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Trying to find $V_{out}(t)$ given the following circuit:

enter image description here

I wasn't sure how to do it, but my approach was basically find the equivalent capacitance and $v(t) = \frac{It}{C_{eq}}$. $C_4$ and $C_3$ are in series, so I combined them with the series equation. Then that equivalent caapacitor is in parallel with $C_2$, so I just added them. $C_1$ doesn't really matter. But then, I get the wrong answer for $V_{out}$ which is supposed to be as written below

$$\frac{C_3I_st}{C_2C_3+C_2C_4+C_3C_4} $$

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  • $\begingroup$ There's no equivalent resistance for a capacitor, so what you have tried is wrong. Consider the charges on the capacitors. $\endgroup$
    – Massimo Ortolano
    Nov 2, 2019 at 18:30
  • $\begingroup$ oops i meant to say equivalent capacitance $\endgroup$
    – Melanie
    Nov 2, 2019 at 18:39

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You basically got the idea right, but after you get the equivalent capacitance you have to backtrack again to get Vout.

Let Cz = C3 parallel C4 --> $C_z = \frac{C_3 C_4}{C_3 + C_4}$ and let Cx = Cz series C2 --> $C_x = \frac{C_3 C_4}{C_3 + C_4} + C_2$ -> $C_x = \frac{C_3 C_4 + C_2 C_3 + C_2 C_4}{C_3 + C_4}$

Now first lets start with the source in series with Cx and C1, Cx will be getting a curret Is so $ I_s = C_x \frac {dV_x}{dt}$ so $ \frac {dV_x}{dt} = I_s / C_x$.

Now lets split Cx back into Cz in parallel with C2, they will both have the same voltage Vx. To ge the current in Cz: $I_z = C_z \frac {dV_x}{dt} = C_z I_s/Cx$

Finally lets split Cz back into C3 and C4. The current Iz passes through C4. $$ Vout = \int \frac{I_z}{C_4}dt = \frac{I_zt}{C_4}$$

substitute and simplify and you'll get your answer.

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