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 ?
- 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