Find an expression for the equivalent capacitance of a combination of three capacitors in series .

Explanation

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))`