1 answer

The figure shows four capacitors with CA = 5.00 µF, CB = 3.00 µF, CC =...

Question:

The figure shows four capacitors with CA = 5.00 µFThe figure shows four capacitors with CA = 5.00 µF, CB = 3.00 µF, CC = 6.00 µF, and CD = 4.00 µF connected across points a and b, which have potential difference ?Vab = 12.0 V.

(a) What is the equivalent capacitance of the four capacitors?

(b) What is the charge on each of the four capacitors?


Answers

cc and cd are in series ccd = cc*cd/cc+cd = 6*4/10 = 2.4*10^-6 f

cb and ccd are in parallel = cbcd = 3+2.4 = 5.4*10^-6 f

ca and cbcd are in series ctot = 5*5.4/10.4 = 2.596*10^-6 f

b) total charge = ctot*v = 2.596*10^-6*12 = 31.15*10^6 C

ca and cbcd are in series so in series charge must be same

qa = qbcd = 31.15*10^6 C

vbcd = 31.15*10^6 C/5.4*10^-6 f = 5.77 v

cb and ccd are in parallel so voltage same vb = vcd = 5.77 v

qb = 5.77*3*10^-6 = 17.3*10^-6 c

qcd = 2.4*10^-6*5.77 = 13.85*10^-6 C

cc and cd are in series so qc=qd = 13.85*10^-6 C

.

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