Message Boards

WOLFRAM COMMUNITY

8123 Views

8 Replies

5 Total Likes

View groups...

Follow this post

Share this post:

GROUPS:

How do I solve and plot the equations for an RC circuit?

Jake Trexel

Jake Trexel, n/a I am disabled and no longer work as a professor.

Posted 10 years ago

I am trying to solve the and plot the equations for an RC circuit. What I have so far is v'[t] == (1/c)*(i[t] - v[t]/r) Any help would be appreciated. Thanks

POSTED BY: Jake Trexel

8 Replies

Sort By:

Bill Simpson

Posted 10 years ago

I am guessing from your wording that something didn't work, but I can't guess what. Can you be much much more specific about what you want, what you did and what went wrong? Or, if you start Mathematica fresh, type in the simplest example of exactly what you are trying to do, don't edit or replace or try again and again, just type (or paste, but even that can be more problematic) your problem, evaluate the whole notebook once, immediately save the notebook and attach that notebook to a post along with a clear simple verbal description of exactly what you were trying to accomplish and why it doesn't seem to be the result you were looking for then I'll give another try at what I hope will help you get where you want to go.

POSTED BY: Bill Simpson

David Keith

Posted 10 years ago

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

POSTED BY: David Keith

David Keith

Posted 10 years ago

Hi again -- I attach the notebook. It assumes we will know v[t] and want i[t] for each. Note that in the parallel case we don't need to solve a diffeq to get that. In the series case we do, and Mathematica supplies integration constants which can be used to accommodate initial conditions. Attachments:

POSTED BY: David Keith

Bill Simpson

Posted 10 years ago

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...

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...

POSTED BY: Bill Simpson

Jake Trexel

Jake Trexel, n/a I am disabled and no longer work as a professor.

Posted 10 years ago

Bill, this gave me more problems than I had!

POSTED BY: Jake Trexel

David Keith

Posted 10 years ago

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]\)\)}}

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]\)\)}}

POSTED BY: David Keith