A series RCL circuit has a resonant frequency of 1500 Hz. When operating at a frequency other than 1500 Hz, the circuit has a capacitive reactance of 5.0 Ù and an inductive reactance of 30.0 Ù. What are the values of (a) L and (b) C?
I know I should be using these 2 equations:
Xc= 1/(2*pi*f*C)
Xl= 2*pi*f*L
Since each have 2*pi*f in them, with rearranging and substituting, you get:
C= L/(Xl*Xc)
However, with this it leaves the 2 unknowns, and I'm not sure what other equations I could use?
f=1500 Hz, X(C)=5Ω, X(L)=30 Ω
X(C)= 1/ωC,…..(1)
X(L)=ωL, ……..(2)
Multiply (1) by (2):
X(C) •X(L)= ωL/ ωC =L/C.
C=L/X(C) •X(L) …..(3)
Resonant frequency is
f=1/2π√(LC) =>
C=1/4•π²•f²•L ….(4)
Equate (3) and (4)
L/X(C) •X(L) =1/4•π²•f²•L,
L=sqrt{X(C) •X(L}/2•π•f=
=sqrt(5•30)/2π•1500=1.3•10^-3 H=
=1.3 mH,
C= L/X(C) •X(L)=
=1.3•10^-3/5•30=8.67•10^-6 =F=
= 8.67 μF
To solve for the values of L and C in the given series RCL circuit, you are correct that we can use the equations:
Xc = 1/(2*pi*f*C)
Xl = 2*pi*f*L
where Xc is the capacitive reactance and Xl is the inductive reactance.
Given that the circuit has a resonant frequency of 1500 Hz, we can substitute this value into the above equations:
Xc = 1/(2*pi*1500*C)
Xl = 2*pi*1500*L
We also have the additional information that the circuit has a capacitive reactance of 5.0 Ω and an inductive reactance of 30.0 Ω when operating at a frequency other than 1500 Hz. Let's use these values to solve for C and L:
1. Simplify the equation for Xc:
5.0 = 1/(2*pi*1500*C)
2. Rearrange the equation to solve for C:
C = 1/(2*pi*1500*5.0)
Now, substitute the given values into the equation:
C ≈ 1 / (2 * 3.1416 * 1500 * 5.0)
C ≈ 1.0607 x 10^(-7) Farads (F)
3. Simplify the equation for Xl:
30.0 = 2*pi*1500*L
4. Rearrange the equation to solve for L:
L = 30.0 / (2*pi*1500)
Now, calculate the value:
L ≈ 9.548 x 10^(-4) Henrys (H)
So, the values of L and C in the series RCL circuit are approximately:
(a) L ≈ 9.548 x 10^(-4) H
(b) C ≈ 1.0607 x 10^(-7) F
To find the values of L and C in the RCL circuit, you can use the resonant frequency and the given values of capacitive reactance (Xc) and inductive reactance (Xl).
Let's start by rearranging the equation for capacitive reactance:
Xc = 1 / (2πfC)
We can substitute the given value of Xc = 5.0 Ω and the resonant frequency f = 1500 Hz:
5.0 Ω = 1 / (2π * 1500 Hz * C)
Now, rearrange this equation to solve for C:
C = 1 / (5.0 Ω * 2π * 1500 Hz)
C ≈ 2.12 × 10^(-7) F
Next, rearrange the equation for inductive reactance:
Xl = 2πfL
Again, substitute the given value of Xl = 30.0 Ω and the resonant frequency f = 1500 Hz:
30.0 Ω = 2π * 1500 Hz * L
Now, rearrange this equation to solve for L:
L = 30.0 Ω / (2π * 1500 Hz)
L ≈ 2.00 × 10^(-3) H
Therefore, the values of (a) L ≈ 2.00 × 10^(-3) H and (b) C ≈ 2.12 × 10^(-7) F in the RCL circuit.