differential equation in electricity

[Kowalski]

Can anyone assist in this differential equation : Ldi/dt + Ri = Usin(wt).

 

Thanks a lot in advance

[timo]

The corresponding homogeneous differential equation (deq) to your problem would be [math] L \frac{dI}{dt} = -R I [/math], which is well known to have solutions of the type [math] I = e^{\lambda t}[/math]. The deq you present is inhomogeneous, as there is an additional driving term of type [math] \sin( \omega t) [/math]. The solutions of an inhomogeneous differential equation are the solutions of the corresponding homogeneous deq plus an arbitrary solution of the inhomogeneous one. From physical intuition, what I'd expect is that due to the damping term RI, the initial oscillation current is suppressed over time, and only the oscillation driven by the [math]\sin(\omega t)[/math] term remains. To make my point: I'd try [math] I(t) = e^{\lambda t} + \alpha \cos ( \beta t ) [/math] as an Ansatz (possibly even chosing [math] \beta = \omega[/math] from the very start on). No guarantee that it's going to work, though - I didn't test it.

Edited November 30, 2010 by timo

[TonyMcC]

Is the generator part of the circuit showing a step in voltage so that after a time equal to 5L/R the current will simply equal V/I amps? Or are we considering what happens as the input voltage rises at a constant rate from one value to another? I'm not really into differential equations but, without capacitance, I'm a bit puzzled by the reference to an oscillation being suppressed over time. Sorry if I'm out of my depth.

[timo]

I meant "initial current" (or more abstract: initial state), not "initial oscillation". Corrected. Don't take that sentence too serious, anyways. It's just a rough idea that's supposed to lead to an Ansatz, not a well-formulated result.

Edited November 30, 2010 by timo