# Does the backward emf reach the Source emf ?

when turning on a d.c circuit connected to a very simple solenoid, does the backward emf ever reach the source emf ? if it does then how does the current continue to flow ?

K^2
It affects the entire circuit, but what you should be thinking about are the potential differences. Lets take a very simple circuit consisting of a battery, a resistor, an inductor coil, and a switch. For simplicity, lets assume that only resistor and inductor have resistances, labeled R1 and R2 rsepectively, and only inductor has inductance.

Once with close the circuit with the switch, current is going to start flowing. The increasing magnetic field in the inductor will induce electric field, which will cause back EMF. What does that mean? It means that the total voltage drop across the inductor is going to be VI = VEMF + I R2. Resistor, on the other hand, simply follows Ohm's law, and VR = I R1. We know that current I is the same everywhere. And we know that total of all voltage drops around the circuit must cancel out. (Kirchoff's Laws.) So V - VR - VI = 0, where V is voltage across battery.

These equations are pretty easy to solve. Just substituting the voltages we know, we get the equation: V - VEMF = I (R1+R2). So the current in the entire circuit is going to be affected by the amount of back EMF. And as the current changes, voltage across the resistor is going to change as well. And so the back EMF affects the entire circuit. Furthermore, if VEMF > V, it can get the current to flow the wrong way. This will never happen in this particular circuit, but can happen in general.

aha \
just a couple of questions
shouldn't Vi = Vemf - IR2 ? since these potential differences are opposite * like two opposing batteries *
now the important one , how much of the original V does the induced Emf reach ? when does V = Vemf ?

K^2