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Obtain the formulafor the effective capacitance of the parallel combinationof different n capacitors. |
Answer» Solution :FIGURE shows n capacitors of capacitance `C_(1),C_(2), cdots , C_(n)` are arranged in parallel.  Here potential difference across each capacitor is same but the plate charges are different to each capacitor. Suppose, `Q_(1),Q_(2),cdots, Q_(n)` are the charges on the capacitors `C_(1),C_(2), cdots, C_(n)` respectively. The total charge `Q= Q_(1)+Q_(@)+cdots, Q_(n)` But `Q_(1)=C_(1)V_(1), Q_(2)=C_(2)V_(2),cdots, Q_(n)=C_(n)V` `:. Q=C_(1)V+C_(2)V+cdos+C_(n)V` `:. (Q)/(V)=C_(1)+C_(2)+cdots C_(n)` If `(Q)/(V)` is the EFFECTIVE capacitance of combination then `C=C_(1)+C_(2)+ cdotsC_(n)` The effective capacitance of n numbers of capacitors in parallel is equal to the ALGEBRAIC sum of the individual capacitance of capacitors. The effective capacitance of the capacitors in parallel is more than capacitance of any one of the parallel connected capacitors. |
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