Charge or Discharge: Calculating Current with i(t)=Vth/RTh

[influx]

http://photouploads.com/images/zxwx.png
Why are they using the i(t) = Vth/RTh ... formula to work out the current? I mean in the image on the right (in the above link), the inductor is de-energising through the resistors so why are they using an equation that is used when energising?

 

[NascentOxygen]

They are showing the general equation. If the network included a voltage source, then that equation would apply for the current through the inductor. It simplifies here because the voltage source is zero.

While you can say the inductor is de-energizing, you could equally consider it to be energizing to a new steady state current (of V_TH/R_TH), and so keep the formula as general as possible until substituting data values at the last line.

 

[gneill]

http://photouploads.com/images/zxwx.png



Why are they using the i(t) = Vth/RTh ... formula to work out the current? I mean in the image on the right (in the above link), the inductor is de-energising through the resistors so why are they using an equation that is used when energising?

My guess would be that somewhere in your course materials they derive an equation that uses the Thevenin equivalent as part of a general solution.

In RL or RC circuits the responses of voltages and currents always follow an exponential curve, one that either decays from one level to another, or increases from a starting level to a higher level (plateau). With appropriate signed constants a general equation produce both "versions".