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Find the total energy stored in the capacitors in the given network of Fig. |
Answer» Solution : In fig , capacitors `C_4` and `C_5` of 2` mu F ` each are in series , their equivalent capacitance `C. = (C_(4) C_(5))/(C_(4) + C_(5)) = (2 xx 2)/( 2 + 2) = 1 mu F ` Now C. and `C_2` are in parallel , hence their equivalent capacitance `C.. = C_(2) + C. = 1 + 1 = 2 mu F` Further C.. and `C_3` are in series , their equivalent capacitance `C_(0) = (C.. C_(3))/(C.. + C_(3)) = (2 xx 2)/(2 + 2) = 1 mu F ` Now C. and `C_2` are in parallel , hence their equivalent capacitance `C.. = C_(2) + C. = 1 + 1 = 2mu F ` Further C.. and `C_3` are in series , their equivalent capacitance `C_(0) = (C.. C_(3))/(C.. + C_(3)) = (2 xx 2)/(2 + 2) = 1 mu F` Finally , `C_0` is in parallel with `C_1` , hence net resultant capacitance `C = C_(0) + C_(1) = 1 + 1 = 2 mu F ` `therefore` Total energy stored in the capacitors in the GIVEN network `U = (1)/(2) C V^(2) = (1)/(2) xx 2 xx 10^(-6) xx (6)^(2) = 3.6 xx 10^(-5) J` |
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