A) Calculate the charge on capacitor C1. B)Calculate the potential difference across capacitor C1. C) Calculate the charge on capacitor C2

D ) Calculate the potential difference across capacitor C2. E) Calculate the charge on capacitor C3. F) Calculate the potential difference across capacitor C3. G) Calculate the charge on capacitor C4. H) Calculate the charge on capacitor C4. I) Calculate the potential difference between points a and

d. Constants In the figure (Figure 1), each capacitor has 4.50 AF and Vab 34.0 V Figure 1 of 1 C1 C2 c .

Remember:

For parallel combination

Ceq = C1 + C2 + C3 +……………

for series combination

1/Ceq = 1/C1 + 1/C2 + 1/C3 + …………

for 2 capacitors in series it will be

Ceq = C1*C2/(C1+C2)

Using this Information:

C1 and C2 are in series, So

C12 = 4.5*4.5/(4.5 + 4.5) = 2.25 uF

C12 and C3 are in parallel, So

C123 = 2.25 + 4.5 = 6.75 uF

C123 and C4 are in series, So

Ceq = C123*C4/(C123 + C4)

Ceq = 6.75*4.5/(6.75 + 4.5) = 2.7 uF

Now remember in capacitors parallel combination voltage

distribution in each part will be same and in series combination

charge distribution in each capacitor will be same.

Qeq = Ceq*Veq = 2.7*34.0 = 91.8 uC

Since C123 and C4 are in series, So

Q123 = Q4 = Qeq

Q123 = 91.8 uC

V123 = Q123/C123 = 91.8/6.75 = 13.6 V

Now Since C12 and C3 are in parallel, So

V12 = V3 = V123

V3 = 13.6 V (Part F)

Q3 = C3*V3 = 4.5*13.6 = 61.2 uC (Part E)

V12 = 13.6 V

Q12 = C12*V12 = 2.25*13.6 = 30.6 uC

Since C1 and C2 are in series, SO

Q1 = Q2 = Q12

Q1 = 30.6 uC = 30.6*10^-6 C (Part A)

V1 = Q1/C1 = 30.6/4.5 = 6.8 V (Part B)

Q2 = 30.6 uC = 30.6*10^-6 C (Part C)

V2 = Q2/C2 = 30.6/4.5 = 6.8 V (Part D)

From above

Q3 = C3*V3 = 4.5*13.6 = 61.2 uC (Part E)

V3 = 13.6 V (Part F)

Q4 = Qeq

Q4 = 91.8 uC = 91.8*10^-6 C (Part G)

V4 = Q4/C4 = 91.8/4.5 = 20.4 V (Part H)

Potential difference between a and d will be equal to V123 (from

given circuit)

Vad = V123 = 13.6 V (Part I)

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