In the figure (Figure 1) , C 1 = C 5 = 8.7 F and C 2= C 3 = C 4 = 4.9 F . The ap

Question

In the figure (Figure 1) , C1 = C5 = 8.7 F and C2= C3 = C4 = 4.9 F . The applied potential is Vab = 220 V

Part A

What is the equivalent capacitance of the network between points a and b?

Express your answer using two significant figures.

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Part B

Calculate the charge on each capacitor and the potential difference across each capacitor.

Express your answer using two significant figures.

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Part C

Express your answer using two significant figures.

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Part D

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Part E

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Part F

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Explanation / Answer

a)

Ceq = C1 || C5 || ((C3||C4) + C2)

= ( 8.7 || 8.7 ) || (4.9 || 4.9 + 4.9)

= 8.7/2 || (4.9/2 + 4.9)

= 4.35 || (7.35)

= 4.35*7.35/(4.35 + 7.35)

= 2.73 uF

b)

Charge on C1 , Q1 = 2.73*10^-6*220 = 6*10^-4 C

Voltage across C1 , V1 = Q1/C1 = 6*10^-4/(8.7*10^-6) = 69 V

Charge on C5 , Q5 = 6*10^-4 C

Voltage across C5 , V5 = Q5/C5 = 6*10^-4/(8.7*10^-6) = 69 V

Voltage across C2 = 220 - 69*2 = 82 V

So, charge on C2, Q2 = 82*4.9*10^-6 = 4*10^-4 C

Voltage across C3 , V3 = 82/2 = 41 V

Charge on C3, V3 = 41*4.7*10^-6 = 2*10^-4 C

Voltage across C4 , V4 = 82/2 = 41 V

Charge on C4 , Q4 = 41*4.7*10^-6 = 2*10^-4 C

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