capacitordctimevoltage
Here is a diagram of the circuit, I am confused about problems a and b.
How do I go about solving these two problems? Thanks!
By the way you don't have to tell me the answer, I just want to know how I should think about it, or any helpful hints would be appreciated. Thanks.
Considering the top circuit :
Applying KCL at the nodes :
A : (Vd - Va) / R1 = ( Va / sC1) + ( (Va - Vb) / sCx)
B : ( (Va - Vb) / sCx ) + ( (Vd - Vb) / sC2) = ( Vb - VL) / R2
L : ( (Vb - VL) / R2 ) = VL / sCL
3 equations for 3 unknowns.
( Don't forget to put Laplace transform of input ie of Vd).
(Did I miss something ?)
Here, let me set up the equations for you:
simulate this circuit – Schematic created using CircuitLab
You have five unknowns — three currents and two voltages. We'll need five equations.
KCL tells us that the sum of currents into any node must be zero:
$$I_{L1} + I1 - I_{L2} = 0$$
$$I_{L2} - I2 - I_{L3} = 0$$
$$-I_{L1} - I1 + I2 + I_{L3} = 0$$
Note that the third equation tells us nothing new; it is simply a combination of the first two.
KVL tells us that the sum of voltages around a loop must be zero, which is really just another way of saying that the voltage across I1 must equal the voltage across R1, etc. We simply apply Ohm's Law to the resistors:
$$V1 = -I_{L1} \cdot R1$$
$$V1 - V2 = I_{L2} \cdot R2$$
$$V2 = I_{L3} \cdot R3$$
Now we have five independent equations that we can solve for our five unknowns.
As Alfred Centauri says, the principle of superposition makes this problem almost trivial to solve, but you may not have been exposed to that yet.
Best Answer
Hints:
A capacitor's voltage-current response is governed by: \begin{equation} I = C \frac{dV_c}{dt} \end{equation}
At \$t = 0\$: think about what happens when you integrate in the limit as \$dt \rightarrow 0\$ (i.e. what is a capacitor unable to do with respect to really fast transients?)
\begin{equation} \lim_{\delta \rightarrow 0^+} \int_{0^-}^{\delta} I dt = \lim_{\delta \rightarrow 0^+} C \int_{0^-}^{\delta} \frac{d V_c}{dt} dt \end{equation} Note that this equation does not imply \$I(0^+) = 0\$; it only provides a mathematical "weak" guarantee that it's integral doesn't change over the infinitesimal.
At \$t \rightarrow \infty\$, think of what happens when all transient behavior goes away (\$\frac{d <value>}{dt} = 0\$).