What is the value of the total capacitive reactance in each circuit? A. f = 1 kHz and C = 1.047 uF.
B. f = 1 Hz, C1 = 10 uF and C2 = 15 uF.
C. F = 60 Hz, C1 = 1 uF and C2 = 1 uF.
To determine the total capacitive reactance in each circuit, we can use the formula XC = 1 / (2πfC), where XC represents the capacitive reactance, f is the frequency, and C is the capacitance.
A. For circuit A, given f = 1 kHz and C = 1.047 uF, we can plug these values into the formula:
XC = 1 / (2π * 1 kHz * 1.047 uF)
To calculate the answer, convert the uF to F:
XC = 1 / (2π * 1 kHz * 1.047 * 10^-6 F)
Simplify:
XC = 1 / (2π * 10^3 Hz * 1.047 * 10^-6 F)
XC ≈ 151.16 Ω
Therefore, in circuit A, the total capacitive reactance is approximately 151.16 Ω.
B. For circuit B, given f = 1 Hz, C1 = 10 uF, and C2 = 15 uF, we can use the same formula and sum up the individual capacitive reactance values for C1 and C2:
XC1 = 1 / (2π * 1 Hz * 10 uF) ≈ 1.59 Ω
XC2 = 1 / (2π * 1 Hz * 15 uF) ≈ 1.06 Ω
Summing them up gives us:
Total XC = XC1 + XC2 ≈ 1.59 Ω + 1.06 Ω ≈ 2.65 Ω
Therefore, in circuit B, the total capacitive reactance is approximately 2.65 Ω.
C. For circuit C, given f = 60 Hz, C1 = 1 uF, and C2 = 1 uF, we apply the formula:
XC1 = 1 / (2π * 60 Hz * 1 uF) ≈ 2.65 Ω
XC2 = 1 / (2π * 60 Hz * 1 uF) ≈ 2.65 Ω
Again, summing them up gives us:
Total XC = XC1 + XC2 ≈ 2.65 Ω + 2.65 Ω ≈ 5.30 Ω
Therefore, in circuit C, the total capacitive reactance is approximately 5.30 Ω.