Problem 19 - College Physics

Question

An ac series \(R-L-C\) circuit contains a \(120 \Omega\) resistor, a \(2.0 \mu \mathrm{F}\) capacitor, and a \(5.0 \mathrm{mH}\) inductor. Find (a) the resonance angular frequency and (b) the length of time that each cycle lasts at the resonance angular frequency.

Step-by-Step Solution

Answer

(a) The resonance angular frequency is \( 10^{4} \text{ rad/s} \). (b) Each cycle lasts approximately \( 6.2832 \times 10^{-4} \text{ s} \).

Step 1: Identify the Given Values

We are provided with the values for the resistance \( R = 120 \ \Omega \), the capacitance \( C = 2.0 \ \mu F \), and the inductance \( L = 5.0 \ mH \). We need to find the resonance angular frequency and the time period per cycle at resonance.

Step 2: Use the Formula for Resonance Angular Frequency

The formula for the resonance angular frequency \( \omega_0 \) in an \( R-L-C \) circuit is \( \omega_0 = \frac{1}{\sqrt{LC}} \).

Step 3: Convert Units

Convert \( L = 5.0 \ \mathrm{mH} \) to \( \mathrm{H} \): \( 5.0 \ \mathrm{mH} = 5.0 \times 10^{-3} \ \mathrm{H} \). Convert \( C = 2.0 \ \mu F \) to \( \mathrm{F} \): \( 2.0 \ \mu F = 2.0 \times 10^{-6} \ \mathrm{F} \).

Step 4: Calculate the Resonance Angular Frequency

Substitute the values into the formula: \[ \omega_0 = \frac{1}{\sqrt{(5.0 \times 10^{-3})(2.0 \times 10^{-6})}} \]Calculate: \[ \omega_0 = \frac{1}{\sqrt{10^{-8}}} = \frac{1}{10^{-4}} = 10^{4} \ \text{rad/s} \]

Step 5: Calculate the Period of Each Cycle

The period \( T \) of a cycle is the reciprocal of the frequency. The frequency \( f \) is related to angular frequency \( \omega_0 \) by \( f = \frac{\omega_0}{2\pi} \). Thus, \( T = \frac{1}{f} = \frac{2\pi}{\omega_0} \).Substitute \( \omega_0 = 10^{4} \ \text{rad/s} \) into the period formula:\[ T = \frac{2\pi}{10^{4}} = \frac{2\pi}{10^{4}} \, \text{s} \]

Step 6: Simplify the Period Expression

\[ T = \frac{2\times 3.1416}{10^{4}} \approx 6.2832 \times 10^{-4} \ \text{s} \]

Key Concepts

Resonance Angular Frequency

In an RLC circuit, the resonance angular frequency is a key characteristic that defines the frequency at which the impedance of the circuit is minimized, and thus, the circuit can potentially store and transfer energy most efficiently. This frequency is important because, at resonance, the inductive reactance and the capacitive reactance are equal in magnitude but opposite in phase. Hence, they cancel each other's effects.
To calculate the resonance angular frequency \( \omega_0 \) in a series RLC circuit, we use the formula:

• \( \omega_0 = \frac{1}{\sqrt{LC}} \)

where:

• \( L \) is the inductance in henries (H)
• \( C \) is the capacitance in farads (F)

By substituting the given values of \( L = 5.0 \times 10^{-3} \) H and \( C = 2.0 \times 10^{-6} \) F into the formula, we find the resonance angular frequency to be \( \omega_0 = 10^4 \text{ rad/s} \). This indicates the circuit's natural frequency of oscillation.

Inductive Reactance

Inductive reactance is a measure of the opposition that an inductor offers to the change in current in an AC circuit. It is denoted by \( X_L \) and is calculated using the formula:

• \( X_L = \omega L \)

where:

• \( \omega \) is the angular frequency in rad/s
• \( L \) is the inductance in henries (H)

The unit for inductive reactance is ohms (\( \Omega \)).
Inductive reactance increases linearly with frequency; the higher the frequency, the greater the opposition to the AC current.
In the context of resonance, at the resonance angular frequency, the inductive reactance becomes exactly equal to the capacitive reactance, allowing for minimal impedance overall in the circuit.

Capacitive Reactance

Similar to inductive reactance, capacitive reactance quantifies how much a capacitor resists changes in voltage in an AC circuit. It is represented by \( X_C \) and can be calculated using:

• \( X_C = \frac{1}{\omega C} \)

where:

• \( \omega \) is the angular frequency in rad/s
• \( C \) is the capacitance in farads (F)

Like inductive reactance, capacitive reactance is measured in ohms (\( \Omega \)). However, unlike inductive reactance, capacitive reactance decreases with increasing frequency. This inverse relationship means that, in an AC circuit with higher frequency, a capacitor will let more current through.
At the resonance angular frequency, the capacitive reactance balances out the inductive reactance, thus lowering the circuit's impedance to a minimum.

Alternating Current Circuits

Alternating Current (AC) circuits are those where the current flows back and forth periodically, as opposed to in a single direction like Direct Current (DC) circuits. These circuits are used extensively in power systems due to their efficiency in transmitting electrical power over long distances.
In AC circuits, components like resistors, inductors, and capacitors respond to changes in voltage and current differently compared to DC circuits. Each has its unique effect on the circuit's behavior, especially with respect to voltage and current.

• Resistors: They provide consistent opposition to current, regardless of frequency.
• Inductors: They oppose changes in current and their reactance increases with frequency.
• Capacitors: They oppose changes in voltage and their reactance decreases with frequency.

Understanding these dynamics is crucial when analyzing circuits like the RLC circuit. The interplay between resistance, inductive reactance, and capacitive reactance determines overall circuit behavior, especially around the resonance point.