Understanding Capacitor Networks: 120+(-60), 30+(-90)

AI Thread Summary
The discussion focuses on understanding capacitor networks, specifically the voltage and charge distribution in a circuit with given values of 120+(-60) and 30+(-90). Participants express confusion about how to determine the charge on the capacitors and the placement of a voltmeter in the circuit diagram. It is noted that after equilibrium, no current flows without a resistive path, indicating a need for clarity in the problem's wording. To analyze the circuit, it is suggested to treat the network as two parallel networks and use the formula Q=VC to find charges. Calculating the equivalent capacitance before and after shorting the switch is emphasized as a crucial step to understand charge flow and voltage at each node.
herich
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Your diagram is rather unclear. I don't see any voltmeter. Where is it supposed to be?
 
herich said:
How can I know the top 2 plates are +60 and below are -60?
120+(-60) and 30+(-90)?
And how can I fill in the blanks~?

The statement of the problem is incorrect and I will assume it is sloppy wording, because after equilibrium is reached, there is no current flowing anywhere in the circuit with no restive path between the terminals of the battery.

But taking the awkward wording as meaning they want to know the total flow of charge as a result of shorting the switch, I have to ask where it is you got the figures for the charges on the capacitors before or after the switch is shorted?

Please show your work as to how you arrived at those figures at the points x and y.
 
Actually there are the figures deduced from my friend's tutor.
But I know that first we can treat the network as two parallel networks, and the two sides of the capacitors can simply computed by using the formula Q=VC.

I just doubt how <+60>and <-60> can be found out. And how to deduce in what way the charge will flow.
 
herich said:
Actually there are the figures deduced from my friend's tutor.
But I know that first we can treat the network as two parallel networks, and the two sides of the capacitors can simply computed by using the formula Q=VC.

I just doubt how <+60>and <-60> can be found out. And how to deduce in what way the charge will flow.

Why don't you start out then by calculating the equivalent capacitance of each leg. Then you will know the first part of what you need - the charge on each capacitor - and hence the voltage at each node.

Then you proceed to calculate the equivalent capacitance of the new network after the short is affected. Armed with that you can calculate the effective new charges and then you can describe the difference.
 

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