Hi Jake,

I was mistaken when I said your equation was incorrect. That's the equation I gave for a parallel RC circuit. The only difference being you are taking i[t] as a given and you are solving for v[t]. I didn't notice that they were the same equations in different form.

I think your notebook looks great.

Kind regards,

David

Hi Jake, I think your equation is incorrect. For parallel (trivial) and series RC circuits driven by an arbitrary v[t] I get the solutions below. (Sorry about Out[4] -- this new code system seems awkward at best.)

In[1]:= parallelEq = i[t] == c v'[t] + v[t]/r

Out[1]= i[t] == v[t]/r + c Derivative[1][v][t]

In[2]:= (* trivial *)
DSolve[parallelEq, i[t], t]

Out[2]= {{i[t] -> (v[t] + c r Derivative[1][v][t])/r}}

In[3]:= seriesEq = v'[t] == r i'[t] + i[t]/c

Out[3]= Derivative[1][v][t] == i[t]/c + r Derivative[1][i][t]

In[4]:= DSolve[seriesEq, i[t], t]

Out[4]= {{i[t] -> E^(-(t/(c r))) C[1] + E^(-(t/(c r))) \!\(
\*SubsuperscriptBox[\(\[Integral]\), \(1\), \(t\)]\(\*
FractionBox[
RowBox[{
SuperscriptBox["E", 
FractionBox[
RowBox[{"K", "[", "1", "]"}], 
RowBox[{"c", " ", "r"}]]], " ", 
RowBox[{
SuperscriptBox["v", "\[Prime]",
MultilineFunction->None], "[", 
RowBox[{"K", "[", "1", "]"}], "]"}]}], "r"] \[DifferentialD]K[1]\)\)}}

See if you can modify this

In[1]:= i[t] = 4; r = 2; c = 3;
sol = v[t] /. DSolve[{v'[t] == (1/c)*(i[t] - v[t]/r), v[0] == 0}, v[t], t]

Out[2]= {8 E^(-t/6) (-1 + E^(t/6))}

In[3]:= Plot[sol[[1]], {t, 0, 20}]

Out[3]= ...PlotSnipped...