Find cab

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 V_CD across points C and D is given as

 

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

 

From above expression we form the following equations

 

q₁ - q₂ = q₃  .............................. (1)

 

q₅ - q₄ = q₃  .................... ..........(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