Capacitance in Parallel and in Series

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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:
Equation 1/C_eq = 1/C1 + 1/C2 + 1/C3 + ... for capacitors in series.


The Attempt at a Solution



Handwritten solution on lined paper: a circuit schematic labeled FIGURE with capacitors C1, C2 and C3 in a loop with a battery, then a redrawn circuit with C2 and C3 in parallel, C2+C3=5+4=9, and series combination 1/C_eq = 1/9 + 1/10 = 19/90, giving C_eq = 4.74 nF.


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

in parallel [tex]{C_{eq}[/tex] = [tex]{C_{1}[/tex] + [tex]{C_{2}[/tex] + [tex]{C_{3}[/tex] ... etc

basically the opposite to resistance

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

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

so [tex]C_{eq}[/tex] = 3.16nF

hopefully that helps