Question


In the following figure, C1 = 6.00 mu F, C2 = 3.00


Please explain all calculation

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Answer #2

Q2 = C2*V2

==> V2 = Q2/C2 = 40/3 = 13.33 volts

here V2 = V1


a)
so, Q1 = C1*V1 = 6*10^-6*13.33 = 80 micro C


Q3 = Q1+Q2 = 120 micro C

b) Vab = V1 + V3

= 13.3 + Q3/C3

= 13.33 + 120/5

= 37.33 volts

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