Why are not inductive response seen clearly in equations?

Question

Suppose that we have a simple rc circuit , a dc source (V), a switch, a bobin (L) and a resistance (R) connected serial respectively, in steady state, current is V/R. When we open the switch at t=0, current and voltages will change according to the following equations,

Current =V/R . exp(-t.R/L) and voltage drop across the bobin= - V. Exp(-t.R/L).

As seen very clearly, voltage drop across the bobin can not be higher than source voltage V, but we know very well that the instant we open the switch there are huge voltage spikes (10-20 times as high as the source voltage) across the bobin, so what is the problem with these equations that the spikes are not seen in these equations ?

Answer 1

The equations you post are the not correct. If you close the switch on the circuit you describe, the CURRENT level is given by
i = (V/R)e^-t*R/L. The current starts out at zero, and ends up (after a long time) at V/R. This equation is derived from the fundamental relation V=L*dI/dt. If you open the switch, you must go back to this fundamental relation. If you stop the current instantly, dI/dt = ∞, so the voltage must also go to ∞. Of course this cannot happen. What does happen is that the air in the switch breaks down at some voltage, and that limits the level to which the voltage can rise. There will also be a big spark at the switch.