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1. ### Polynomial approximation to find function values

Polyfit outputs polynomial coefficients, so I think what you'd have to do is plot ln(R) = ln(a) + b/T, and look for b = 0. Now you will know what value a is. From there, divide your polyfit matrix by a, and you will have the coefficients to the power series of eb/T. With the power series...
2. ### Engineering Derivation of Source-Free RC circuit

Now, to elaborate on the above (sorry for separate post) equation of v(t) = Ae(-t/RC) where V(0) = A = V0. Your equation is ln(V) = -t/RC + A. So let's check this out with the info I gave you in my previous post: Remove the natural log: eln(V) = e-t/RC + A Now rewrite: V = e-t/RCeA...
3. ### Engineering Derivation of Source-Free RC circuit

Two things you can do here. You're answer is already correct even though it says "+A" and not "+ln(A)". Let me explain. When you get a constant of integration, it is representative of a single value determined from initial conditions. If no conditions are given, it remains a letter; in...
4. ### Explanation of graphs involving capacitors (charging/discharging)

It's all relative, that's what I'm saying. It isn't charged below 0V with respect to where the switch is initially. It appears charged with a positive V here. It's when you flip the switch to position 2 that it appears to be below 0V. This is because the relative voltage (ground) of the...
5. ### Explanation of graphs involving capacitors (charging/discharging)

Yes that is what appears to be happening. Since C2 is not fully discharged upon altering the switch, it looks to R3 and C1 as having a "negative" voltage. Remember that the reference voltage is 0V. The initial voltage at t=0 in the new configuration goes up to bring the total to 4V across R3...
6. ### Explanation of graphs involving capacitors (charging/discharging)

Are they all supposed to be on the same plot? What I mean is, are they all being plotted simultaneously or are you just trying to superimpose three separate plots as one? If it's the latter, it could be that the Voltage amplitudes don't actually correspond. It would make sense to me that each...