The minimum possible value of the circuit's resistance is approximately 2100 Ω (to two significant figures).
Here's how to find the minimum possible resistance:
Calculate the bandwidth:
The current must be at least 50% of the resonance current within the frequency range of 54-60 MHz. This means the bandwidth is 6 MHz (60 MHz - 54 MHz).
Determine the quality factor (Q):
The Q factor determines the sharpness of the resonance peak. A lower Q means a wider bandwidth.
Use the formula Q = f_res / bandwidth:
Q = 57 MHz / 6 MHz ≈ 9.5
Calculate the minimum resistance:
The Q factor is also related to the resistance by Q = R / (2 * pi * f_res * L).
Rearrange for R:
R = Q * 2 * pi * f_res * L
Plug in values, using the inductance L = 0.56 μH (from your answer to part A):
R = 9.5 * 2 * pi * 57 * 10^6 * 0.56 * 10^-6 ≈ 2050 Ω
Therefore, the minimum possible value of the circuit's resistance is approximately 2100 Ω (to two significant figures).
The probable question can be: Question: A television channel is assigned the frequency range from 54 MHz to 60 MHz. A series RLC tuning circuit in a TV receiver resonates in the middle of this frequency range. The circuit uses a 14 pF capacitor. A) What is the value of the inductor? I found the answer of this to be L = .56 uH I need help with question B) B) In order to function properly, the
A television channel is assigned the frequency range from 54 MHz to 60 MHz. A series RLC tuning circuit in a TV receiver resonates in the middle of this frequency range. The circuit uses a 14 pF capacitor.
A) What is the value of the inductor?
I found the answer of this to be L = .56 uH
I need help with question B)
B) In order to function properly, the current throughout the frequency range must be at least 50% of the current at the resonance frequency. What is the minimum possible value of the circuit's resistance?
Express your answer to two significant figures and include the appropriate units.