In the diagram above, V = 100 volts; C1 = 12 microfarads; C2 = 24 microfarads; R = 10 ohms.
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Please question is not full edit questions and your answers is here a. Q = Cε = 12 μF × 100 V = 1200 μC
b. Connecting the two capacitors puts them in parallel with the same voltage so V1 = V2 and V = Q/C which gives Q1/C1 = Q2/C2 or Q1/12 = Q2/24 and Q2 = 2Q1. We also know the total charge is conserved so Q1 + Q2 = 1200 μC so we have Q1 + 2Q1 = 1200 μC so Q1 = 400 μC
c. V = Q/C = 33.3 V
d. When the battery is reconnected, both capacitors charge to a potential difference of 100 V each. The total charge is then Q = Q1 + Q2 = (C1 + C2)V = 3600 μC making the additional charge from the battery 2400 μC.
b. Connecting the two capacitors puts them in parallel with the same voltage so V1 = V2 and V = Q/C which gives Q1/C1 = Q2/C2 or Q1/12 = Q2/24 and Q2 = 2Q1. We also know the total charge is conserved so Q1 + Q2 = 1200 μC so we have Q1 + 2Q1 = 1200 μC so Q1 = 400 μC
c. V = Q/C = 33.3 V
d. When the battery is reconnected, both capacitors charge to a potential difference of 100 V each. The total charge is then Q = Q1 + Q2 = (C1 + C2)V = 3600 μC making the additional charge from the battery 2400 μC.
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