Circuits capacitors, voltage and charge problem.

[zyphriss2]

A. Suppose in the figure that C1 = C2 = C3 = 54.2 microfarads and C4 = 68.1 microfarads.

If the charge on C2 is Q2 = 62.4 microfarads determine the charge on each of the other capacitors.


B. charge=voltage*capacitance
series equivalent capacitance=1/c1 + 1/c2 + 1/c3 + ...
Parallel equivalent capacitance= c1+c2+c3


I just tried taking the voltage across the C2 then plugging it into find the charge at c4 which i thought was 49.6 microfarads...I knew it was a long shot but its my last problem and I have no clue where to start.

 

[tiny-tim]

Hi zyphriss2!

To find the charge across C4, you obviosly need to know the voltage across C4 …

how would you find that?

 

[nlightner]

A. To solve this problem, we can use the formula Q=CV, where Q is the charge, C is the capacitance, and V is the voltage. We know that C2 has a charge of 62.4 microfarads, and all the capacitors have the same capacitance of 54.2 microfarads except for C4, which has a capacitance of 68.1 microfarads. So we can set up the following equations:

Q1 = 54.2 microfarads * V1
Q2 = 62.4 microfarads * V2
Q3 = 54.2 microfarads * V3
Q4 = 68.1 microfarads * V4

We also know that in a series circuit, the voltage is divided among the capacitors, so we can set up the equation V1 + V2 + V3 + V4 = Vtotal. And in a parallel circuit, the charge is divided among the capacitors, so we can set up the equation Q1 + Q2 + Q3 + Q4 = Qtotal.

Using these equations, we can solve for the charges on each capacitor:

Q1 = (54.2/239.9) * 62.4 microfarads = 14.1 microfarads
Q2 = (62.4/239.9) * 62.4 microfarads = 16.2 microfarads
Q3 = (54.2/239.9) * 62.4 microfarads = 14.1 microfarads
Q4 = (68.1/239.9) * 62.4 microfarads = 17.6 microfarads

Therefore, the charge on each capacitor is Q1 = 14.1 microfarads, Q2 = 16.2 microfarads, Q3 = 14.1 microfarads, and Q4 = 17.6 microfarads.

B. Your approach of using the formula Q=CV is correct. However, in a series circuit, the equivalent capacitance is given by the inverse sum of the individual capacitances, so it should be 1/54.2 + 1/54.2 + 1/54.2 + 1/68.1 = 1/239.9. And in a parallel circuit, the equivalent capacitance