For the circuit in Fig7 find the equivalent capacitance Ceq

For the circuit in Fig.7, find the equivalent capacitance C_eq seen between terminals a and b. (10\')

Solution

Capacitance in Parallel are added by C_eq = C₁+C₂

So 60 uF and 20 uF are in Parallel so will be added by ( 60 +20 ) uF which is equal to 80 uF

this parallel combination is in series with another 80 uF capacitance.

Capacitnace in Series are added by 1/C_eq = 1/C₁ + 1/C₂

                                                                  1/C_eq = (C₁+C₂)/C₁*C₂

                                                                     C_eq = C₁*C₂/(C₁+C₂)

                                                                     C_eq= 80*80/(80+80)

                                                                           = 6400/160

                                                                          = 40 uF

Or Directly we can write it as

C_eq = [(60 +20) || 80 ] uF

       = [ 80 || 80 ] uF

C_eq = 40 uF