calculus question for engineers

'd/dx' is always followed by a variable to which it applies. Later you will run into the concept of a operator, e.g. D=d/dx, and you then can factor out the variable in differential equations leaving an expression in D , but even there it's understood that the 'D' ultimately operates on the variable of the ODE.

You can certainly say 1/(dx/dy) = dy/dx, and considering d/dx as an operator, you could say 1/[ d/dx (y) ], but is this ever written as dy/x? no.

dave y.

dave y. wrote in news: snipped-for-privacy@4ax.com:

OK, thanks, that helps. What about in a case of the inductor equation v=(L) di/dt.

There is no variable after the di/dt. This equation (as simple as it is) confuses me so I don't understand how to apply it (at all) to an inductor.

Thanks again!

L is a constant multipled by di/dt (the rate of change of current with respect to time)

i is the variable after the d/dt.

At this point your problem is probably more to do with your understanding of the physics of inductors. The equation states that the voltage on the terminals of an inductor carrying a current i equals the inductance times the rate of change of the current in the inductor.

Bill

-- Ferme le Bush

Salmon Egg wrote in news:C0F8F073.37043% snipped-for-privacy@sbcglobal.net:

Ok, say I connect the inductor to a 9 volt battery and I remove it. How can I calculate how much voltage will build up inside the inductor?

What is the DC resistance of the inductor? This will set the maximum current of the circuit: i = 9/r. Now you have to guess how fast the current will change when you either connect or disconnect the battery. If you think that it goes from zero to max in 0 time you will get a discontinuity in your calculation and get an infinite voltage as a result.

Lets assume 1 ohm. This will give you a max current of 9 amps at steady state. Assume that from the time you connect your battery to full current is 100 milliseconds. This means that your di is 9 amps and your dt is

100x10-3seconds. If you divide 9 by 100x10-3 you get 90 volts across the inductor for a brief period. This assumes that the battery has zero impedance and all the reactance is inductive.

If my expanation is flawed I am sure that the next person in line will correct me.

Again, your problem is primarily with the physics. (The ambiguity over whether it refers to the inductor or the battery is inconsequential in this case, but you should try to avoid ambiguity when describing technical configurations.) The typical current interruption situation you describe is present in television horizontal sweep circuits. As presented, the problem is not soluble. In practice, there is stray or intentional capacitance across the inductor as well as residual resistance and even radiation resistance. In terms of describing the circuit, with differential equations, you have to add terms corresponding to these complications. You are not there yet.

Bill

-- Ferme le Bush

You have 9 volts across your inductor. That is the terminal voltage so that's how much voltage is built up inside it.

The rate of change of current (di/dt) equals 9/L. Your current starts at zero and increases, in a straight ramp, by 9/L amps per second, forever, or until you remove the battery. As you remove the battery there is an ugly transient that your ideal model can't deal with, i drops to zero immediately, di/dt is (negative) infinite, and voltages blow up. In the real world there is an ugly transient that can result in many kV driving a current arc through air as the current goes to zero in a fast but not immediate way. The distributor and spark plugs in your car may be an example of this. Or unplug a running vacuum from a receptacle a couple times to see it with your own eyes. This hardly counts as calculus in your idealized example it is more like algebra.

j

The concept of limiting the current and stretching out the transient in time is correct but there is a clean solution. If there is an R, then this is a series LR circuit that gives a classic ode, V=Ri+Li'. Current will start at zero and approach V/R exponentially (well, logarithmically maybe) with time constant tau=L/R. i=(V/R)*[1-e^-(t/tau]. The current increases quickly at first, then more and more slowly as it approaches i=V/R.

j

Where can I get a 9V battery that does that? :=) I offer $100 USA on the spot.

Bill

-- Ferme le Bush

------- Not voltage inside- voltage across. What is the resistance of the circuit? You will have a differential equation to solve

9=L(di/dt) +Ri At t=0, i =0, Ldi/dt =9V The current will rise exponentially (i =(9/R)e^-Rt/L) and the voltage across the inductance will be (9-I*R) (note: assuming the inductor has no resistance- which is a lie- and that capacitance is negligable) so as time increases you will end up approaching i=9/R and di/dt =0. Draw a graph. I also suggest getting the basics of calculus which seem to be the root of your problem.

jay"

Only in textbooks

He's talking about what happens when he disconnects the battery. Here's "steady state" prior to disconnect:

+--------+ | | [Batt] [Inductor] | | +--------+

so i = 9/(R_batt + R_inductor) and V (across inductor) = i*R_inductor

At t0, when he disconnects the battery, voltage across the inductor will rise from V = 9 - i(Rinductor) to V = L* di/dt as the collapsing magnetic field around the inductor "attempts" to keep i constant. The inductor, wiring to it and the air gap at the disconnect point form an L R C circuit that will determine dt.

Ed

-------- You are right, I looked only at the connect. As you indicate the unknown RLC circuit when disconnection occurs (unknown in that what is the capacitance and what is the arc resistance?) so what voltage occurs due to Ldi/dt is also unknown. --

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer