Capacitor Circuit: Equivalent Capacitance & Charge

[elleeyeesay03]

Homework Statement

For the capacitor circuit shown below:
A. Find the equivalent capacitance
B. What is the charge on the 3$$\mu$$F capacitor?

Homework Equations

Q= C$$\Delta$$V

Parallel= C1 +C2+C3
Series= (1/C)=(1/C1)+(1/C2)+(1/C3)

C1= 8$$\mu$$F
C2= 5$$\mu$$F
C3= 3$$\mu$$F

The Attempt at a Solution

 

[gneill]

Hi elleeyeesay03, welcome to Physics Forums.

Have you made any attempt to simplify the circuit? Can you identify any serial or parallel combinations of capacitors?

 

[engineertech0]

C2 & C3 are in parallel so they would be equal to 8uf by adding 3uf + 5uf. You will then consider that once capacitor and take it in parallel with C1. Having to 8uf capacitors in parallel would yield you an equivalent of 4uf as the current would be splitting between the two.

 

[gneill]

C2 & C3 are in parallel so they would be equal to 8uf by adding 3uf + 5uf. You will then consider that once capacitor and take it in parallel with C1. Having to 8uf capacitors in parallel would yield you an equivalent of 4uf as the current would be splitting between the two.

While C2 and C3 are in parallel, yielding an 8μF equivalent, the resulting equivalent capacitance is in SERIES with C1, not in parallel with it. The net equivalent capacitance is indeed 4μF though, as you stated.

 

[rwisz]

I would approach this problem by first analyzing the given information and identifying the relevant equations and variables. From the given information, we can see that the circuit consists of three capacitors in parallel, with capacitance values of 8μF, 5μF, and 3μF respectively. To find the equivalent capacitance of capacitors in parallel, we use the formula C_eq = C1 + C2 + C3. Plugging in the values, we get C_eq = 8μF + 5μF + 3μF = 16μF. Therefore, the equivalent capacitance of the circuit is 16μF.

To find the charge on the 3μF capacitor, we can use the equation Q = CΔV, where Q is the charge, C is the capacitance, and ΔV is the potential difference across the capacitor. In this case, the potential difference across all three capacitors should be the same, as they are connected in parallel. Therefore, we can use the voltage across any one of the capacitors, which we can determine by using the equation for equivalent capacitance in parallel, C_eq = Q/ΔV. Rearranging the equation, we get ΔV = Q/C_eq. Plugging in the values, we get ΔV = Q/16μF. We also know that the potential difference across the 3μF capacitor is the same as the overall potential difference, so we can set it equal to ΔV. This gives us the equation 3μF = Q/16μF, which we can solve for Q. Multiplying both sides by 16μF, we get Q = 48μC. Therefore, the charge on the 3μF capacitor is 48μC.

In summary, the equivalent capacitance of the circuit is 16μF and the charge on the 3μF capacitor is 48μC. This approach demonstrates the use of relevant equations and logical reasoning to solve a capacitor circuit problem.