Capacitance in Parallel and in Series

AI Thread Summary
The discussion centers on calculating the equivalent capacitance of capacitors in parallel and series. The user initially calculated the equivalent capacitance as 4.74 nF but later realized the correct answer is 3.16 nF. The correct approach involves using the formulas for series and parallel combinations, specifically noting that for series, the reciprocal formula applies. The user corrected their method by recognizing the need to combine C1 and C2 in parallel before adding C3 in series. This clarification led to a better understanding of the capacitance calculations.
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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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