Capacitor problem lots of help

[supermenscher]

A 0.40E-6F capacitor and a 0.70E-6F capacitor are connected in parallel to a 12V battery. Calculate the potential difference and charge across each capacitor. I tried c1v1=c2v2 but that answer wasn't right. Can someone please help me...I would really appreciate it.

 

[drag]

Capacitors connected in line have the same charge,
but capacitors connected in parallel have the same
potential difference and the charge is calculated with c.
Good Luck.

 

[M98Ranger]

First, let's calculate the total capacitance when two capacitors are connected in parallel. The formula for calculating capacitance in parallel is C = C1 + C2, where C1 and C2 are the individual capacitances. So, in this case, the total capacitance would be:

C = 0.40E-6F + 0.70E-6F = 1.10E-6F

Next, we can use the formula Q = CV to calculate the charge on each capacitor. Q represents the charge, C is the capacitance, and V is the potential difference. Since the capacitors are connected in parallel, they will have the same potential difference (12V) across them, but the charge will be split between them.

For the 0.40E-6F capacitor:

Q1 = (1.10E-6F)(12V) = 0.48E-5C

For the 0.70E-6F capacitor:

Q2 = (1.10E-6F)(12V) = 0.84E-5C

So, the potential difference across each capacitor is 12V, but the charge on each capacitor is different. I'm not sure why your calculation using the equation C1V1 = C2V2 didn't work, but it is possible that you may have made a mistake in your calculations or used the wrong values for the capacitance. I hope this helps and feel free to ask for clarification if needed.