Question

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 120 V supply?

Answer 1

(a) Capacitance of each of the three capacitors, C = 9 pF Equivalent capacitance (Ceq) of the combination of the capacitors is given by the relation, $\frac{1}{Ceq}=\frac{1}{C}+\frac{1}{C}+\frac{1}{C}=\frac{3}{C}=\frac{3}{9}=\frac{1}{3 }=\frac{1}{C_eq}=\frac{1}{3} c_eq=3pF$

Therefore, total capacitance of the combination is 3 pF.

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(b) Supply voltage, V = 100 V Potential difference (V1) across each capacitor is equal to one-third of the supply voltage.

$V1=\frac{V}{3} =\frac{120}{3} =40 V$

Therefore, the potential difference across each capacitor is 40 V.