Capacitance in Parallel and in Series

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SUMMARY

The discussion focuses on calculating the equivalent capacitance of capacitors in series and parallel configurations. The values provided are C1 = 10 nF, C2 = 5 nF, and C3 = 4 nF. The correct calculation for the equivalent capacitance when combining C1 and C2 in parallel, followed by C3 in series, yields an equivalent capacitance of 3.16 nF. The initial incorrect answer of 4.74 nF was due to a misunderstanding of the series and parallel formulas.

PREREQUISITES
  • Understanding of capacitor configurations: series and parallel
  • Knowledge of capacitance formulas: Ceq = C1 + C2 + C3 for parallel, 1/Ceq = 1/C1 + 1/C2 + 1/C3 for series
  • Basic algebra skills for manipulating equations
  • Familiarity with nanofarad (nF) as a unit of capacitance
NEXT STEPS
  • Study the derivation of the formulas for equivalent capacitance in series and parallel
  • Practice problems involving multiple capacitors in various configurations
  • Explore the impact of capacitor values on circuit behavior in RC circuits
  • Learn about the applications of capacitors in electronic circuits
USEFUL FOR

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

islandboy401
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Homework Statement



In the FIGURE (shown below in "attempt at solution" section), find the equivalent capacitance of the combination. Assume that C1=10 nF; C2 = 5 nF; and C3 = 4 nF.


Homework Equations



For parallel capacitors: Ceq = C1 + C2 + C3 +...

For series capacitors:
Series%20Capacitance.JPG


The Attempt at a Solution



number%209.jpg


My answer is 4.74 nF though my answer key says 3.16 nF. Am I doing something wrong?
 
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in series : \frac{1}{C_{eq}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}}... etc

in parallel {C_{eq} = {C_{1} + {C_{2} + {C_{3} ... etc

basically the opposite to resistance

so from your diagram... (C1 // C2) + C3

so it should be \frac{1}{C_{eq}} = \frac{1}{10 + 5} + \frac{1}{4}

so C_{eq} = 3.16nF

hopefully that helps
 
Thanks...I understand what I did wrong.
 

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