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Capacitance in series and parallel circuit

  • #1
[SOLVED] Capacitance in series and parallel circuit

Homework Statement



Determine the equivalent capacitance between A and B for the group of capacitors in the drawing.

Homework Equations



Cseries = 1/Cs = 1/c1 + 1/c2 + ...
Cparallel = Cp = C1 + C2 + ...

The Attempt at a Solution



The 4 and 12 micro farad capacitors are in parallel, so Cp = 16mf.
The other four are in series, so 1/5+1/24 = 1/C = 4.138mf (top)
1/6+1/8 = 1/C = 3.429mf (bottom)
All of these are in series, so 1/4.138+1/3.429+1/16 = 1/c; c = 1.678mf

This is not right. Where is my logic faulty?
 

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Answers and Replies

  • #2
Astronuc
Staff Emeritus
Science Advisor
18,704
1,716
The 24 uF, 12 uF and 8 uF capacitors are in series - since they are end to end between the two nodes in common with the 4 uF capacitor.

So find the capacitance of the series of 24 uF, 12 uF and 8 uF, and that combination is parallel with the 4 uF capacitor. That group is in series with the 5 uF and 6 uF capacitors between nodes A and B.
 
  • #3
Doc Al
Mentor
44,867
1,114
The 4 and 12 micro farad capacitors are in parallel, so Cp = 16mf.
Capacitors are in parallel only if their ends are directly connected by nothing but a wire. That's not the case here, since the 24 and 8 micro farad capacitors are in the way.

Similarly, capacitors are in series only if there's nothing else in the path.

Do it step by step. Find the obvious combinations and replace them with their equivalents. Then redraw it, looking for more combinations.

The first combination that I see are several capacitors in series. Start there.
 
  • #4
Thanks guys; apparently I didn't quite understand how to combine initially. I get it now!
 

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