Problem 48 - College Physics

Question

A series RLC radio receiver circuit with an inductance of \(1.50 \mu \mathrm{H}\) is tuned to an FM station at \(98.9 \mathrm{MHz}\) by adjusting a variable capacitor. When the circuit is tuned to this station, (a) what is its inductive reactance? (b) What is its capacitive reactance? (c) What is its capacitance?

Step-by-Step Solution

Answer

Inductive reactance: \( X_L = 930.3 \Omega \). Capacitive reactance: \( X_C = 930.3 \Omega \). Capacitance: \( C = 1.12 \times 10^{-12} F \).

Step 1: Understand the Problem

We have a radio receiver circuit which is RLC (Resistor, Inductor, Capacitor) type. It is tuned to a frequency of 98.9 MHz using a variable capacitor. We need to find the inductive reactance, capacitive reactance, and the capacitance in the circuit.

Step 2: Calculate Angular Frequency

Convert the given frequency to angular frequency. The angular frequency \( \omega \) is calculated using the formula:\[\omega = 2\pi f\]where \( f = 98.9 \times 10^6 \) Hz (given). Substitute this value into the formula to find \( \omega \).

Step 3: Find Inductive Reactance (a)

The inductive reactance \( X_L \) is calculated using the formula:\[ X_L = \omega L \]where \( L = 1.50 \times 10^{-6} \) H (given). Substitute \( \omega \) found in Step 2 into this equation to compute \( X_L \).

Step 4: Calculate Capacitive Reactance (b)

In a resonant circuit, the capacitive reactance \( X_C \) can be found from the relationship:\[ X_C = X_L \]Since the circuit is in resonance at \( 98.9 \) MHz, use the value of \( X_L \) from Step 3 to state that \( X_C = X_L \).

Step 5: Determine Capacitance (c)

Capacitive reactance is given by the formula:\[ X_C = \frac{1}{\omega C} \]Rearrange this equation to solve for the capacitance \( C \):\[ C = \frac{1}{\omega X_C} \]Substitute for \( \omega \) from Step 2 and \( X_C \) from Step 4 into this formula to find \( C \).

Key Concepts

Inductive Reactance

Inductive reactance is a vital concept in alternating current (AC) circuits, particularly when dealing with inductors. It refers to the resistance faced by an AC current due to the presence of an inductor in the circuit. The inductive reactance, represented by the symbol \(X_L\), depends on both the frequency of the AC signal and the inductance of the inductor. It can be calculated using the formula:\[ X_L = \omega L \]Where:

• \(X_L\) is the inductive reactance, in ohms (\(\Omega\)).
• \(\omega\) is the angular frequency, given by \(2\pi f\).
• \(L\) is the inductance, in henries (H).

As the frequency \(f\) or the inductance \(L\) increases, the inductive reactance \(X_L\) also increases. This implies that the inductor opposes higher frequency signals more than lower frequency signals. It's important to note that inductive reactance only applies in AC circuits and becomes zero in DC circuits, as DC has zero frequency. This concept plays a crucial role in tuning circuits to specific frequencies, such as in the RLC circuit problem specified for receiving FM signals.

Capacitive Reactance

Capacitive reactance is quite similar in its role to inductive reactance but is related to capacitors in an AC circuit. It provides the resistance to the change in voltage across a capacitor due to changing AC signals.The capacitive reactance is represented by \(X_C\) and is given by the formula:\[ X_C = \frac{1}{\omega C} \]Where:

• \(X_C\) is the capacitive reactance, in ohms (\(\Omega\)).
• \(\omega\) is the angular frequency, calculated by \(2\pi f\).
• \(C\) is the capacitance, in farads (F).

Interestingly, \(X_C\) is inversely proportional to both the frequency \(f\) and the capacitance \(C\). As the frequency or capacitance increases, \(X_C\) decreases. This behavior allows capacitors to pass signals with higher frequencies while blocking lower frequencies. In a resonant RLC circuit, the inductive and capacitive reactance often balance each other, leading to specific applications like filters and tuners in electronics.

Capacitance Calculation

Capacitance is a measure of a capacitor's ability to store charge per unit voltage across its plates. It’s vital in determining the behavior of an AC circuit, particularly regarding its frequency response.In the context of the exercise, calculating the capacitance \(C\) enabled the circuit to tune into a specific radio frequency, like the FM station at 98.9 MHz. The capacitance can be derived from the capacitive reactance in a resonant circuit using:\[ C = \frac{1}{\omega X_C} \]Where:

• \(C\) is the capacitance, in farads (F).
• \(\omega\) is the angular frequency, \(2\pi f\).
• \(X_C\) is the capacitive reactance, derived from \(X_L\) in resonance.

In such circuits, the precise adjustment of capacitance is crucial to ensuring the circuit resonates at the desired frequency, effectively filtering or enhancing signals at that frequency. This tuning process is fundamental in radio receivers, allowing them to select different stations by varying the capacitance.