Three capacitors each of capacitance 9 pF are connected in series.
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?
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(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, 1/Ceq=1/C+1/C+1/C=3/C=3/9=1/3
⇒1/Ceq=1/3⇒Ceq=3 pF
Therefore, total capacitance of the combination is 3 pF.
(b) Supply voltage, V = 100 V
Potential difference (V1) across each capacitor is equal to one-third of the supply voltage. ∴V1=V/3=120/3=40V
Therefore, the potential difference across each capacitor is 40 V.
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