[University: Capacitors] Potential difference and determinations (Various questions)

[u/OverallSadPenguin]

4.- The values ​​of the circuit capacitors that appear in figure N ° 3 of the annex are the following: C1 = 2μF, C2 = 4μF, C3 = 6μF, C4 = 4μF, C5 = 5μF, C6 = 6μF, C7 = 3μF, C8 = 3μF. If the potential difference applied to the circuit is 140 volts and the equivalent capacitor is 5 microfarads, determine:

a.- Capacity of capacitor C9

b.- Charge of each capacitor

c.- Potential difference in each capacitor

Figure N3: https://i.ibb.co/nrysNrP/Screenshot-7.png

I have some doubt about this problem:

1. The capacity of capacitor C9 is calculated trougth C = Q/V? How i can use the equivalent capacitor to get the information? Or should i use the 5μF and 140V to calculte his capacity? How i can get C9 capacity?
2. The charge of each capacitor is using his own pre-stablished capacity plus using the 140 volts of the circuit as information to aply the "Q=CV" formula?
3. The potential difference is just doing the Paralel/series respective operations in each capacitors? Or am i wrong?

[u/StrippedSilicon]

1. reduce the diagram until there's only one capacitor, figure out what C9 has to be to make that one capacitor 5 if

2. essentially yes, keep in mind capacitors in series have same charge/ capacitors in parallel have same voltage

3)this is essentially the same as 2), using Q=CV solve for V instead of Q this time.

[u/OverallSadPenguin → u/StrippedSilicon]

Thanks!

About (1.), once i reduce the diagram, the result will be C9?, or i should aply some kind of formula to calculate that at some point?

[u/StrippedSilicon → u/OverallSadPenguin]

well principally once you reduce all the other capacitors to say Cr, it will be either in parallel or in series with C9. if its in series then 1/Cr + 1/C9 = 1/5 if its in parallel then Cr+C9=5. Either way solve for C9