# Find an Expression For the Voltage

1. Jan 29, 2010

### mmmboh

Hey, so I have an assignment and this is one of the questions on it, I did it but I am not sure if what I did is right.

What I did was find Ceq=(C1C2)/(C1+C2))

So then Q=CV(t)=(C1C2)/(C1+C2))V(t)

Then I found the voltage across C2=Q/C2=(C1C2)/(C1+C2))V(t)/C2=(C1)/(C1+C2))V(t)

And then V0=V(t)-(C1)/(C1+C2))V(t)

Can anyone tell me if what I did is right or if I am completely off?

2. Jan 29, 2010

### Leptos

For starters, shouldn't it be Q = CVT = (C1C2VT)/(C1+C2)?

3. Jan 29, 2010

### xcvxcvvc

You could have just found the voltage across $$c_1$$ like you did for $$c_2$$ and stopped there.: $$\frac{q}{c_1}=v_o=v(t) \frac{c_2}{c_1+c_2}$$
Your expression you found simplifies to that anyway:
$$v_o=v(t)[1 - \frac{c_1}{c_1+c_2}]$$
$$v(t)[\frac{c_1+c_2}{c_1+c_2} - \frac{c_1}{c_1+c_2}]= v(t) \frac{c_1+c_2 - c_1}{c_1+c_2}=v(t) \frac{c_2}{c_1+c_2}$$
I don't know if you know about complex impedance yet but you can also use that to arrive to the same answer:
$$Z_T = Z_C_1 + Z_C_2$$
$$Z_T = \frac{1}{\omega c_1}+\frac{1}{\omega c_2}$$
These impedances work like resistance with v = IR. Therefore,voltage division works too.
$$v_o = v(t) \frac{ Z_C_1}{Z_T}$$
$$v_o = v(t) \frac{ \frac{1}{\omega c_1}}{\frac{1}{\omega c_1}+\frac{1}{\omega c_2}}$$
Omegas cancel and you come to the same answer.

Last edited: Jan 29, 2010
4. Jan 31, 2010

### Redbelly98

Staff Emeritus
Just for the record: no, Q=CV for a capacitor, as the OP said.