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Find the equivalent capacitance of the given figure. |
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Answer» SOLUTION :` (1)/(dc) =(y)/(in _0 K_1bdx) +(d_0-y)/( in _0 k_1 bdx )+ (d_0 -y)/( in _0 k_2 bdx) ` ` (1)/(dc)=(d_0k_1+y (k_2-k_1)/( in _0 k_1k_2bdx) ` ` dc = (in _0 k_1k_2 bdx)/( d_0k_1+y(k_2-k_1))` All these capacitors (small )are PARALLEL so ` C_(AQ) = INT _0^(C_aq) dC =int _0^(a) ( in _0 k_1k_2bdx)/( d_0 k_1+y(k_2-k_1)) ` Now `""(d_0)/(a) =(y)/(x) rArr C_(aq)=int _0^(a) (in_0 k_1 k_2bdx)/(d_0 k_1+(d_0)/( a) (k_2-k_1) )` ` C_(aq) =(ain _0k_1k_2b)/( d_0( k_2-k_1))1n [((k_2-k_1))/( k_1) ]` 
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