I just have a quick question; I was reading an Engineering book, and I came across the following statement:If the rise in time of the signal is so fast that the current through the inductor increases very rapidly, producing appreciable voltage drop.I know that the voltage across an inductor is: V L(di/dt), so why would a smaller amount of time create a voltage drop? According to my understanding of that equation, shouldn't the voltage increase?Perhaps it has been too long since I have had Calc I.Thank you.
Yes the Voltage will increase. Consider the following: The engineering book should have stated If the rise time of the input signal Voltage is so fast that the current through the inductor increases very rapidly it will produce an appreciable Voltage drop across the inductor.The statement in the engineering book was referring to rise time (not a rise in time) which is a change in current with respect to time or as you stated di/dt. A faster rise time means that a smaller change in time (dt) is required for a given amount of rise (increase) in current (di). This results in a larger value for di/dt. Of course this would cause an increase in L di/dt which as you know is the Voltage drop across the inductor that the engineering book was referring to.
Your book is not very good. I'm not sure what it means by rise in time. If the current is increasing swiftly in shorter intervals of time then the voltage increases dramatically. You are correct in your thinking. lim dt--0 (as di--large values) di/dt infinity
This Site Might Help You. RE: Voltage Drop Across An Inductor? I just have a quick question; I was reading an Engineering book, and I came across the following statement: If the rise in time of the signal is so fast that the current through the inductor increases very rapidly, producing appreciable voltage drop. I know that the voltage
Voltage Drop Across Inductor
