Homework: DC circuit with capacitors

[gruba]

Homework Statement

In the given DC circuit, while switches S1,S2 are open, conductor C1 is charged with Q10, and conductor C2 is not charged. After the switch S1 is closed, through conductors flows the amount of charge q1=40µC. After the switch S2 is closed, through conductors flows the amount of charge q2= -50µC. Find E3 and Q10.

E5=15 V

Ig=0.5 A

R1=200 Ω

R2=100 Ω

R3=100 Ω

R4=60 Ω

R5=30 Ω

R6=40 Ω

C1=2 µF

C2=5 µF

Homework Equations

- DC circuit analysis methods

The Attempt at a Solution

I tried using node voltage method, but it gives three equations with four variables.
How to find voltage in the branch where capacitors are (for every case)?

 

[mfb]

I tried using node voltage method, but it gives three equations with four variables.

Please show your work, otherwise it is impossible to tell what you missed.

 

[gruba]

Please show your work, otherwise it is impossible to tell what you missed.

Circuit has four nodes. In the first case (switches are open), setting referent potential V₃=0 and applying voltage node method for nodes 1,2,4 gives:

1: \frac{V_1}{20}-\frac{V_4}{20}=-1 \Rightarrow V_1=V_4-20

2: \frac{V_2}{50}-\frac{V_4}{100}=\frac{E_3}{100} \Rightarrow 2V_2-V_4-E_3=0

4: \frac{117V_4}{1800}-\frac{V_1}{20}-\frac{V_2}{100}=\frac{1}{2}-\frac{E_3}{100}

This system can't be solved.

 

[mfb]

There is no voltage drop at the amperemeter.

It's hard to follow your equations if you directly plug in numbers.

 

[gruba]

There is no voltage drop at the amperemeter.

It's hard to follow your equations if you directly plug in numbers.

Do you have any suggestions what method to use in this problem?

 

[mfb]

There is no voltage drop at the amperemeter.

You can directly find V1 without calculations, then you are down to three unknowns for three equations.

 

[gruba]

You can directly find V1 without calculations, then you are down to three unknowns for three equations.

How do you find potential V₁?

 

[mfb]

With the hint I posted twice:
There is no voltage drop at the amperemeter.