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Three capacitors each of capacitane 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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Answer Text

Solution :
`(1)/(C_(s)) = (1)/(9) + (1)/(9) + (1)/(9) = (3)/(9) = (1)/(3), C_(s) = 3pF` <br> p.d. across each capacitor `= (V)/(3) = (120)/(3) = 40V`

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