A series RCL circuit has a resonant frequency of 1470 Hz. When operating at a frequency other than 1470 Hz, the circuit has a capacitive reactance of 5.0 Ù and an inductive reactance of 29.0 Ù.
(a) What is the value of L?
1 H
(b) What is the value of C?
2 F
F = Fr*sqrt(Xl/Xc)
F = 1470*sqrt(29/5) = 3540.23 Hz
a. WL = 29
L = 29/W = 29/22244 = 1.3*10^-3 H=1.3mH
b. 1/WC = 5
WC = 1/5 = 0.2
C = 0.2/22244 = 9*10^-6F = 9 uF.
NOTES
1. Sqrt = Square root
2. Fr = Resonant Freq.
3. W = 2PI*F.
To find the value of L and C in the series RCL circuit, we can use the following formulas:
Resonant Frequency (fr):
fr = 1 / (2π√(LC))
Capacitive Reactance (Xc):
Xc = 1 / (2πfC)
Inductive Reactance (Xi):
Xi = 2πfL
Given that the resonant frequency (fr) is 1470 Hz, the capacitive reactance (Xc) is 5.0 Ω, and the inductive reactance (Xi) is 29.0 Ω, we can solve for L and C using the above formulas.
(a) Value of L:
Using the formula for inductive reactance:
Xi = 2πfL
Substituting the given values:
29.0 Ω = 2π(1470 Hz)L
Simplifying the equation:
L = 29.0 Ω / (2π(1470 Hz))
Calculating the value of L:
L ≈ 2.4307 x 10^(-4) H
Rounding to 1 significant figure, L ≈ 0.0002 H or 2 x 10^(-4) H
Therefore, the value of L is 0.0002 H or 2 x 10^(-4) H.
(b) Value of C:
Using the formula for capacitive reactance:
Xc = 1 / (2πfC)
Substituting the given values:
5.0 Ω = 1 / (2π(1470 Hz)C)
Simplifying the equation:
C = 1 / (2π(1470 Hz)5.0 Ω)
Calculating the value of C:
C ≈ 6.8023 x 10^(-7) F
Rounding to 1 significant figure, C ≈ 7 x 10^(-7) F
Therefore, the value of C is 7 x 10^(-7) F.
To find the value of L (inductance) and C (capacitance), we can use the formulas for reactance in an RCL circuit:
Xl = 2πfL, where Xl is the inductive reactance, f is the frequency, and L is the inductance.
Xc = 1/(2πfC), where Xc is the capacitive reactance, f is the frequency, and C is the capacitance.
Given that the resonant frequency, f, is 1470 Hz, we can use this frequency to find values of L and C corresponding to the given reactance values.
(a) To find L:
At resonant frequency, the circuit is purely resistive, so Xl = 0.
Using Xl = 2πfL, we can set Xl to 0 and solve for L:
0 = 2π(1470)L
0 = 2940πL
L = 0
So the value of L is 0 H.
(b) To find C:
Given that Xc = 5.0 Ω at a frequency other than the resonant frequency, we can use Xc = 1/(2πfC) to find C:
5.0 Ω = 1/(2πfC)
C = 1/(2πfXc)
C = 1/(2π(1470)(5.0))
C ≈ 1.7 x 10^(-5) F
So the value of C is approximately 1.7 x 10^(-5) F.