Potential Difference across Capacitor.

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SUMMARY

The discussion focuses on calculating the potential difference across capacitors in a parallel configuration, specifically involving a 6 microfarad capacitor (C1) and a 2 microfarad capacitor (C2). Participants clarified that the voltage across both capacitors remains constant at 90 volts, as voltage is not divided in parallel circuits. Key equations discussed include V = Q / C for voltage calculation and PE = (0.5)(C)(ΔV²) for energy stored in capacitors. The correct approach involves using known charge and capacitance values to determine voltage.

PREREQUISITES
  • Understanding of capacitor configurations (series and parallel)
  • Familiarity with the equations V = Q / C and PE = (0.5)(C)(ΔV²)
  • Basic knowledge of microfarads as a unit of capacitance
  • Ability to perform calculations involving microCoulombs and microfarads
NEXT STEPS
  • Learn about series and parallel capacitor combinations and their voltage behavior
  • Study the concept of energy stored in capacitors using PE = (0.5)(C)(ΔV²)
  • Explore practical applications of capacitors in electronic circuits
  • Investigate the effects of varying capacitance values on circuit performance
USEFUL FOR

Students studying electrical engineering, electronics enthusiasts, and anyone looking to deepen their understanding of capacitor behavior in circuits.

Aznhmonglor
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Homework Statement



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Homework Equations



Not sure which equations to use but I know that we'll have to use the answers from part a and b to answer the questions.



The Attempt at a Solution



What I did was that I said that C1 and the 6 microfarads are parallel to C2 and the 2 microfarads capacitor and said that the voltage would just be halved. Really confused on what to do.
 
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remember that voltage is not 'split up' to components in parallel, this only occurs to components in series, which is the second half of the problem
 
Chewy0087 said:
remember that voltage is not 'split up' to components in parralel, both the top & the bottom will receive 90V (separateley) and the 90 is shared between the two capacitors.

hmmm thanks that helps to clarify that point. So its 90 volts to C1 and the 6 microfarads capacitor and to the C2 and 2 microfarads capacitor. What equation do I need to use? I have a couple here. Looking at PE=(.5)(C)(deltaV^2) is that the right one? I think the main problem here is that I don't know what equation to use or how to look at the problem.
 
hmm I'm afraid not, that's the equation for working out the energy stored in the capacitor, so you'll need it later, however

V = Q / C voltage = charge / capacitance , you know the Q & C for each capacitor, so try to work out V
 
Chewy0087 said:
hmm I'm afraid not, that's the equation for working out the energy stored in the capacitor, so you'll need it later, however

V = Q / C voltage = charge / capacitance , you know the Q & C for each capacitor, so try to work out V

hmmm...Im still getting some wrong answers. So let's say for C2, I know the charge on that capacitor is 120 microCouloumbs. I am not sure what the Capacitance for it is. Would the capacitance for C2 be the sum of (1/C1)+(1/C2)=(1/Ctot)? so it would be that V=120/1.5?

Just figured out this part. Now onto the total energy stored by group?

the equation used would be PE=(.5)(C)(deltaV^2) ?
 
Last edited:
correcto, but look

let's take C1, you worked out earlier that charge was 120 * 10^-6
and you also know that it's capacitance is 2 * 10^-6 -it tells you on the top of the picture!

so now you have Q and C, and V = Q / C , they seem to work, try them again
 
Problem solved!

Thanks a million.
 

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