Charge and potential differnecr in a mixed circuit

[Laxman2974]

Homework Statement

a potential difference V = 100 V is applied across a capacitor arrangement with capacitances C1 = 11.0 µF, C2 = 7.00 µF, and C3 = 3.00 µF. Find the following values Charge on each capacitor, potential difference on each capacitor, and stored energy in Joules.

The circuit is set up with battery and then a mixed series and parallel. C1 and C2 are in a series on a branch and C3 is on a separate branch in parallel with both C1 and C2.

Homework Equations

In a series I find charge with (1/c1 + 1/c2) = q/v looking for q. I plugged in (1/11 + 1/7) = q/100. SOlved and got a charge of 23.4 microCoulombs each on C1 and C2.

So now I am stuck I thought I solved the series first and then used that combine capacitance to find the charge on C3. Help

 

[LowlyPion]

I would suggest finding the equivalent capacitance of all of them before applying voltage.

As I read your problem you have C1 and C2 in series with each other but taken together in parallel with C3. With the value of the effective capacitance you can calculate total stored charge in the system. Then maybe work backwards splitting the charge between the two branches could give you some insight?

 

[Moetasim]

I would like to clarify that potential difference and charge are two different concepts in a circuit. Potential difference, also known as voltage, is the difference in electric potential energy per unit charge between two points in a circuit. On the other hand, charge is the amount of electric charge stored in a capacitor.

To find the charge on each capacitor, we can use the equation Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference. Plugging in the values given in the problem, we can find the charge on each capacitor:

Q1 = (11.0 µF)(100 V) = 1100 µC
Q2 = (7.00 µF)(100 V) = 700 µC
Q3 = (3.00 µF)(100 V) = 300 µC

To find the potential difference on each capacitor, we can rearrange the equation to V = Q/C. Using the values we calculated for Q, we get:

V1 = 1100 µC/11.0 µF = 100 V
V2 = 700 µC/7.00 µF = 100 V
V3 = 300 µC/3.00 µF = 100 V

Since the circuit is set up in a mixed series and parallel configuration, the total capacitance of the circuit can be found by calculating the equivalent capacitance using the equations for series and parallel capacitors. The equivalent capacitance for C1 and C2 in series is 4.4 µF, and when this is added in parallel with C3, the total equivalent capacitance is 7.4 µF.

To find the stored energy in the circuit, we can use the equation W = 1/2CV^2, where W is the stored energy, C is the equivalent capacitance, and V is the potential difference. Plugging in the values, we get:

W = 1/2(7.4 µF)(100 V)^2 = 3700 µJ or 3.7 mJ

I hope this helps clarify the concepts of potential difference and charge in a circuit, and how they are related to capacitance. It is important to note that the charge on each capacitor is not affected by the configuration of the circuit, but the potential difference and stored energy can change depending on the arrangement of the capac