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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  • 1 month ago
    Best 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

  • 1 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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