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A network of four capacitors each of 12 mu Fcapacitance is connected to a 500 V supply as shown in the Fig . Determine equivalent capacitance of the network . |
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Answer» Solution :GIVEN that `C_1 = C_2 = C_3 = C_4 = 12 mu F ` and supply voltage V = 500 V Since capacitors `C_1 , C_2` and `C_3`are connected in series , their combined capacitance `C_(123)` is given by `(1)/(C_(123)) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3)) = (1)/(12) + (1)/(12) + (1)/(12) = (3)/(12) = (1)/(4) implies C_(123) = 4 mu F ` As `C_(123)` has been connected in parallel to `C_(4)` , hence equivalent capacity of the network `C_(EQ) = C_(123) + C_(4) = (4 + 12) mu F = 16 mu F ` |
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