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A system of capacitors , connected as shown , has a total energy of 160 mJ stored in it. Obtain the value of the equivalent capacitance of this system and the value of x . |
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Answer» Solution :In the arrangement of capacitors shown , capacitors of capacitances `C_2` and `C_3` are JOINED in parallel and their equivalent capacitance `C_(23) = 7 + 3 = 10 mu F ` Now `C_(1) , C_(23)` and `C_(4)` are joined in series , hence their equivalent capacitance C. is given by `(1)/(C.) = (1)/(C_(1)) + (1)/(C_(23)) + (1)/(C_(4)) = (1)/(10) + (1)/(10) + (1)/(15) = (8)/(30)` `IMPLIES C. = (30)/(8) mu F = 3.75 mu F ` As `C_(5)` is in parallel to C. , hence equivalent capacitance `C_(eq)` of the arrangement is `C_(eq) = C. + z = (3.75 + z) mu F ` `therefore` Energy of the system `U = (1)/(2) C_(eq) , V^(2)` , where U = 160 mJ = 0.16 J and V = 200 V `therefore 0.16 = (1)/(2) XX C_(eq) (200)^(2) implies C_(eq) = (0.16 xx 2)/((200)^(2)) = 8 xx 10^(-6) F = 8 mu F ` As `C_(eq) = (3.75 + z) = mu F = 8 mu F ` , hence we have z = `4.25 mu F ` |
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