How to Calculate Charge and Potential Difference in Parallel Capacitors?

  • Thread starter Thread starter supermenscher
  • Start date Start date
  • Tags Tags
    Capacitor
AI Thread Summary
When two capacitors, a 0.40E-6F and a 0.70E-6F, are connected in parallel to a 12V battery, the total capacitance is calculated as 1.10E-6F. Each capacitor experiences the same potential difference of 12V. The charge on the 0.40E-6F capacitor is 0.48E-5C, while the charge on the 0.70E-6F capacitor is 0.84E-5C. The initial confusion arose from using the incorrect formula for capacitors in parallel, which do not share the same charge but do share the same voltage. Understanding these principles clarifies the calculations for charge and potential difference in parallel capacitors.
supermenscher
Messages
47
Reaction score
0
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.
 
Physics news on Phys.org
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.
 


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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top