Four capacitors are connected as shown in Figure P1633 C 14

Four capacitors are connected as shown in Figure P16.33 (C = 14.0 µF).

(a) Find the equivalent capacitance between points a and b.
1 µF

(b) Calculate the charge on each capacitor if a 18.5 V battery is connected across points a and b.
the 20.0 µF capacitor
2 µC
the 6.00 µF capacitor
3 µC
the 3.00 µF capacitor
4 µC
capacitor C
5 µC

.

Solution

capacitance:

    C1 = 14*3/14+3 = 2.47 uF

Capacitance:

   C2 = 2.47 + 6 = 8.47 uF

Hence, the equivalent capacitance:

      C = 8.47*20/8.47+20 = 5.95 uF

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the charge on the 20.0 µF capacitor:

   Q = CV = 5.95 µF * 18.5 = 110.075 uC

the charge on the 6 µF capacitor:

  Q = CV = 6 µF * [18.5 - (110.075/20) ] = 77.98 uC

the charge on the 3 µF capacitor:

  Q = 110.075 uC - 77.98 uC = 32.095 uC

the charge on the 14 µF capacitor:

  Q = 110.075 uC - 77.98 uC = 32.095 uC