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Three capacitors each of capacitance 9 pF are connected in series. (a) What is the total capacitance of the combination ? (b) What is the potential difference across each capacitor if the combination is connected to a 120V supply.

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Solution :
(a) Capacitance of each of the three capacitors , C=9 pF <br> equivalent capacitance (C) of the combination of the capacitors is given by the relation, <br> `1/(C')=1/C+1/C+1/C` <br> `=1/9+1/9+1/9=3/9=1/3` <br> `:. C'=3mu F ` <br> Therefore, total capacitance of the combination is `3muF` <br> (b) supply voltage, V=100V <br> potential difference (V) across each capacitor is equal to one-third of the supply voltage. <br> `:. V'=V/3=120/3=40 V` <br> therefore, the potential difference across each capacitor is 40 V.

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