Thank you for the explanations in this thread and also the derivations

I have managed to get to grips with how to derive an expression for I(t) and then finding the maximum.

Really my goal is now to justify that if both the peak current and voltage swing are known, R=V/I can approximate small changes (mOhm level) in the resistance in the circuit.
This is what I was really doing - trying to get an expression for the peak current so that I could justify removing I and C and just using R=V/I

I now seem to understand that this is more-so the case when the circuit is overdamped. What is the explanation for why when the circuit is overdamped that the inductance has little influence on the peak current so that you can approximate it with V=IR? Obviously it is something to do with the damping factor.

In fact the trouble I had explaining some real life measurements I think was due to the fact that some of my configurations were underdamped and using R=V/I isn't that accurate in this case. I have to make sure all my circuits have low inductance.

Here I have attached simulation of an RLC circuit

I have also plotted the real resistance in the circuit, as well as the 'calcresistance' where I have estimated it by dividing the voltage by the peak current (V=IR ohms law).
I have also plotted the 'ratio' between these

As we can see, when the resistance is small (circuit underdamped), my calcresistance using V=IR is several times the actual resistance in the circuit. However as the resistance increases (circuit overdamped) this ratio tends closer and closer to 1

Really I need to find a concise explanation for this.

 

Hello,

First, the determination of whether or not the circuit is under or over damped does not come from the resistance alone. It comes from the discriminant. If we get a negative discriminant then the solution contains sines and or cosines, but if positive then it is overdamped. The expression is:
C*R^2-4*L
If that is positive then it is overdamped, but if negative then underdamped. But you can see it contains R as well as C and L.

Second, if the inductance is small you can see that expression will be positive. But also, when L is small that means that it's current can follow another elements current much faster than if it were large. If it were large it would take longer to build up the current. if it is small then it can follow the capacitor current pretty closely and therefore does not influence the total response as much.
It's almost like having a 10 ohms resistor in series with a 0.1 ohm resistor: the 0.1 ohm resistor has little influence on the total current while the 10 ohm resistor determines the most.

 

This is equivalent to requiring that \(\frac{4L}{CR^2}>1\) which I already pointed out in post #12.

But besides that, the expressions I've given work for both the underdamped and overdamped cases (just not the critically damped case) provided the ArcTan of complex numbers is handled properly. Many people may not have a calculator that can evaluate ArcTan for complex arguments, so they will only be able to use the expressions I've given for the underdamped case, but for those with advanced calculators, the given expressions work for both cases.

Hi
Could you please provide expression for overdamped cases. 4L/R^2C < 1
i.e 0.45 in my exercise. R=600mohms, L = 22uH ., c=540uF
Thanks in advance

 

Hi
Could you please provide expression for overdamped cases. 4L/R^2C < 1
i.e 0.45 in my exercise. R=600mohms, L = 22uH ., c=540uF
Thanks in advance

The expression for both the underdamped and overdamped cases is provided in post #12.