Find an Expression For the Voltage

[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 C_eq=(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?

 

[Leptos]

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 C_eq=(C₁C₂)/(C₁+C₂₎)

So then Q=CV(t)=(C₁C₂)/(C₁+C₂₎)V(t)

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

 

[xcvxcvvc]

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 C_eq=(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?

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.

 

[Redbelly98]

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

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