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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