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.
(V/R.(1-exp(-t.R/L) ). 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 ? Of course one can say that “voltage across a bobin is L. di/dt at any time, you begin to open the switch at t=0, and the switch is open at t=0+ , so dt = 0+ - 0 which is very very small, and so L.di/dt is very very high.” Yes. But the equations given before is valid from t=0, right? I think Laplace transfroms is not enough to solve this problem.
2006-11-09 18:24:36 · 2 answers · asked by akg_dnn 2 in Science & Mathematics ➔ Engineering
First of all, the Laplace transform applies to systems in steady-state. The condition you are talking about is transient.
Secondly, the equation you give applies when the circuit is closed. When you open the circuit, there is no longer a loop, so that equation doesn't apply any more.
What happens when you open the switch is that the simple mathematical model of the system is no longer valid. You have, mathematically, an infiniite resistance (the open switch) and a non-zero current (the steady-state current in the inductor) and that's a combination that is outside the limits of the model.
In reality there will be a spark or leakage current (neither of which is possible in the usual linearized model) so that the inductor can have a continuous decay of current. That's why you get the spike of voltage across the inductor that you are familiar with.
2006-11-10 16:10:27 · answer #1 · answered by AnswerMan 4 · 1⤊ 0⤋
You err when you assume the voltage across the bobbin cannot exceed the source voltage. Actually the Laplace transform explains this phenomenon quite well
2006-11-09 19:28:27 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋