Question

when simplifying Vo sinwt = iR+ L.di/dt ,to undergo first order linear differention, why is Vo divided by R instead of L?

APPLICATION FIRST ORDER LINEAR Given: a resistor inductor circuit driven by a sine wave generator with resistance R ohms, L Henrys. The generator voltage is given by V = Vo sin wt AA R Vo sin (w t) (Yeah, that's an inductor let's see you do better!) di The differential equation for the circuit (from Kirchoff's voltage law) is Vo sin(wt) = i R + L dt Solve this differential equation for the current. Then write an expression for the voltage across the resistor. Graph the equation you get -you may use a TI calculator but do it ultimately on MS EXCEL. When graphing the equation you will have to pick values for R, L and Vo and omega. You can pick whatever values you want, but I recommend using the value of 1 for all these parameters. By looking at the graph, state when is the voltage across the resistor highest and lowest (for what values of frequency)? This is an example of a low pass filter.

Answer 1

This is because the general form is y'+a(t) y=b(t)

The differential equation here is in i so to have the term as i'(t) we divide by term R to get it in the general linear form