Question

Series-connected 11-pF and 21-pF capacitors are placed in parallel with series-connected 22- pF and 30-pF capacitors. Determine the .equivalent capacitance​

Answer 1

19.9 pF

Step-by-step explanation:

Given that:

Series connection :

11pF and 21pF

C1 = 11pF ; C2 = 21pF

Cseries = (C1*C2)/ C1 + C2

Cseries = (11 * 21) / (11 + 21)

Cseries = 7.21875 pF

C1 = 22pF ; C2 = 30pF

Cseries = (C1*C2)/ C1 + C2

Cseries = (22 * 30) / (22 + 30)

Cseries = 12.6923 pF

Equivalent capacitance is in parallel, thus,

7.21875pF + 12.6923 pF = 19.91105 pF

= 19.9 pF