Potential Difference across capacitors in parallel.

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SUMMARY

The discussion focuses on the analysis of two capacitors, C1 = 20 micro Farads and C2 = 5 micro Farads, connected in parallel to a 12-V battery. The equivalent capacitance is calculated as Ceq = C1 + C2, resulting in 25 micro Farads. The charge on each capacitor is determined using the formula Q = CV, yielding Q1 = 2.4x10^-4 C for C1 and Q2 = 6x10^-5 C for C2. The potential difference across each capacitor is confirmed to be equal to the battery voltage of 12V, as capacitors in parallel share the same voltage.

PREREQUISITES
  • Understanding of capacitor fundamentals, including capacitance and charge.
  • Familiarity with the formula Q = CV for calculating charge.
  • Knowledge of equivalent capacitance in parallel circuits.
  • Basic principles of energy stored in capacitors, including Ec = QV/2.
NEXT STEPS
  • Study the concept of energy stored in capacitors using Ec = (C(V^2))/2.
  • Learn about the implications of connecting capacitors in series versus parallel.
  • Explore practical applications of capacitors in electronic circuits.
  • Investigate the effects of varying capacitance values on overall circuit performance.
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Students studying electrical engineering, electronics enthusiasts, and anyone looking to deepen their understanding of capacitor behavior in parallel circuits.

OUmecheng
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Homework Statement



Two capacitors, C1 = 20 micro Farads and C2 = 5 Micro Farads are connected in parallel and the resulting combination is connected to a 12-V battery. Find (a) the Equivalent capacitance of the combination, (b) the electric charge on each capacitor, (c) the potential difference across each capacitor, and (d) the energy stored in each capacitor.

Homework Equations



Ceq = C1 + C2

Q = CV

Ec = QV/2 = (C(V^2))/2 = (Q^2)/2C

The Attempt at a Solution



(a) 2.0x10^-6 F + 5x10^-6 F = 2.5x10^-5 F(b) Q = CV,

Q1 = (20x10^-6 F)(12V) = 2.4x10^-4 C

Q2 = (5x10^-6 F)(12V) = 6x10^-5 C(c) Not quite sure what this is asking for?(d) Ec = CV^2/2

Ec1 = c1V^2 / 2 = 0.00144 F*V^2

Ec2 = c2V^2 / 2 = 3.6x10^-4 F*V^2

As you can see I've worked most of this out, seems simple enough, but I don't know what to do for (c)?
 
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It's simpler than you think. Draw yourself the circuit and look at it.
 
kuruman said:
It's simpler than you think. Draw yourself the circuit and look at it.

errrr charge over capacitance?
 
Last edited:
How did you do part (b)? What voltage did you assume across each capacitor? :rolleyes:
 
OUmecheng said:
errrr charge over capacitance?
Please don't edit previous posts. Just post a new response.

Your "attempt at a solution for (b) says


(b) Q = CV,

Q1 = (20x10^-6 F)(12V) = 2.4x10^-4 C

Q2 = (5x10^-6 F)(12V) = 6x10^-5 C

Tell me in English what do the numbers highlighted in red represent?
 
kuruman said:
Please don't edit previous posts. Just post a new response.

Your "attempt at a solution for (b) says


(b) Q = CV,

Q1 = (20x10^-6 F)(12V) = 2.4x10^-4 C

Q2 = (5x10^-6 F)(12V) = 6x10^-5 C

Tell me in English what do the numbers highlighted in red represent?

Battery Voltage/Potential Difference?
 
OUmecheng said:
Battery Voltage/Potential Difference?
It is not a ratio. In plain English the equation Q = CV means

The charge on a capacitor is the same as the capacitance multiplied by the potential difference (or voltage) across the terminals of the capacitor.

Can you answer (b) now?
 
kuruman said:
It is not a ratio. In plain English the equation Q = CV means

The charge on a capacitor is the same as the capacitance multiplied by the potential difference (or voltage) across the terminals of the capacitor.

Can you answer (b) now?

So the potential difference is the same as charge on a capacitor divided by the capacitance.

V1= Q/C1
V2= Q/C2

Thanks!
 
Actually, you are doing it backwards. Memorize this definition

"Two circuit elements in parallel share the same potential difference (or voltage) across their terminals"

and use it constructively.

Cheers.
 

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