FM radio stations use radio waves with frequencies from 88.0 to 108 MHz to broadcast their signals. Assuming that the inductance has a value of 4.60 x 10^-7 H, determine the range of capacitance values that are needed so the antenna can pick up all the radio waves broadcasted by FM stations.
_____F (lowest acceptable capacitance)
____F (highest acceptable capacitance)
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Physics please help! - bobpursley, Wednesday, October 17, 2007 at 3:07pm
Find Xl (2PI f L)
then find C such that
Xc= 1/2PI*f*C where Xc= Xl
do it at each frequency end.
Please tell me where i went wrong.
Xl =(2*pi*f*L)
Xl = 2*3.142*(88.0x10^6)(4.60x10^-7)
Xl = 254.37632
I substitute this into the formula Xc= 1/2PI*f*C to find the value of C.
Xc = 1 / (2*3.1428(88.0x10^6)C)
254.37632 = 1 / 552992000 C
C = 7.11 x10^-12
According to the system this answer is incorrect for the lowest acceptable capacitance. Please tell me where I went wrong!
To determine the range of capacitance values needed for the antenna to pick up all the radio waves broadcasted by FM stations, we need to calculate the lowest and highest acceptable capacitance values.
We know that the FM radio stations use radio waves with frequencies from 88.0 to 108 MHz. We can convert these frequencies to Hz by multiplying them by 10^6.
The lowest frequency in Hz is 88.0 MHz = 88.0 x 10^6 Hz
The highest frequency in Hz is 108 MHz = 108 x 10^6 Hz
Now let's calculate the lowest and highest values of inductive reactance (Xl) using the formula Xl = 2πfL, where f is the frequency in Hz and L is the inductance value given (4.60 x 10^-7 H).
For the lowest frequency:
Xl (lowest) = 2π * (88.0 x 10^6 Hz) * (4.60 x 10^-7 H)
Xl (lowest) ≈ 1.028 Ω
For the highest frequency:
Xl (highest) = 2π * (108 x 10^6 Hz) * (4.60 x 10^-7 H)
Xl (highest) ≈ 1.262 Ω
Now let's calculate the lowest and highest acceptable capacitance values (C) using the formula Xc = 1/(2πfC), where Xc is the capacitive reactance.
For the lowest acceptable capacitance:
Xc (lowest) = Xl (lowest)
1/(2π * (88.0 x 10^6 Hz) * C (lowest)) = 1.028 Ω
C (lowest) ≈ 1.93 x 10^-12 F (Farads)
For the highest acceptable capacitance:
Xc (highest) = Xl (highest)
1/(2π * (108 x 10^6 Hz) * C (highest)) = 1.262 Ω
C (highest) ≈ 1.57 x 10^-12 F (Farads)
So, the range of capacitance values needed for the antenna to pick up all the radio waves broadcasted by FM stations is approximately:
Lowest acceptable capacitance (C): 1.93 x 10^-12 F
Highest acceptable capacitance (C): 1.57 x 10^-12 F
To determine the range of capacitance values needed for the antenna to pick up all the radio waves broadcasted by FM stations, we need to find the lowest and highest acceptable capacitance values.
First, let's calculate the lowest acceptable capacitance (F):
1. Use the formula Xl = 2πfL, where Xl is the inductive reactance (impedance), f is the frequency, and L is the inductance.
Xl = 2π(88.0x10^6)(4.60x10^-7)
Xl = 1.0767 Ω (approximately)
2. The impedance of the capacitor (Xc) should be equal to the inductive reactance (Xl) at the lowest frequency.
Xc = Xl
1 / (2π(88.0x10^6)C) = 1.0767
C = 1 / (2π(88.0x10^6)(1.0767))
C ≈ 1.859 x 10^-12 F
Therefore, the lowest acceptable capacitance is approximately 1.859 x 10^-12 F.
Now, let's calculate the highest acceptable capacitance (F):
1. Follow the same steps as above but use the highest frequency (108 MHz):
Xl = 2π(108x10^6)(4.60x10^-7)
Xl = 1.3617 Ω (approximately)
2. The impedance of the capacitor (Xc) should be equal to the inductive reactance (Xl) at the highest frequency.
Xc = Xl
1 / (2π(108x10^6)C) = 1.3617
C = 1 / (2π(108x10^6)(1.3617))
C ≈ 9.142 x 10^-13 F
Therefore, the highest acceptable capacitance is approximately 9.142 x 10^-13 F.
To summarize:
Lowest acceptable capacitance (F) ≈ 1.859 x 10^-12 F
Highest acceptable capacitance (F) ≈ 9.142 x 10^-13 F