- #1

diredragon

- 323

- 15

## Homework Statement

##E_1=6V##

##E_3=5V##

##R_1=150Ω##

##R_2=R_3=100Ω##

##R_4=50Ω##

##R_5=300Ω##

##C_1=1.5 mF##

##C_2=0.5mF##

Text:

In the first condition when the switches 1 and 2 are open the capacitor 2 has charge on it. First the switch 1 closes and the charge flow of ##q_1=1 mC## is established to flow through the branch as shown in the picture. Then the switch 2 closes and the flow is now ##q_2=-1.4mC##. Find the initial charge on capacitor 2 (##Q_{20}##).

## Homework Equations

3. The Attempt at a Solution [/B]

My only problem here is with the charge division. Correct me if I'm wrong here:

When the switch 1 closes the voltage across 1 and two must be the same so we have:

##q_{11}+q_{12}=q_1##

##\frac{q_{11}}{C_1}=\frac{Q_{20}+q_{12}}{C_2}##

I can't find the voltage they're equal to because i don't have ##E_2## but i do have the change when the switch 2 closes. So:

##\frac{q_{11}+q_{21}}{C_1}=\frac{Q_{20}+q_{12}+q_{22}}{C_2}##

The change is then just:

##ΔU_{c1}=\frac{q_{21}}{C_1}##

##ΔU_{c2}=\frac{q_{22}}{C_2}##

And ##q_{12}+q_{21}=q_2## and from there i can find the ##q_{12}## and ##q_{22}## and have the change in the voltage while my Δ-circuit looks like this:

Where finding the current is trivial. Is this the right track?

Thanks