# Four capacitors A, B, C, and D have capacitances of 9, 13, 7, and 21 µF, respectively.?

Capacitors A and B are connected in parallel. The combination is then connected in series with C and D. What is the effective capacitance?

Answer is 4.24 µF but how to find it ?

### 2 Answers

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- nyphdinmdLv 71 month agoBest Answer
Connect A to B in parallel. The effective capacitance is Cab = CA +CB

Connect Cab to C and D in series --> 1/Ceff = 1/Cab + 1/Cc + 1/Cd

1/Ceff = (Cc*Cd +Cab*Cd + Cab*Cd)/(Cab*Cc*Cd)

Ceff = (Cab*Cc*Cd)/(Cc*Cd +Cab*Cd + Cab*Cd)

Cab = 22 uF so Ceff = 4.24 uF

- OutlierLv 41 month ago
In parallel, the effective capacitance is given by the sum of the capacitances of the capacitors, so that's 9 + 13 = 22 µF

In series, the effective capacitance E is the inverse of the sum of the inverse capacitances of the capacitors.

That is, 1/E = 1/22 + 1/7 + 1/21

=> 1/E = 109/462

=> E = 462/109 = 4.24 µF

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