Current Through Inductor in Series RL circuit

Question

I Was reading an article on LR Series Circuit on Electronics Tutorials.

The excerpt from that article is as follows:

Assume that the switch S is open until it is closed at a time $t = 0$, and then remains permanently closed producing a “step response” type voltage input. The current, i begins to flow through the circuit but does not rise rapidly to its maximum value of $I_\mathrm{max}$ as determined by the ratio of $V / R$ (Ohms Law).

This limiting factor is due to the presence of the self induced emf within the inductor as a result of the growth of magnetic flux, (Lenz’s Law). After a time the voltage source neutralizes the effect of the self induced emf, the current flow becomes constant and the induced current and field are reduced to zero."

I want to understand this neutralization process by which current rises. Or Why did the current in circuit rise even though it was strongly opposed at beginning by 'Self Induced Emf'?

Answer 1

The time-dependent current $i(t)$ in your circuit obeys the equation $$ v(t) = R\:i(t) + L\:\frac{\mathrm di}{\mathrm dt}(t), $$ coming from the resistor and the inductor, respectively.

When the voltage is switched on at $t=0$, the current is $i(0)=0$, so the first term vanishes $-$ and therefore the whole voltage must be borne by the self-inductance term. This then determines the rate of change of the current after the switch has been turned on, as $$ \frac{\mathrm di}{\mathrm dt}(0) = \frac{V_0}{L}. $$

However, this is not a sustainable situation, because the self-inductance term can only happen if the current is growing, and if it is growing then the current itself must change, so the resistive term $R\:i(t)$ will start to contribute. This reduces the rate of growth, so the current continues to increase, but it does so ever more slowly.

This 'feedback loop' will then continue until the current saturates at the steady-state value, and the current's rate of change decreases (in an exponential decay) down to zero.