Three capacitors of capacitances 2pF, 3pF and 4pF are connected in parallel.

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asked Jan 3, 2018 in Physics by sforrest072 (157,439 points) 63 451 1316

Three capacitors of capacitances 2pF, 3pF and 4pF are connected in parallel.
(a) What is the total capacitance of the combination?
(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

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answered Jan 3, 2018 by mdsamim (213,225 points) 5 10 24
selected Jan 3, 2018 by sforrest072
 
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(a) Capacitances of the given capacitors: C1 = 2 pF, C2 = 3 pF and C3 = 4 pF
For the parallel combination of the capacitors, equivalent capacitor is given by Ceq the algebraic sum,
Therefore Ceq = C1 + C2 + C3 = 2 + 3 + 4 = 9 pF

Therefore, total capacitance of the combination is 9 pF.

(b) Supply voltage, V = 100 V
The voltage through all the three capacitors is same = V = 100 V
Charge on a capacitor of capacitance C and potential difference V is given by the relation, q = VC
For C = 2 pF, charge = VC = 100 × 2 = 200 pC = 2 × 10–10 C
For C = 3 pF, charge = VC = 100 × 3 = 300 pC = 3 × 10–10 C
For C = 4 pF, charge = VC = 100 × 4 = 400 pC = 4 × 10–10 C

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