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# Calculate Equivalent Capacitance

Junior Member
Join Date: Mar 2012
Posts: 1

 03-02-2012, 11:53 AM
Hi,

I would like to know the solution for the question attached in the attachement.Can anyone help me please.
Attached Thumbnails

Super Moderator
Join Date: Nov 2011
Location: Germany
Posts: 2,191

 03-04-2012, 08:49 AM
Start by simplyfying the circuit:
the two lower left 1F capacitors can be replaced by asingel cap. As can the 1F and 3F cap. in the top middle.

Harald

Senior Member
Join Date: Apr 2010
Location: Tombstone, Arizona
Posts: 540

 03-04-2012, 08:48 PM
You can never go wrong by writing the node equations, but first simplify the circuit by combining parallel capacitors. In the attached drawing I have added a voltage source (V) from which to calculate the current (I). This will be the constraint equation V/I=Zc that the input terminals to the network show a capacitor impedance of 5 farads. So once the equations are done, I let MathCAD crank out the answer. Of course, the final answer has been redacted in accordance with the rules of the forum.
Attached Files
 Capacitance-Bridge.pdf (28.9 KB, 71 views)

Senior Member
Join Date: Mar 2012
Location: Richmond, California
Posts: 577

 03-10-2012, 01:55 AM
What I did was convert all the capacitance into Xc's so that I could convert a Y network into a delta networks and treat them exactly like resistors and convert them back. Then I combined the capacitors except for the 3F and the C and subtracted the number from 5F and got 2.625F. Then I tried took the 3F and C series and tried the answers until I god 2.625F. The result was 21F. There are many ways of looking at this circuit.You can treat the circuit as an AC circuit and do nodal analysis, or thevenize, or do mesh currents, and more. The important thing is to remember exactly what a capacitor is and how it works. The rest is just numbers.

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