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Question
Calculate equivalent capacitance of the circuit shown in the Figure given below:
Solution
In the given figure,
C1 and C2 are connected in parallel so
C = C1 + C2
= 50 μF + 50 μF
= 100 μF
Here C is required with C3 in series, hence
`1/"C"_"eq" = 1/"C" + 1/"C"_3`
`= 1/100 + 1/25`
`= (1 + 4)/100`
`= 5/100`
`= 1/20`
⇒ Ceq = 20 μF
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