Question

Calculate equivalent capacitance of the circuit shown in the Figure given below:

Answer 1

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`

⇒ C_eq = 20 μF