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Answer to Question #12555 in Electric Circuits for Nathan

Question #12555
Four Capacitors are connected in Series. They have capacities of 2.2uF, 3.3uF, 4.7uF and 5.6uF. The Series combination is connected to a 200 Volt DC supply. Determine:

a) The total charge
b) The charge across each capacitor
c) the voltage across each capacitor


* Thank you very much in advance if this can be answered. Will help me understand more*
Expert's answer
1/C = 1/C1 + 1/C2 + 1/C3 + 1/C4
C = 1 / (1/C1 + 1/C2 + 1/C3 + 1/C4)
C = 1
/ (1/(2.2*10^-6) + 1/(3.3*10^-6) + 1/(4.7*10^-6) + 1/(5.6*10^-6))
C = 10^-6 /
(1/2.2 + 1/3.3 + 1/4.7 + 1/5.6) = 0.87*10^-6 F

a) The total charge
q =
C * U = 0.87*10^-6 * 200 = 1.74*10^-4 C
b) The charge across each
capacitor
qi = (C / Ci) * q
q1 = (0.87 / 2.2) * 1.74*10^-4 = 0.69*10^-4
C
q2 = (0.87 / 3.3) * 1.74*10^-4 = 0.46*10^-4 C
q3 = (0.87 / 4.7) *
1.74*10^-4 = 0.32*10^-4 C
q4 = (0.87 / 5.6) * 1.74*10^-4 = 0.24*10^-4 C
c)
the voltage across each capacitor
Ui = qi / Ci
U1 = 0.69*10^-4 / 2.2*10^-6
= 31.36 V
U2 = 0.46*10^-4 / 3.3*10^-6 = 13.94 V
U3 = 0.32*10^-4 /
4.7*10^-6 = 6.81 V
U4 = 0.24*10^-4 / 5.6*10^-6 = 4.29 V

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