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 .

## Answer

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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