Please assist thank you.
henry2,
answered
R = 3300 ohms.
Xl = 2pi*F*L = 628 ohms.
Xc = 1/(2pi*F*C) = 7958 ohms.
Z = 3300+j(628-7958)
Z = 3300-j7330 = 8038 ohms[-65o].
henry2,
answered
X = Xl*Xc/(Xl-Xc) = 628*7958/(-7958+628) = -j682 ohms. = 682[-90].
Z = R*X/(R-jX) = 3300*682[-90]/(3300-j682)
Z = 3300*682[-90]/3370[-12o] = 668 ohms[-78o].
Explain Bot
answered
1/Z = 1/R + 1/Zc + 1/Zl
Where Zc represents the impedance of the capacitor and Zl represents the impedance of the inductor. Let's calculate each impedance step by step:
1. Calculate the impedance of the resistor (Zr):
Zr = R = 3.3 kΩ
2. Calculate the impedance of the capacitor (Zc):
Zc = 1 / (jωC), where j is the imaginary unit (√(-1)), ω is the angular frequency, and C is the capacitance.
ω = 2πf, where f is the frequency.
Given the frequency f = 1 kHz = 1 x 10^3 Hz, we can calculate ω:
ω = 2π(1 x 10^3 Hz) = 2π x 10^3 rad/s
Now, we can calculate the impedance of the capacitor:
Zc = 1 / (j(2Ï€ x 10^3 rad/s)(0.02 x 10^-6 F))
Zc = 1 / (j(4Ï€ x 10^-3) F)
Zc = 1 / (0 - j(4Ï€ x 10^-3) F)
Zc = 1 / (-j(4Ï€ x 10^-3) F)
Zc = -j / (4Ï€ x 10^-3) F
3. Calculate the impedance of the inductor (Zl):
Zl = jωL, where j is the imaginary unit (√(-1)), ω is the angular frequency, and L is the inductance.
Given the inductance L = 100 mH = 100 x 10^-3 H, we can calculate the impedance of the inductor:
Zl = j(2Ï€ x 10^3 rad/s)(100 x 10^-3 H)
Zl = j(200Ï€ x 10^-3) H
Zl = j(0.2Ï€) H
4. Substitute the values of Zr, Zc, and Zl into the formula:
1/Z = 1/R + 1/Zc + 1/Zl
1/Z = 1/(3.3 kΩ) + 1/(-j(4π x 10^-3) F) + 1/(j(0.2π) H)
Now, we need to calculate the reciprocal of the sum on the right side of the equation to find 1/Z:
1/Z = 1/(3.3 kΩ) + j(1/(0.2π) H) - j(1/(4π x 10^-3) F)
Finally, we can take the reciprocal of both sides to find Z:
Z = 1/(1/(3.3 kΩ) + j(1/(0.2π) H) - j(1/(4π x 10^-3) F))
Performing the calculations will give you the final impedance value for the parallel alarm system at a frequency of 1 kHz.
