(8e25p65&66) A capacitance C1-10.9 F is connected in series with a capacitance C

Question

(8e25p65&66) A capacitance C1-10.9 F is connected in series with a capacitance C2 = 4.7 F, and a potential difference of 175 V is applied across the pair. Calculate the equivalent capacitance 3.44 uF Submit Answer Incorrect. Tries 1/5 Previous Tries What is the charge on C1? Submit Answer Tries o/s What is the charge on C2? Tries o/5 Submit Answer What is the potential difference across C1? Submit Answer Tries 0/5 What is the potential difference across C2? Submit Answer Tries o/5 (c25p72) Repeat for the same two capacitors but with them now connected in parallel. Calculate the equivalent capacitance Submit Answer Tries 0/s What is the charge on C1? Submit Answer Tries o/5 What is the charge on C2? Submit Answer Tries 0/5 What is the potential difference across C1? Submit Answer Tries 0/5 What is the potential difference across C2?

Explanation / Answer

in series

Ceq = (C1*C2)/(C1+C2) = [(10.9*4.7)/(10.9+4.7)]*10^-6 = 3.28 uF

charge on C1 is Q1 = Ceq*V = 3.28*10^-6*175 = 574*10^-6 C

charge on C2 is Q2 = Q1 = 574*10^-6 C

potential difference across C1 is V1 = Q1/C1 = (574*10^-6)/(10.9*10^-6) = 52.66 V

potential difference across C2 is V2 = V-V1 = 175-52.66 = 122.34 V

in parallel

Ceq = C1+C2 = (10.9+4.7)*10^-6 = 15.6*10^-6 F

charge on C1 is Q1 = C1*V = 10.9*10^-6*175 = 1.9*10^-4 C

charge on C2 is Q2 = C2*V = 4.7*10^-6*174 = 8.178*10^-4 C

potential difference across C1 is V1 = 175 V

potential difference across C2 is V2 = 175 V

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