A capacitance C1 = 7.0x10^-6F is connected in series with a capacitance C2 = 4.0
ID: 2288631 • Letter: A
Question
A capacitance C1 = 7.0x10^-6F is connected in series with a capacitance C2 = 4.0x10^-6 F, and a potential difference of 275 V is applied across the pair. Calculate the equivalent capacitance. Incorrect. Tries 3/5 Previous Tries What is the charge on C1? Tries 0/5 What is the charge on C2? Tries 0/5 What is the potential difference across C1? Tries 0/5 What is the potential difference across C2? Tries 0/5 (c25p72) Repeat for the same two capacitors but with them now connected in parallel. Calculate the equivalent capacitance. Tries 0/5 What is the charge on C1? Tries 0/5 What is the charge on C2? Tries 0/5 What is the potential difference across C1? Tries 0/5 What is the potential difference across C2?
Explanation / Answer
A capacitance C1 = 7.0x10^-6F is connected in series with a capacitance C2 = 4.0x10^-6 F, and a potential difference of 275 V is applied across the pair. Calculate the equivalent capacitance.
equivalent capacitance = C1*C2 /(C1+C2) = 7*4 / (7+4) = 2.5455 * 10^-6 F
now total charge flowing in circuit
Q = C_eq *V = 2.5455*10^-6* 275 = 7*10^-4 coulombs
What is the charge on C1?
since C1 and C2 are in series the charge Q is same in both C1 and C2
so Q_C1 = 7*10^-4
What is the charge on C2?
Q_C2 = Q_C1= 7*10^-4
What is the potential difference across C1?
V = Q/C = 7*10^-4/ (7*10^-6) = 100 volts
What is the potential difference across C2?
V = Q/C = 7*10^-4 /(4*10^-6) = 175 volts
Repeat for the same two capacitors but with them now connected in parallel.
Calculate the equivalent capacitance.
C_eq = C1+C2 = 7*10^-6 F+4*10^-6 F = 11*10^-6 F
What is the charge on C1?
parallel connection so V is same for C1 and C2; V_C1 = V_C2 = 275 volts
Q_C1 = C1* V = 7*10^-6*275 = 1.925*10^-3 coulombs
What is the charge on C2?
Q_C2 = 4*10^-6*275 = 1.1*10^-3 coulombs
What is the potential difference across C1?
V_C1 = 275 volts
What is the potential difference across C2?
V_C2 = 275 volts
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