Three capacitors 15µf, 18µf and 12µf are connected in a circuit. Find equivalent capacitance when they are connected in – 1) Series 2) Parallel

1 Answer

Answer :

Ans: Value of equivalent capacitance: Given: C1= 15μF, C2=18 µF, C3= 12µF 

i)For Series combination of capacitors: 1/Cs = (1/C1)+( 1/C2) +(1/C3) = (1/15)+( 1/18)+( 1/12) 1/Cs = 0.0666+0.0555+0.0833 1/Cs = 0.2054 Cs = 4.868 µF 

 ii) For parallel combination of capacitors: Cp = C1 + C2 + C3 = 15 +18+ 12 = 45 µF 

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