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Find an expression for the equivalent capacitance of a combination of three capacitors in series . |
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Answer» Solution :Consider a SERIES combination of three capacitors of CAPACITANCES `C_(1) , C_(2)` and `C_(3)` , respectively . Obviously charge on each plate of each CAPACITOR has same magnitude and charges on two plates of a capacitor has opposite signs . If potential differences across the three capacitors be `V_1 , V_2` and `V_3` respectively then from the relation . Q = V C , we have `V_(1) = (Q)/(C_(1)) , V_(2) = (Q)/(C_(2)) ` and `V_(3) = (Q)/(C_(3))` The total potential difference across the whole circuit `V = V_(1) + V_(2) + V_(3) = (Q)/(C_(1)) + (Q)/(C_(2)) + (Q)/(C_(3))` If C be the resultant capacitance in series arrangement then `Q = C V ` or `V = (Q)/(C)` `THEREFORE(Q)/(C) = (Q)/(C_(1)) + (Q)/(C_(2)) + (Q)/(C_(3)) implies (1)/(C) = (1)/(C_(1)) + (1)/(C_(2)) + (1)/(C_(3))` 
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