Problem 43 - University Physics with Modern Physics

Question

An \(L-R-C\) series circuit has \(C=4.80 \mu \mathrm{F}, L=0.520 \mathrm{H}\) and source voltage amplitude \(V=56.0 \mathrm{V}\) . The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude \(80.0 \mathrm{V},\) what is the value of \(R\) for the resistor in the circuit?

Step-by-Step Solution

Answer

The resistance \( R \) is approximately 71.5 ohms.

Step 1: Identify the Formula to Use

At resonance, the impedance of the circuit is minimized and equals the resistance, so the formula to use is the impedance calculation: \[ Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \]. However, at resonance, \(Z=R\). The current \(I\) at resonance is \(I = \frac{V}{R}\).

Step 2: Determine the Resonant Frequency

The resonant frequency \( \omega \) is the frequency at which the circuit's inductive reactance equals its capacitive reactance. It can be found using the formula:\[ \omega = \frac{1}{\sqrt{LC}} \].Plug in the values for \(L\) and \(C\):\( \omega = \frac{1}{\sqrt{0.520 \times 4.80 \times 10^{-6}}} \approx 2,040 \text{ rad/s} \).

Step 3: Calculate Current in the Circuit

Use the given voltage across the capacitor to find the current. The voltage across the capacitor at resonance is given by \(V_C = \frac{I}{\omega C}\), where \(V_C = 80.0 \text{ V}\).Solve for \(I\):\[ I = V_C \cdot \omega C = 80.0 \cdot 2,040 \times 4.80 \times 10^{-6} \approx 0.783 \text{ A} \].

Step 4: Find the Resistance \(R\)

Since current at resonance is \( I = \frac{V}{R} \), we can rearrange this to find \( R \):\[ R = \frac{V}{I} = \frac{56.0}{0.783} \approx 71.5 \text{ ohms} \].

Key Concepts

Resonance Frequency

The resonance frequency in an L-R-C circuit is a fundamental concept that relates to the frequency at which the circuit naturally oscillates. This frequency is significant because it is where the inductive reactance (\( \omega L \)) and capacitive reactance (\( \frac{1}{\omega C} \)) are equal, effectively canceling each other out. At this point, the circuit's impedance is purely resistive, meaning the circuit operates at maximum efficiency.
To calculate the resonance frequency, use the formula:

• \( \omega = \frac{1}{\sqrt{LC}} \)

By substituting the inductance \(L = 0.520 \text{ H}\) and capacitance \(C = 4.80 \mu\text{F} = 4.80 \times 10^{-6} \text{ F}\), you find that:
\[ \omega = \frac{1}{\sqrt{0.520 \times 4.80 \times 10^{-6}}} \approx 2,040 \text{ rad/s} \].
Understanding this frequency is key to controlling how the circuit responds to alternating current and achieving optimal performance.

Impedance

In the context of L-R-C circuits, impedance represents the total opposition a circuit presents to the flow of alternating current (AC). Unlike resistance, which only considers resistive elements, impedance also accounts for the effects of capacitors and inductors within the circuit.
At resonance, impedance and resistance become equal, as the effects of inductive and capacitive reactances cancel each other out. The formula to determine impedance in an L-R-C circuit is:

• \( Z = \sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2} \)

At resonance, this simplifies to \(Z = R\), as the reactance terms balance out. Understanding impedance helps in predicting how the circuit will respond at different frequencies.
This insight is crucial for designing circuits with specific AC characteristics.

Current Calculation

Calculating the current in an L-R-C circuit, especially at resonance, involves understanding the relationship between the voltage across a component and the overall circuit impedance. At resonance, the formula used is \( I = \frac{V}{R} \), where \(I\) is the current, \(V\) is the total voltage supplied, and \(R\) is the resistance.
In this problem, we also use the given voltage across the capacitor to determine the current. The formula used here is:

• \( V_C = \frac{I}{\omega C} \)

Given the capacitor voltage \(V_C = 80.0 \text{ V}\), we can calculate current \(I\) as:
\[ I = V_C \cdot \omega C = 80.0 \cdot 2,040 \times 4.80 \times 10^{-6} \approx 0.783 \text{ A}\].
With this current, understanding how changes in voltage or circuit parameters affect the current is easier, allowing better control and optimization of circuit performance.

Resistive Circuit Analysis

Resistive circuit analysis focuses on understanding how resistors influence circuit behavior, especially when combined with other components like capacitors and inductors. In an L-R-C circuit operating at resonance, this analysis means primarily focusing on the resistance since the inductive and capacitive reactance cancel out.
The resistance \(R\) in this circuit is found by rearranging the formula for current at resonance, \( I = \frac{V}{R} \), to solve for \(R\):
\[ R = \frac{V}{I} = \frac{56.0}{0.783} \approx 71.5 \text{ ohms}\].
This resistance value is crucial as it directly influences how much current flows through the circuit at its resonance state.

• Analyzing resistive elements in this way can help tailor circuits for desired electrical behavior under specific circumstances.

Gaining insights into resistive circuit analysis allows you to predict and manipulate how the circuit will perform with alternating currents.