Please solve and explain problem.
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.
Calculate the charge on C2.
Now assume that the two capacitors are connected in parallel. Calculate the voltage difference across C1.
Calculate the charge on C2.© BrainMass Inc. brainmass.com December 24, 2021, 8:38 pm ad1c9bdddf
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