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Asked by vigyants | 05 May, 2022, 02:59: PM
Given network of  capacitors of equal capacitance C which is in a triangle shape is shown in left side.
Plates of capacitors are numbered so that we can draw the regular shpe circuit as shown in right side of figure
and calculate the equivalenet capacitance across the points A and B.

Let us apply a potential difference V acorss A and B .

Let us consider the charge accumulated in each capacitors as shown in figure.

Potential difference VCD across points C and D is given as

VCD = (-q1 / C ) + ( q2 / C ) = (-q3 / C ) = ( q4 / C ) - (q5 / C )

From above expression we form the following equations

q1 - q2 = q.............................. (1)

q5 - q4 = q .................... ..........(2)

At node C , sum of charges equal zero. Hence we have

q1 + q3 = q4 ..........................(3)

At node D , sum of charges equal zero. Hence we have

q2 =  q5 + q3 ..........................(4)

By solving eqn.(1) and eqn.(3) , we get

q1 = (1/2) ( q4 + q2 )  ........................... (5)

q3 = (1/2) (q4 - q2 ) ............................(6)

By solving eqn.(2) and eqn.(4) , we get

q5 = (1/2) ( q2 + q4 )........................(7)

q3  = (1/2) ( q2 - q4 ) ....................(8)

from eqn.(6) and eqn.(8) , we get q3 = 0 .

Hence middle capacitor shown in the right side figure is not charging .
Hence the middle capacitor can be removed from the circuit.

Modified circuit after removing the middle capacitor is shown below.

From above circuit shown above, we can easily calculate the equivalent capacitance as C

Answered by Thiyagarajan K | 06 May, 2022, 07:49: PM

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