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

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60 V

40 V

50 V

80 V

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{C_{s}} = \frac{1}{C_{1}} + \frac{1}{C_{2}} + \frac{1}{C_{3}} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} \\ \frac{1}{C_{s}} = \frac{3}{9} \Rightarrow C_{s} = 3 pF \end{array}$

Charge, $q = C_{s} V = 3 \times 120 = 360 pC$

Potential difference across

$\begin{array}{l} C_{1}, (V_{1}) = \frac{q}{C_{1}} \\ = \frac{360}{9} = 40 V = V_{2} = V_{3} \end{array}$

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