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# Find c equivalent? Asked by npriya1297 8th October 2019, 7:05 PM Capcitance values of each capacitor is not given. It is assumed all capacitors have equal capacitance.

The given network of capacitors is redrawn in step-1. Markings A, B, D, E, F , G , H and I are done .

It is required to find the effective capacitance between points A and B.

In Step-1, we see that, in the loop D-E-I-F, two capacitors are in series and this series combination is parallel to another capacitor.

Capacitance C' of series combination of two capacitors is given by,   C' = ( C×C)/ (C+C) = C/2.
If this capacitance C' is parallel to another capacitance C, then effective capacitance is [ C/2 + C ] = (3/2)C

Hence all the capcitors in the loop D-E-I-F are replaced with a single capacitor of capacitance (3/2)C in step-2

Now, we see that in step-2, in loop A-D-F-G , two capacitors of capacitance C and (3/2)C are in series
and this series combination is parallel to another capacitance C.

Capacitance C'' of series combination of (3/2)C and C is  given by,   C'' = [ (3/2)C × C ] / [ (3/2)C + C ] = (3/5) C
If this capacitance C'' is parallel to another capacitance C, then effective capacitance is [ (3/5)C + C ] = (8/5)C

Hence all the capcitors in the loop A-D-F-G are replaced with a single capacitor of capacitance (8/5)C in step-3

In step-3, we end up with only two capacitors in series. Hence effective capacitance is [ (8/5)C × C ] / [ (8/5)C + C ] = (8/13) C

Hence effective capacitance of given network of capacitors is  (8/13)C
Answered by Expert 9th October 2019, 8:58 AM
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