Problem 25 - College Physics

Question

\(\bullet\) In a series \(R-L-C\) circuit, \(L=0.200 \mathrm{H}, C=80.0 \mu \mathrm{F},\) and the voltage amplitude of the source is 240 \(\mathrm{V}\) . (a) What is the resonance angular frequency of the circuit? (b) When the source operates at the resonance angular frequency, the current amplitude in the circuit is 0.600 A. What is the resistance \(R\) of the resistor? (c) At the resonance frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

Step-by-Step Solution

Answer

(a) 250 rad/s; (b) 400 Ω; (c) Inductor: 30 V, Capacitor: 30 V, Resistor: 240 V.

Step 1: Find the Resonance Angular Frequency

For a series \(R-L-C\) circuit, the resonance angular frequency \(\omega_0\) is given by the formula: \[\omega_0 = \frac{1}{\sqrt{LC}}\]Substitute the given values: \(L = 0.200 \mathrm{H}\) and \(C = 80.0 \mu \mathrm{F} = 80.0 \times 10^{-6} \mathrm{F}\) into the equation:\[\omega_0 = \frac{1}{\sqrt{0.200 \times 80.0 \times 10^{-6}}} = \frac{1}{\sqrt{16.0 \times 10^{-6}}}\]\[\omega_0 = \frac{1}{4.0 \times 10^{-3}} = 250 \text{ rad/s}\]The resonance angular frequency is \(250 \text{ rad/s}\).

Step 2: Calculate the Resistance R

At resonance, the impedance of the circuit is purely resistive, so \(R = \frac{V}{I}\), where \(V\) is the voltage amplitude and \(I\) is the current amplitude.The voltage amplitude \(V = 240 \text{ V}\) and the current amplitude \(I = 0.600 \text{ A}\).Substitute these values into the formula:\[R = \frac{240}{0.600} = 400 \Omega\]The resistance \(R\) is \(400 \Omega\).

Step 3: Determine the Peak Voltage Across the Inductor

At resonance, the peak voltage across the inductor \(V_L\) is given by:\[V_L = I \cdot \omega_0 \cdot L\]Substitute the known values \(I = 0.600 \text{ A}\), \(\omega_0 = 250 \text{ rad/s}\), and \(L = 0.200 \text{ H}\):\[V_L = 0.600 \times 250 \times 0.200 = 30 \text{ V}\]The peak voltage across the inductor is \(30 \text{ V}\).

Step 4: Determine the Peak Voltage Across the Capacitor

The peak voltage across the capacitor \(V_C\) is given by:\[V_C = \frac{I}{\omega_0 C}\]Substitute the known values \(I = 0.600 \text{ A}\), \(\omega_0 = 250 \text{ rad/s}\), and \(C = 80.0 \times 10^{-6} \text{ F}\):\[V_C = \frac{0.600}{250 \times 80.0 \times 10^{-6}} = \frac{0.600}{0.02} = 30 \text{ V}\]The peak voltage across the capacitor is \(30 \text{ V}\).

Step 5: Determine the Peak Voltage Across the Resistor

The peak voltage across the resistor \(V_R\) is simply the current amplitude \(I\) times the resistance \(R\):\[V_R = I \times R\]Substitute the known values, \(I = 0.600 \text{ A}\) and \(R = 400 \Omega\):\[V_R = 0.600 \times 400 = 240 \text{ V}\]The peak voltage across the resistor is \(240 \text{ V}\).

Key Concepts

Resonance Angular Frequency

In a series RLC circuit, the concept of resonance angular frequency is worth grasping. Resonance occurs when the inductive and capacitive reactances are equal in magnitude but opposite in phase. This balancing act makes the circuit's impedance purely resistive, and the circuit resonates at its natural frequency, known as the resonance angular frequency.

The resonance angular frequency \( \omega_0 \) is mathematically expressed by the formula:

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

where:

• \( L \) is the inductance in henrys
• \( C \) is the capacitance in farads

This formula implies that the resonance angular frequency depends inversely on the square root of the product of the inductance and capacitance.

In our example, substituting the given values \(L = 0.200 \, \mathrm{H}\) and \(C = 80.0 \, \mu\mathrm{F} = 80.0 \times 10^{-6} \, \mathrm{F}\), we find \( \omega_0 = 250 \, \mathrm{rad/s} \). This means that the circuit reaches resonance at an angular frequency of 250 radians per second, thus minimizing the impedance and increasing the current flow.

Peak Voltage Calculations

In an RLC circuit at resonance, determining peak voltages involves understanding how voltage divides among different components. Let us break down how to calculate them for the inductor, capacitor, and resistor.

**Inductor (\(V_L\))**: The peak voltage across the inductor is calculated with \(V_L = I \cdot \omega_0 \cdot L\).
Given the current amplitude \(I = 0.600 \mathrm{A}\) and the resonance angular frequency \(\omega_0 = 250 \, \mathrm{rad/s}\), substituting into the formula yields \(V_L = 30 \, \text{V}\).

**Capacitor (\(V_C\))**: For the capacitor, the peak voltage is calculated by \(V_C = \frac{I}{\omega_0 C}\).
Again using the known values, this results in \(V_C = 30 \, \text{V}\).

**Resistor (\(V_R\))**: The peak voltage across the resistor is straightforward: \(V_R = I \times R\).
With a resistance \(R = 400 \, \Omega\), the voltage across the resistor calculates to \(V_R = 240 \, \text{V}\).

Together, these calculations demonstrate the energy distribution across each circuit component at the point of resonance.

RLC Circuit Analysis

Analyzing a series RLC circuit involves gathering insights into how components interact, especially at resonance. At resonance, the inductive and capacitive reactances cancel each other, leading to purely resistive impedance. This condition maximizes the current for a given voltage source.

When the circuit operates at its resonance angular frequency, as we have calculated \(\omega_0 = 250 \, \mathrm{rad/s}\), the impedance is significantly minimized. This is because the reactive components (inductor and capacitor) counteract each other, leaving only the resistance to oppose current flow.

Using Ohm's Law \(R = \frac{V}{I}\), where \(V = 240 \, \text{V}\) and \(I = 0.600 \, \text{A}\), we calculated the resistance to be \(400 \, \Omega\). This value of resistance at resonance highlights the role of resistance in controlling circuit current.

Overall, understanding the behavior of RLC circuits at resonance is crucial for efficient circuit design and operation, as it allows for maximum current with minimal impedance.