In the solution of this problem they wrote alpha= R/2L. That formula applies when its a series RLC. What elements of the circuit are we talking about when we're saying series/parallel. I used to think it was the inductance and the capacitor. Does opening the switch making the inductance capacitor resistor a separate circuit has anything to do with it ?




Here's how I think if i'm asked to find i(t) -current through the inductance - for t > or = 0

I get i(0) so I get a relationship like this A1+A2 =0

then I derive the general equation of the case.
=> di/dt = ...
then I think that di/dt isn't "anything" so I remember that the voltage of a inductance is V= L di/dt. so I multiply both sides of the newly derived equation to get V= ...
In this circuit V(0) would be 0 because the inductance became a wire.

Is this a wrong way to think ? It's never working for me.

Back to the solution I posted. How did they find di(0)/dt ?

 

In the solution of this problem they wrote alpha= R/2L. That formula applies when its a series RLC. What elements of the circuit are we talking about when we're saying series/parallel. I used to think it was the inductance and the capacitor. Does opening the switch making the inductance capacitor resistor a separate circuit has anything to do with it ?

Yes - this is a series circuit. There is only one path through which current may flow around the remaining RH loop when the switch opens - hence no parallel paths exist.

Back to the solution I posted. How did they find di(0)/dt ?

If you consider the initial circuit conditions - starting with the general DE for vo(t) in terms of the LH side of the loop

\(v_{o}(t)=Ri_{o}(t)+L\frac{di_{o}(t)}{dt}\)

Hence

\(\frac{di_{o}(t)}{dt}=\frac{1}{L}(v_{o}(t)-Ri_{o}(t))\)

At t=0

\(v_{o}(0)=75V\)

\(i_{o}(0)=15mA\)

\(\frac{di_{o}(0)}{dt}=\frac{1}{L}(v_{o}(0)-Ri_{o}(0))\)

\(\frac{di_{o}(0)}{dt}=\frac{1}{1}(75-5000*15e^{-3})=0\)

 

How is your last equation equal to zero ? It's equal to -3659

 

\(15e^{-3}\)

is meant to be scientific format notation - for the 15mA value. I'm surprised you didn't realise that.

perhaps I should have either written more correctly [by convention]

\(15E^{-3}\)

or

\(15*10^{-3}\)

I tend to use notations interchangeably & somewhat ambiguously perhaps.

In any case the solution I gave is correct if you allow my intended meaning.

 

ok That makes sense. Thank you.