How will capacitors C1 and C2 charge in this circuit?

I am learning about RC circuits and came across this question:



At t=0, both the capacitors are short circuited. I am confused what happens after that.

By KCL, all 5A passes through C2. Therefore:

$$ \begin{aligned} V_O &= V_{C2} \\ \\ &= \frac{1}{C_2}\int{i_{C2}\cdot dt} \\ \\ &= \frac{1}{C_2}\int{5\cdot dt} \\ \\ &= \frac{5}{C_2}t \end{aligned} $$

This is the case regardless of the presence or state of R and C1, and it is a linear and never-ending rise of \$V_{C2}\$.

For R and C1, initially there's no voltage across R, so no current through it, and all 5A passes through C1:

$$ \begin{aligned} i_{C1}(0) &= 5A \\ \\ i_R(0) &= 0A \\ \\ \end{aligned} $$

Voltage \$V_{C1}\$ across both C1 and R starts at zero, but rises over time as \$i_{C1}\$ flows and C1 charges. More and more of the 5A total flows through R, less and less through C1. Eventually:

$$ \begin{aligned} i_{C1}(\infty) &= 0A \\ \\ i_R(\infty) &= 5A \\ \\ \end{aligned} $$

The final value of \$V_{C1}\$, for which all 5A flows through R, is found with Ohm's law:

$$ \begin{aligned} V_{C1}(\infty) &= i_R(\infty) \times R \\ \\ &= 5R \\ \\ \end{aligned} $$

Until that happens, \$V_{C1}\$ follows the familiar exponential rise:

$$ \begin{aligned} V_{C1} &= V_{C1}(\infty) \left(1 - e^{\frac{-t}{RC_1}} \right) \\ \\ &= 5R\left(1 - e^{\frac{-t}{RC_1}} \right) \end{aligned} $$

The current through capacitor C2 will remain unchanged; so C2 will charge linearly and the voltage across it will increase indefinitely.

Capacitor C1 will initially charge up to the voltage across the resistor R, which will then remain constant since the current through R is constant.

Since there's a constant current, you can analyze the two capacitors individually, they don't interact.

C2 charges linearly from 0V at a rate proportional to current and inversely proportional to capacitance. You can write the equation for Vc2(t).

C1 starts out with 0V and ends up with V = 5A*R across it. After a very long time you just consider the capacitors as open circuits (and any inductors as shorts). The curve is exponential since the R has no effect at t= 0 and takes all the current at t = \$\infty\$.

For C1, you can convert the current source and resistor to the Thevenin equivalent voltage source and resistor and you have the familiar case of a capacitor charging towards a fixed voltage through a resistor, which you've no doubt solved before, so you can also write the equation for the voltage across that capacitor as a function of time.

The analysis is in an earlier answer. Regarding the four numbered questions:

Q1: The RC value is unclear, which R and which C? There are two C's. As for the R: just note that a current source has a large internal resistance. If you assume an "infinite" internal resistance, you could compute an infinite RC value if you wish.

But it is also unclear what you mean with "initially short circuited". You probably mean that at T=0 both capacitors are uncharged.

Q2: Yes, you already gave the explanation.

Q3: Yes, why not? Some current will go through C1.

Q4: With steady state you probably mean at T -> infinite. Then, the current in C1 will be zero, the voltage over R and C1 will be constant at 5 A times R. C2 will continue charging with 5 A so its voltage will increase each second with 5 A times C. Not quite a "steady" state.