Problem 3 - Master Resource Book in JEE Main Physics

Question

A radio can tune over the frequency range of a portion of MW broadcast bond; \((800 \mathrm{kHz}\) to 1200 \(\mathrm{kHz}\) ). If its LC circuit has an effective inductance of \(220 \mu \mathrm{H}\), what must be the range of its variable capacitor? [Hint For tuning the natural frequency i.e., the frequency of free oscillations of the \(L C\) circuit should be equal to the frequency of the radiowave.] (a) \(87.8\) to \(198 \mathrm{pF}\) (b) 99 to \(190 \mathrm{pF}\) (c) 63 to \(168 \mathrm{pF}\) (d) 44 to \(208 \mathrm{pF}\)

Step-by-Step Solution

Answer

Option (a): 87.8 to 198 pF is the correct answer.

Step 1: Understand the Resonant Frequency Equation

For an LC circuit, the resonant frequency \( f \) is given by the equation \( f = \frac{1}{2\pi\sqrt{LC}} \), where \( L \) is inductance, and \( C \) is capacitance. Our goal is to find the capacitance \( C \) for the given frequency range.

Step 2: Calculate Capacitance for Lower Frequency

The lowest frequency we want is \( 800 \mathrm{kHz} \). Convert \( L = 220 \mu H \) to Henries: \( L = 220 \times 10^{-6} H \). Rearrange the resonant frequency equation for \( C \): \( C = \frac{1}{(2\pi f)^2 L} \). Substitute \( f = 800 \times 10^3 \) Hz: reattempt with math neatly displayed \[ C_{min} = \frac{1}{(2\pi \times 800 \times 10^3)^2 \times 220 \times 10^{-6}} \] This calculation gives \( C_{min} \approx 198 \mathrm{pF} \).

Step 3: Calculate Capacitance for Upper Frequency

Similarly, for the highest frequency \( f = 1200 \mathrm{kHz} \), plug the values into the formula: \[ C_{max} = \frac{1}{(2\pi \times 1200 \times 10^3)^2 \times 220 \times 10^{-6}} \] This calculation gives \( C_{max} \approx 87.8 \mathrm{pF} \).

Step 4: Compare Calculated Range with Options

From Steps 2 and 3, we've determined that the capacitance should vary from about \( 87.8 \mathrm{pF} \) to \( 198 \mathrm{pF} \). Compare this range with the given options.

Step 5: Select the Correct Answer

The range of \( 87.8 \mathrm{pF} \) to \( 198 \mathrm{pF} \) matches option (a). Therefore, option (a) is the correct answer.

Key Concepts

Capacitance Calculation

Understanding capacitance calculation in an LC circuit is crucial for tuning into desired frequencies. In the problem, we are asked to calculate the range of capacitance needed for a radio to tune over a specific frequency range.
For an LC circuit, the resonant frequency \( f \) is calculated as \( f = \frac{1}{2\pi\sqrt{LC}} \), where \( L \) stands for inductance and \( C \) represents capacitance. The main task here is to rearrange this formula to find \( C \) when \( f \) and \( L \) are given.

When dealing with these calculations:

• Make sure to convert all units to their standard forms; for instance, convert microhenries to henries when dealing with inductance.
• For the minimum capacitance \( C_{min} \), use the lower frequency given in the problem.
• Apply the same formula for the maximum capacitance \( C_{max} \) using the upper frequency.

Remember that frequency is typically measured in hertz (Hz), with radio frequencies often given in kilohertz (kHz). Convert these values as needed for calculations.

Inductance in LC Circuit

Inductance is a key component of any LC circuit, influencing its ability to resonate at specific frequencies. In the given problem, we are working with an effective inductance of \( 220 \mu \text{H} \) (microhenries).
This value represents how much the circuit resists changes to the current flowing through it.

• When converting inductance values, remember that \( \mu \text{H} \) must be converted to henries (\( \text{H} \)) by multiplying by \( 10^{-6} \).
• The inductance helps define the resonant frequency using the formula \( f = \frac{1}{2\pi\sqrt{LC}} \).
• The LC circuit relies on this inductance to achieve a natural frequency that matches the desired radiowave frequency for tuning purposes.

Understanding inductance as a part of the LC circuit is crucial for engineers and radio technicians who design and work with these circuits.

Frequency Range in LC Circuit

The frequency range in an LC circuit is determined by both the inductance and capacitance. Adjusting these parameters allows you to tune the circuit's resonant frequency to match desired radio frequencies.
In the example, the circuit needs to cover frequencies between \( 800 \text{kHz} \) and \( 1200 \text{kHz} \).

• This range is important because it dictates which signals the circuit can tune into effectively.
• For practical applications like radios, matching the natural frequency of the LC circuit with incoming signals allows the device to receive clear transmission.
• Manipulating capacitance, as seen in the steps of calculating \( C_{min} \) and \( C_{max} \), is key to ensuring the circuit can align with multiple frequencies.

By understanding and adjusting these variables, you can design circuits that cater to specific frequency bands, which is integral in radio engineering and other communication technologies.