# Analyzing voltage divider circuit with capacitors

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

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

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

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

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. $$Vout = \int \frac{I_z}{C_4}dt = \frac{I_zt}{C_4}$$