Problem 3(15 pts The figure below shows a network of capacitors connected to a b

Question

Problem 3(15 pts The figure below shows a network of capacitors connected to a battery. The values of the and of the battery's emf are indicated in the figure. The switch is Initially in a) weh the switch n the open position, calculate the charge stored in each capacitor and the voltage across each capacitor . Record your answers in the table below, and show your work on te sheet of paper Assume that the circuit has been connected long enough for all the to be fully charged. b) The switch Provide quanttative s now closed. What effect does that have on the charge stored by capacitor C? ave answers (Le. provide an explanation AND numerical values, don't just say "it ?.-8.00 F Charge (uc) Voltage (V) 3.22 V Ca 8.00 ?F 16-1NC 3.23y

Explanation / Answer

for the condition, when the switch is open
C4 is not part of the circuit
C1, C2 are in series, C3, C5 are in parallel
hence charge on C1, Q1 = Charge on C2 = Q2 = Q3 + Q5

hence
10 = Q1/C1 + Q2/C2 + Q3/C3
and
Q3/C3 = Q5/C5

Q3/2 = Q5/5

10 = Q1/5 + Q1/10 + Q3/2

now, Q = Ceff*10

and Ceff = (1/C1 + 1/C2 + 1/(C3 + C4))^-1 = 2.258064 uF
hence
Q = 22.580645 uC = Q1 = Q2
hence
Q3 = 6.4516129032 uC
Q5 = 16.12903225806 uC

hence
Q1 = 22.59 uC, V1 = Q1/C1 = 4.518 V
Q2 = 22.59 uC, V2 = 2.259 V
Q3 = 6.45 uC , V3 = 3.225 V
Q4 = 0 C, V4 = 0 V
Q5 - 16.129 uC, V5 = 3.225 V

b) now the switch is closed
so ceff = (1/5 + 1/10 + 1/(8 + 2 + 5)) = 2.7272727272 uF
hence, Ceff increases, so Q1 increases, and Q2 increases
so Q1 = Ceff*10 = 27.2 uF

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