HRW 27-53

I will start with the junction and two loop (left and right) equations:

$$\begin{eqnarray*} I_1-I_2-I_3 &=& 0 \\ \mathscr{E} -R_1I_1-R_2I_2 &=& 0 \\ -R_3I_3-\frac{Q}{C}+R_2I_2 &=& 0. \end{eqnarray*}$$

Note that $I_3$ is the current through the capacitor, so

$$I_3=\frac{dQ}{dt}.$$

Thus, we want to end up with a equation involving only $Q$ and $I_3$. Thus we solve the first equation for $I_1$ and substitute into the second equation to get

$$\mathscr{E}-R_1(I_2+I_3)-R_2I_2=0,$$

which we then solve for $I_2$ to get

$$I_2=\frac{\mathscr{E}-R_1I_3}{R_1+R_2}.$$

Now we substitute this into the third equation to remove $I_2$, which, after collecting terms, becomes

$$-\left( R_3 +\frac{R_1R_2}{R_1+R_2}\right)I_3 -\frac{Q}{C}+\frac{R_2\mathscr{E}}{R_1+R_2}=0.$$

At first glance this looks entirely intractable, but notice that everything besides $I_3$ and $Q$ is a constant! So, let's turn them into the more simplistic (and suggestive) constants:

$$\begin{eqnarray*} \left( R_3 +\frac{R_1R_2}{R_1+R_2}\right) &\to& R'\\ \frac{R_2\mathscr{E}}{R_1+R_2} &\to& \mathscr{E}', \end{eqnarray*}$$

which yields the equation

$$-R'\frac{dQ}{dt}-\frac{Q}{C}+\mathscr{E}'=0,$$

which should look very familiar. Since this is exactly the usual RC loop equation, we immediately obtain the solution

$$Q(t)=C\mathscr{E}'\left(1-e^{-t/R'C}\right),$$

in terms of our primed quantities. Thus the capacitor sees an effective EMF given by $\mathscr{E}'$, and this arrangement is known as a voltage divider (something you'll see a lot of in a circuits class...); and the capacitor sees an effective resistance given by $R'$, which is written as $R_3$ in parallel with the series combination $R_1$ and $R_2$, which is the equivalent resistance if we remove the EMF! Thus, this is the justification for the statements made earlier.

Finally, let us find the voltage $V_2$. This quantity is the same as the voltage through the parallel branch that includes the capacitor, thus we have

$$\begin{eqnarray*} V_2 &=& V_3+V_C\\ &=& I_3R_3+\frac{Q}{C}, \end{eqnarray*}$$

and we have

$$I_3=\frac{dQ}{dt}=\frac{\mathscr{E}'}{R'}e^{-t/R'C},$$

so putting things together we obtain

$$\begin{eqnarray*} V_2(t) &=& \frac{R_3}{R'}\mathscr{E}'\left(e^{-t/R'C}\right)+ ... ...thscr{E}'\left[1-\left(1-\frac{R_3}{R'}\right)e^{-t/R'C}\right], \end{eqnarray*}$$

Note that $R_3<R'$ so the term in the parenthesis is positive, and the function increaes in time as we know it must. Now, if we substitute in our values we have $\mathscr{E}'=\frac{1}{2}\mathscr{E}={\rm 600V}$ and $R/R'=2/3$, and thus

$$V_2(t)=({\rm 600V})\left( 1-\frac{1}{3}e^{-t/\tau}\right),$$

which yields the limiting values

$t$	$V_2(t)$
0	400V
$\infty$	600V

Moreover, the functional form of $V_2(t)$ has exponential dependence, just as we assumed.