Question

the frequency at the other end of the broadcast band is 0.542 mhz . what is the maximum capacitance cmax of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?

Answer 1

The maximum capacitance (Cmax) for an LC circuit to cover an AM broadcast range of 540 to 1600 kHz, with an inductance of 2.5 mH, is calculated using the resonant frequency formula. By rearranging to solve for C and substituting the given values, Cmax corresponds to the lowest frequency of 540 kHz.

Step-by-step explanation:

To determine the maximum capacitance (Cmax) required for an LC circuit in an AM tuner to be adjustable over the AM broadcast band range of 540 to 1600 kHz, we can use the formula for the resonant frequency of an LC circuit:
$$
f = \frac{1}{2\pi\sqrt{LC}}
$$

Where f is the frequency, L is the inductance and C is the capacitance. Since we are looking for the maximum capacitance and the frequency at the other end of the broadcast band is 0.542 MHz (or 542 kHz), we rearrange the formula to solve for C:
$$
C = \frac{1}{{(2\pi f)}^2 L}
$$
Given that the inductance L is 2.5 mH (or 0.0025 H), we can insert these values into the equation to find Cmax:
$$
Cmax = \frac{1}{{(2\pi \cdot 542000)}^2 \cdot 0.0025}
$$

Calculating this value will give us the maximum capacitance needed for the tuner to cover the entire broadcast band. Note that Cmax corresponds to the lowest frequency of 540 kHz, since the capacitance needed for resonance increases as the frequency decreases.

Answer 2

The maximum capacitance Cmax for the variable capacitor to cover the AM broadcast band from 540 kHz to 1600 kHz with an inductor of 2.5 mH is approximately 137.4 nF.

Step-by-step explanation:

To determine the maximum capacitance Cmax for a variable capacitor in an LC circuit within an AM tuner, applicable to the AM broadcast band frequency range from 540 kHz to 1600 kHz, we utilize the formula for the resonant frequency of an LC circuit: f = 1 / (2π√(LC)), where f is the frequency, L is the inductance, and C is the capacitance. Given the inductance is 2.5 mH, we can rearrange the formula to solve for C when f is at the ends of the broadcast band. The maximum capacitance Cmax will correspond to the minimum frequency of 540 kHz.

First, convert the frequency from MHz to Hz (540 kHz = 540,000 Hz) and the inductance from mH to H (2.5 mH = 0.0025 H). Then, insert the values into the rearranged formula C = 1 / (4π²f²L) to find Cmax.

Cmax = 1 / (4π²(540,000)²(0.0025)) ≈ 0.0000001374 F or 137.4 nF (nanofarads).