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 13 pF capacitor.

A.) What is the value of the inductor?

Express your answer to two significant figures and include the appropriate units.

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.

Answer 1

A)

f_H = higher limit of frequency = 60 MHz

f_L = lower limit of frequency = 54 MHz

middle frequency is hence given as

f = (f_H + f_L)/2 = (60 + 54)/2 = 57 MHz

f = 57 x 10⁶ Hz

C = capacitance = 13 pF = 13 x 10^-12 F

L = Inductance = ?

resonance frequency is given as

f = 1/(2 sqrt(LC))

57 x 10⁶ = 1/(2(3.14) sqrt(L(13 x 10^-12))

L = 6.0 x 10^-7 H

B)

I_r = current at resonance

I = current at other frequency in the given frequency range

V = supply voltage

z_r = Impedance at resonance frequency = R

z = Impedance at other frequency in the given frequency range = sqrt(R² + (X_L - X_C)²)

for constant supply voltage, we have

I 1/z

So

I/I_r = z_r/z

Given that :

I = (0.5) I_r

I/I_r = 0.5

So

z_r/z = 0.5

R = (0.5) sqrt(R² + (X_L - X_C)²)

Squaring both side

R² = (0.25) (R² + (X_L - X_C)²)

3 R² = (X_L - X_C)²

3 R² = ((2fL) - 1/(2fC))²

at f = 60 x 10⁶ Hz

3 R² = ((2(3.14)(60 x 10⁶)(6 x 10^-7)) - 1/(2(3.14)(60 x 10⁶)(13 x 10^-12)))²

R = 13 ohm