Problem 63 - University Physics with Modern Physics

Question

In an \(L-R-C\) series circuit, the source has a voltage amplitude of \(120 \mathrm{V}, R=80.0 \Omega,\) and the reactance of the capacitor is 480\(\Omega .\) The voltage amplitude across the capacitor is 360 \(\mathrm{V}\) . (a) What is the current amplitude in the circuit? (b) What is the impedance? (c) What two values can the reactance of the inductor have? (d) For which of the two values found in part (c) is the angular frequency less than the resonance angular frequency? Explain.

Step-by-Step Solution

Answer

(a) 0.75 A; (b) 160 Ω; (c) 618.56 Ω, 341.44 Ω; (d) 341.44 Ω yields a smaller angular frequency.

Step 1: Calculate Current Amplitude

Given the voltage across the capacitor (\( V_C = 360 \, \text{V} \)) and the capacitive reactance (\( X_C = 480 \, \Omega \)), use the relation \( V_C = I X_C \) to solve for the current amplitude (\( I \)):\[ I = \frac{V_C}{X_C} = \frac{360 \, \text{V}}{480 \, \Omega} = 0.75 \, \text{A} \]

Step 2: Calculate the Impedance

The impedance (\( Z \)) of an L-R-C circuit is calculated using \( Z = \sqrt{R^2 + (X_L - X_C)^2} \), where \( X_L \) is the inductive reactance. First, solve for \( Z \) using the current amplitude (\( I \)) formula \( V = IZ \) where \( V \) is the voltage amplitude:\[ 120 = 0.75 Z \]\[ Z = \frac{120}{0.75} = 160 \, \Omega \]

Step 3: Find Reactance of the Inductor

To find \( X_L \), use \( Z = \sqrt{R^2 + (X_L - X_C)^2} \) as follows:\[ 160 = \sqrt{80^2 + (X_L - 480)^2} \]Square both sides and solve for \( X_L \):\[ 160^2 = 80^2 + (X_L - 480)^2 \]\[ 25600 = 6400 + (X_L - 480)^2 \]\[ 19200 = (X_L - 480)^2 \]\[ \sqrt{19200} = |X_L - 480| \]\[ \approx 138.56 = |X_L - 480| \]Thus, the two possible values are \( X_L = 480 + 138.56 = 618.56 \, \Omega \) and \( X_L = 480 - 138.56 = 341.44 \, \Omega \).

Step 4: Determine Angular Frequency's Relationship to Resonance

Resonance occurs when \( X_L = X_C \). For the calculated values, resonance would be closer to \( 480 \, \Omega \).The resonance angular frequency (\( \omega_0 \)) is \( \omega_0 = \frac{1}{\sqrt{LC}} \). A smaller \( X_L \) compared to \( X_C \) implies a smaller angular frequency according to \( X_L = \omega L \).Since \( 341.44 \, \Omega < 480 \, \Omega \), it indicates that the angular frequency is lower for \( X_L = 341.44 \, \Omega \).

Key Concepts

Impedance

In an L-R-C series circuit, impedance is a crucial factor that represents the circuit’s opposition to the flow of alternating current (AC). Unlike simple resistance, impedance combines resistance (R) and reactance (X) in a complex manner. It is expressed as a single value and calculated using the relation:

• \( Z = \sqrt{R^2 + (X_L - X_C)^2} \)

Where:

• \( R \) is the resistance
• \( X_L \) is the inductive reactance
• \( X_C \) is the capacitive reactance

The impedance is given in ohms (\( \Omega \)) and affects how easily current flows through the circuit.
In the example provided, the circuit's impedance was calculated to be \( 160 \, \Omega \). It's important to note that this value is necessary for determining other parameters such as current amplitude in the circuit.

Current Amplitude

Current amplitude in an L-R-C series circuit relates to the maximum value of alternating current flowing through it. It is influenced by the total voltage and the impedance of the circuit, given by the formula:

• \( I = \frac{V}{Z} \)

Where:

• \( I \) is the current amplitude
• \( V \) is the total voltage provided by the source
• \( Z \) is the impedance

This is deeply connected to Ohm’s law, extended for AC circuits. For the exercise example, the current amplitude was found using:

• \( I = \frac{360 \, \text{V}}{480 \, \Omega} = 0.75 \, \text{A} \)

Knowing the current amplitude is vital for assessing how the circuit behaves, especially in terms of energy delivery and efficiency.

Inductive Reactance

Inductive reactance is a measure of a coil's opposition to changes in current due to its inductance in AC circuits. It is calculated with:

• \( X_L = \omega L \)

Where:

• \( X_L \) is the inductive reactance
• \( \omega \) is the angular frequency
• \( L \) is the inductance

Inductive reactance causes the current to lag behind the voltage, which can affect the phase difference in the circuit.
In the exercise, potential inductive reactance values were calculated as \( 618.56 \, \Omega \) and \( 341.44 \, \Omega \), showing how changing parameters can shift circuit dynamics.

Resonance Angular Frequency

The resonance angular frequency is the frequency at which an L-R-C series circuit naturally oscillates, reaching a point of minimum impedance and maximum current amplitude. This frequency occurs when:

• \( X_L = X_C \)

And it can be expressed as:

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

Resonance is a special condition where energy exchange between inductive and capacitive elements is maximized without net loss, leading to greater vibrational energy. It's pivotal for applications requiring stable frequencies, such as radio and TV tunings.
When examining the exercise's findings, the angular frequency for \( X_L = 341.44 \, \Omega \) was identified to be below resonance, highlighting the critical role of precise tuning in maintaining desired circuit performance.