NEET Class neet Answered
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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