Equivalent Capacitance in a circuit

  • Thread starter carpelumen
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  • #1
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Homework Statement


In the circuit shown in the figure the voltage of the battery is 15.0 V, and the capacitors have the following capacitances: C1 = 2.32 mF, C2 = 2.25 mF, C3 = 4.85 mF, C4 = 4.89 mF
What is the equivalent capacitance of the four capacitors?

http://tinypic.com/r/24dn5oi/8
Two sets of series capacitors that are parallel to each other

Homework Equations


Ceq=(1/C1+1/C2+1/C3...)^-1 ---- series
Ceq=(C1+C2+C3...) --- parallel

The Attempt at a Solution


Ceq=(C1+C3)(C2+C4)/(C1+C2+C3+C4)
=(7.17)(7.14) / 14.31
= 3.58 mF

According to online hw system, it's wrong. Is my equation incorrect?
 

Answers and Replies

  • #2
phinds
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I think you've got the series/parallel actions backwards. Do capacitors add directly when in series or when in parallel?
 
  • #3
CWatters
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Perhaps show us how you got...

Ceq=(C1+C3)(C2+C4)/(C1+C2+C3+C4)
 
  • #4
CWatters
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I think you've got the series/parallel actions backwards. Do capacitors add directly when in series or when in parallel?

I don't think that's his problem because his relevant equations..

Ceq=(1/C1+1/C2+1/C3...)^-1 ---- series
Ceq=(C1+C2+C3...) --- parallel

... are correct. Capacitors add in parallel.
 
  • #5
phinds
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... are correct. Capacitors add in parallel.

Right, which is what led me to wonder how he got a term (c1+c3)
 
  • #8
phinds
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Even closer to 3.11 :)
 

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