SammyS said:
What does That refer to?
In the OP, your statement of the problem is not clearly stated. It's split between 1. The problem statement, and
Homework Equations
.
Then, in Post #20, you add to the problem statement.
So I repeat ehild's request for a complete statement of the problem.
Where does this last part come from? I see it mentioned nowhere else. Did I miss it?
So I repeat ehild's request for a complete statement of the problem.
Do you want me to post it as a separate reply or as a part of the original post? Also the R0=R1=R2=R is an edit, looking at post #21 you can see it was originally written as part of the problem.
In any case here is the full description as a reply:
"We consider the circuit shown in Figure 2, where E is an emf source, R0, R1, and R2 are resistances and L is an ideal coil. The switch is turned on at the time t=0 and the currents i1 and i2 run through the circuit as indicated by the figure.
a.) Determine i1 and i2 right after the switch is turned on at the time t=0 as well as for t => infinity.
ANSWERED
b.) Show by using Kirschhoff's law at the time t>=0 that
\begin{equation}
\frac{di_1}{dt} + \frac{\tilde{R}}{L}i_1 = \frac{\tilde{\mathcal{E}}}{L}
\end{equation}
and determine R~ and E~.
Found R~ and E~ by using Thevenin's theorem.
\begin{equation}
\tilde{\mathcal{E}} = \mathcal{E}\frac{R_2}{R_0 + R_2}
\end{equation}
\begin{equation}
\tilde{R} = \frac{R_0R_1 + R_2R_0 + R_1R_2}{R_0 + R_2}
\end{equation}
Not quite shown the equation yet.We now assume that R0=R1=R2=R.
c.) Determine i1(t) and i2(t) for t >= 0 and plot them as functions of t.
This should just be inserting R into the place of R0, R1, and R2 in part a.
Short term:
\begin{equation}
i_1(t) = \frac{\mathcal{E}}{3R}(1-e^{-\frac{2R}{3L}t})
\end{equation}
\begin{equation}
i_2(t) = \frac{\mathcal{E}}{3R}(1-\frac{1}{2}e^{-\frac{2R}{3L}t})
\end{equation}
Long term:
\begin{equation}
i_1 = \frac{\mathcal{E}}{3R}
\end{equation}
\begin{equation}
i_2 = \frac{\mathcal{E}}{3R}
\end{equation}