C 1 = 16.60µF and C 2 = 10.60 µF 6.00 microF --II-- ------l l------ l --II-- l C

Question

C1 = 16.60µF and C2 = 10.60 µF

                      6.00 microF
                        --II--
  ------l       l------
                l      --II--       l
         C1 =         C2      = C1
   l                    l
                --------l/---------
                         9.00 V
                       (battery)

symbols: = means capacitor and II also meanscapacitor


Find the following:

(a) the equivalent capacitance of the capacitors in the figureabove?
1      Answer in µF

(b) the charge on each capacitor (answers in microC)
on the right 16.60 µFcapacitor ?
on the left 16.60 µFcapacitor?
on the 10.60 µFcapacitor?
on the 6.00 µF capacitor?

(c) the potential difference across each capacitor (answers inV)
on the right 16.60 µFcapacitor?
on the left 16.60 µFcapacitor? on the 10.60 µFcapacitor? on the 6.00 µF capacitor?

on the right 16.60 µFcapacitor ?
on the left 16.60 µFcapacitor?
on the 10.60 µFcapacitor?
on the 6.00 µF capacitor?

Explanation / Answer

total capacitance... . first put   6.00 and C2 inparallel,   total of    6.00 + 10.60 =   16.60 . Then this is in series with the other two... .         1 / total =   1 / 16.60    +   1 /16.60   +   1 / 16.60   .      total   =  5.5333 uF . Total charge = CV = 5.5333 *9 =    49.8 uC . Charge on    16.60:    49.8 uC Charge on    16.60:    49.8 uC Charge on    10.60:    31.8 uC Charge on     6.00:     18.0 uC . Potential differenceacross every capacitor is   3 volts,    due to symmetry (i.e. 16.60   16.60 16.60)

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