Explore BrainMass

# Physics: Potential Difference of a Capacitor

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Part 1:

A 0.410 mF (C1) and a 0.630 mF (C2) capacitor are connected in series to a 12.00 V battery. Calculate the potential difference across C1.

Part 2:

Calculate the charge on C2.

Part 3:

Now assume that the two capacitors are connected in parallel. Calculate the voltage difference across C1.

Part 4:

Calculate the charge on C2.

https://brainmass.com/physics/capacitors/physics-potential-difference-capacitor-296220

## SOLUTION This solution is FREE courtesy of BrainMass!

1. The total capacitance in series is:
C = (0.41mF)(0.63mF)/(0.41mF+0.63mF)
C = 0.248mF
The total stored charge is:
q = CVbatt
q = (0.248mF)(12V)
q = 2.976mC
The voltage at C1 is:
V1 = q/C1
V1 = 2.976mC / 0.410mF
V1 = 7.259 V

2. Since the charge in series is constant, and we have solved it in the previous problem, the charge at C2 is:
q2 = q = 2.976mC

3. When the capacitors are connected in parallel, the voltage between 2 capacitors are constant. So we have:
V1 = Vbatt = 12V

4. The charge across C2 is:
q2 = C2 Vbatt
q2 = (0.630mF)(12V)
q2 = 7.56mC

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!