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## Homework Statement

In the following diagram, The voltmeter reading is 5 volts when the switch K is open and C1 is not charged.

When you close K calculate the following:

1- Voltmeter reading

2- Charge of each capacitor

3- Voltage of each capacitor

## Homework Equations

Usual electronics equation

## The Attempt at a Solution

Well, I was looking through my friend's private tutor papers and I found this question. My solution is different that his so I thought I could bring it here so someone can check it and see if I am missing something.

C1 is not charged thus it doesn't contribute to the voltage of the voltmeter (assuming actual current flowing or assuming it is negligible) so the voltage reading is purely C2's voltage

Now instead of assuming Q as transferred charged, you could choose a much simpler way.

C1 and C3 (it didnt say if C3 is charged so I assumed it is not) are in series then the resultant is parallel with C2

So

##\frac{Q_t}{C_{eq}} = V_{after~opening~ the switch} ##

you get ## V = 4.25 volts ##

The Charge for any of C1 and C3 capacitors is the voltage times the equivalent capacitance of C1 and C3 which is in series

## V ~ ( \frac{1}{C_1} + \frac{1}{C_3} ) = Q_{3 ~ or ~ 1} ##

## Q = 9 uC ##

the charge for C2 is simply

## Q_2 = C_2 V ##

## Q_2 = 51 uC ##

Now the first requirement needs the voltmeter reading which is:

##V = V_{C_2} - V_{C_1} ##

## V = 4.25 - Q/C_1 ##

## V = 4.25 - 2.25 ##

## V = 2 ## which he found to be the same

the rest is straight forward.

For the charges and voltages, he got it different because he assumed that C2 and C1 are connected in series which is wrong because they definitely don't have the same charge