Question

A television channel is assigned the frequency range from 54 MHz
to 60 MHz. A series *R**L**C*tuning circuit in
a TV receiver resonates in the middle of this frequency range. The
circuit uses a 21 pF capacitor.

**Part A**

What is the value of the inductor?

Express your answer to two significant figures and include the
appropriate units.= 0.37 *μ*H

I found this answer.

**Part 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.

Need help with part B but part A should give you a start on Part B.

Answer #1

A) Resonant frequency, f_{o} = (54 + 60) / 2 = 57
MHz

Also, f_{o} = 1/2π(LC)^{1/2}

=> L = 1/(2πf_{o})^{2}C = 1/[(2π * 57 *
10^{6})^{2} * (21 * 10^{-12})] = 0.37
μH

B) Current, I = V / [R^{2} +
X^{2}]^{1/2}

Current at resonant frequency, I_{o} = V/R

For I ≥ 0.5I_{o},

[R^{2} + X^{2}]^{1/2} ≤ 2R

=> X^{2} ≤ 3R^{2}

=> X ≤ 3^{0.5}R

=> R ≥ X/3^{0.5}

=> R_{min} = X_{max}/3^{0.5}

Now, X = ωL - 1/ωC

Now, dX/dω = L + 1/ω^{2}C > 0

So, X is a strictly increasing function,

So, X_{max} = (2π * 60 * 10^{6} * 0.37 *
10^{-6}) - 1/[(2π * 60 * 10^{6} * 21 *
10^{-12}] = 13.17 ohm

So, R_{min} = 13.17 / 3^{0.5} = 7.6 ohm

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