3. For the circuit below, obtain the differential equation that governs $v_2(t)$.
This can be obtained relatively easily by thinking in terms of KCL applied to the $v_1$ node.
Express all the currents in terms of the voltages. Then, consider the current $i_2$. There are two
ways to express this current. One way involves Ohm's law. The other way employs the $i-v$
relationship for the capacitor $C_2$. Equating these two expressions for $i_2$ allows you to obtain
an expression for $v_1$ in terms of $v_2$. Using that (and returning to the KCL expression) allows
you to obtain a differential equation that only involves $v_2$.
You should obtain a second-order differential equation with constant coefficients. We could,
if we so desired, write the equation in the form:
$$\frac{d^2v_2(t)}{dt^2} + 2\alpha \frac{dv_2(t)}{dt} + \omega_0^2 v_2(t) = 0.$$
The $\alpha$ and $\omega_0$ depend on $R_1$, $R_2$, $C_1$, and $C_2$ (in a way that is distinct from what we have
with parallel or serial $RLC$ circuits), but the way in which $v_2(t)$ behaves would be no dif-
ferent from what we've seen before. It would be either underdamped, critically damped, or
underdamped (depending on the relative sizes of $\alpha$ and $\omega_0$).
Finally, you can be fairly confident that you have the right differential equation if, when you
set $R_2$ to zero, you get a first-order differential equation with an equivalent capacitance of
$C_1+C_2$ (i.e., the time constant is $R_1(C_1+C_2)$ if $R_2$ goes to zero).
$i_0$
$v_1$ $i_2$
$R_2$
$v_2$
$i_1$
+
+
$R_1$
$C_1$
$v_1$
$C_2$
$v_2$
