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capacitance
Asked by aakashneet33 | 21 Feb, 2024, 07:01: PM

At points between A and F equivalent capacitance C/2 replaces the two capacitors of  equal capacitance C connected in series .

Similarly , at points between C and D equivalent capacitance C/2 replaces the two capacitors of  equal capacitance C connected in series .

Let us give a potential difference V  at point A with respect to point D .

Let us consider the charge flow through all capacitors as shown in figure,

By allpying Kirchoff's voltage law to the loop ABEFA , we get

Hence we get ,  q1 + q3 = 2 q2 ....................... (1)

By allpying Kirchoff's voltage law to the loop BCDEB , we get

Hence we get, q3 + q5 = 2 q4 ...........................(2)

By charge conservation at junction B

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

By charge conservation at junction E

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

By substituting q2 from eqn.(4), we rewrite eqn.(1) as

q1 + 3 q3 = 2 q5  ...................(5)

By substituting q4 from eqn.(3), we rewrite eqn.(2) as

2q1 - 3 q3 =  q5  ...................(6)

By solving eqn.(5) and (6) , we get

q1 = q5

q3 = (1/3) q5

By substituting q3 in eqn.(4) , we get , q2 = (2/3)q5

By substituting q3 in eqn.(3) , we get , q4 = (2/3)q5

By adding potenial differences along the line ABCD , we get

By substituting q1 and q4 in above expression, we get , q5 = (3/7) C V

Hence we get the other charges as

q1 = (3/7) CV    ;   q2 = (2/7) CV ;    q3 = (1/7) CV   ;  q4 = (2/7) CV

From figure we see that total charge Q supplied by potential difference V is

Q = q1 + q2 = (5/7) C V

Equivalent capacitance =   Q/V = (5/7)C

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Answered by Thiyagarajan K | 22 Feb, 2024, 05:00: PM
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