Question

The AM radio band covers the frequency range from \(520 \mathrm{kHz}\) to \(1610 \mathrm{kHz}\). Assuming a fixed inductance in a simple LC circuit, what ratio of capacitance is necessary to cover this frequency range? That is, what is the value of \(C_{\mathrm{h}} / C_{\mathrm{l}}\) where \(C_{\mathrm{h}}\) is the capacitance for the highest frequency and \(C_{1}\) is the capacitance for the lowest frequency? a) 9.59 b) 0.104 c) 0.568 d) 1.76

Short answer

Answer: The ratio of capacitance is 0.104.

Step-by-step solution

Find the formula relating capacitance and frequency

In a simple LC circuit, the resonance frequency is given by: $$f = \frac{1}{2\pi \sqrt{LC}}$$ We need to make capacitance, C, the subject of the formula: $$C = \frac{1}{(2\pi f)^2 L}$$ We now have a formula relating capacitance and frequency in an LC circuit.

Calculate the capacitance for the highest frequency

We are given the highest frequency, 1610 kHz, and we need to find the corresponding capacitance \(C_\mathrm{h}\). Using the formula from Step 1, we have: $$C_\mathrm{h} = \frac{1}{(2\pi 1610 * 10^3)^2 L}$$

Calculate the capacitance for the lowest frequency

Likewise, we are given the lowest frequency, 520 kHz, and need to find the corresponding capacitance \(C_\mathrm{l}\). Using the formula from Step 1, we have: $$C_\mathrm{l} = \frac{1}{(2\pi 520 * 10^3)^2 L}$$

Find the ratio of capacitance \(C_\mathrm{h}/C_\mathrm{l}\)

Now, we need to find the ratio of the capacitances. Divide \(C_\mathrm{h}\) by \(C_\mathrm{l}\): $$\frac{C_\mathrm{h}}{C_\mathrm{l}} = \frac{\frac{1}{(2\pi 1610 * 10^3)^2 L}}{\frac{1}{(2\pi 520 * 10^3)^2 L}}$$ Simplifying the expression, we get: $$\frac{C_\mathrm{h}}{C_\mathrm{l}} = \frac{(2\pi 520 * 10^3)^2 L}{(2\pi 1610 * 10^3)^2 L}$$ Now, we can cancel out the L from the numerator and the denominator, which leaves us with: $$\frac{C_\mathrm{h}}{C_\mathrm{l}} = \frac{(2\pi 520 * 10^3)^2}{(2\pi 1610 * 10^3)^2}$$

Calculate the value of the capacitance ratio

After the simplification, we can now plug the values into the formula to find the final ratio of capacitance: $$\frac{C_\mathrm{h}}{C_\mathrm{l}} = \frac{(2\pi 520 * 10^3)^2}{(2\pi 1610 * 10^3)^2} = 0.104$$ Hence, the correct answer is (b) 0.104.

Key concepts

Resonance Frequency

In an LC circuit, the concept of resonance frequency is fundamental. The resonance frequency is the specific frequency at which the circuit's inductive and capacitive reactances are equal in magnitude but opposite in phase. This condition causes the circuit to oscillate at its natural frequency with minimal impedance, allowing maximum current to flow. The formula to find the resonance frequency \( f \) is: \[ f = \frac{1}{2\pi \sqrt{LC}} \] Where:

• \( L \) is the inductance
• \( C \) is the capacitance
• \( \pi \) is the constant Pi, approximately equal to 3.14159

Knowing the resonance frequency is crucial for tuning circuits like those used in radios. By adjusting either the inductance or capacitance, we can modify the circuit to resonate at the desired frequency. Changing these components allows the radio to switch from one station to another, picking signals across different frequencies.

Capacitance Ratio

The capacitance ratio in an LC circuit is defined as the ratio of capacitance values needed to cover different frequency ranges. In the context of our provided example, it's specifically the ratio \( \frac{C_{\mathrm{h}}}{C_{\mathrm{l}}} \) which relates to the highest and lowest frequencies, respectively.To understand why this ratio is significant, remember that capacitance in an LC circuit is inversely related to the square of the frequency: \[ C = \frac{1}{(2\pi f)^2 L} \]This means that higher frequencies correspond to lower capacitance and vice versa. The calculation of this ratio helps in designing circuits that can operate over a range of frequencies, like a radio tuned to different stations. By calculating \( \frac{C_{\mathrm{h}}}{C_{\mathrm{l}}} \), engineers can determine the necessary adjustments in capacitance to cover a given frequency band, such as the AM radio range, ultimately affecting the tuning capability of the device.

AM Radio Frequency Range

The AM radio frequency range is an essential part of the electromagnetic spectrum used for broadcasting audio signals. It spans from 520 kHz to 1610 kHz. AM radio stations across the world use this range to modulate signals so they can be broadcast over long distances, especially at night when AM radio waves can bounce off the ionosphere. To utilize this range effectively, designers use LC circuits to tune into specific frequencies within this band. Each radio station transmits a signal at a unique frequency, and by adjusting the LC circuit's components (inductance \( L \) and capacitance \( C \)), a radio can "select" a station to listen to by resonating at that station's frequency.Designing a tuning circuit involves finding the right capacitance ratio so the circuit can cover the full AM frequency range. By adjusting the capacitor, the circuit can dynamically alter its resonance frequency, allowing the listener to move from one frequency (or station) to another smoothly.