and is increasing at a rate of 3.31 A/s, what is the potential difference across the coil at this moment?

Homework Equations

V = IR
Induced emf = -L * dI/dt

The Attempt at a Solution

It is easy to calculate that V = 14.381, induced = 1.3571.

But since it is an increasing current, it seems like lenz's law should imply that the induced emf opposes V, i.e. Voverall = V - induced.

But the correct answer is that Voverall = V + induced = 15.7 V, and I don't understand why that is.

 

[member 553083]

Can someone explain this?I think the reason for this is that V is the sum of two parts: the IR drop, and the induced emf. Since V = IR and the IR drop is always in the same direction as the current, the IR drop will have the same sign as the current. The induced emf, however, is always in the opposite direction as the current (due to Lenz's Law). So, when the current is increasing, the induced emf will be positive and will add to the IR drop.