Question

In an \(L-R-C\) series circuit the source is operated at its resonant angular frequency. At this frequency, the reactance X\(_C\) of the capacitor is 200 \(\Omega\) and the voltage amplitude across the capacitor is 600 V. The circuit has \(R\) = 300 \(\Omega\). What is the voltage amplitude of the source?

Short answer

The voltage amplitude of the source is 900 V.

Step-by-step solution

Understanding Resonance in L-R-C Circuits

In an L-R-C series circuit operating at resonance, the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \). Consequently, the impedance \( Z \) of the circuit is minimal and equals only the resistance \( R \). At resonance, the voltage across the entire circuit, \( V_s \), is equal to the current \( I \) multiplied by the resistance \( R \).

Determine Current in the Circuit

The current \( I \) in the circuit can be found from the voltage across the capacitor. Using \( V_C = I \cdot X_C \), where \( V_C = 600 \, V \) and \( X_C = 200 \, \Omega \), we solve for \( I \): \( I = \frac{V_C}{X_C} = \frac{600}{200} = 3 \, A \).

Calculate Voltage Amplitude of the Source

The voltage amplitude of the source \( V_s \) is given by the product of the current \( I \) and the circuit resistance \( R \). Thus, \( V_s = I \cdot R = 3 \, A \times 300 \, \Omega = 900 \, V \).

Key concepts

Resonant Frequency

In an L-R-C series circuit, resonance occurs when the angular frequency of the AC source matches the circuit's natural frequency. At this resonant frequency, a fascinating phenomenon takes place where the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \). This balance effectively cancels out their otherwise opposing effects. As a result, the impedance of the circuit is minimized to just the resistance \( R \) of the resistor.
Resonance is important because it allows the circuit to operate at its most efficient state, with maximum current flow for a given voltage. It's like finding that sweet spot where the circuit can "breathe" easily, with minimal resistance holding it back.

Capacitive Reactance

Capacitive reactance is a measure of how much a capacitor opposes the flow of alternating current (AC) due to its storage of energy in the form of an electric field. It is denoted as \( X_C \), and its value is inversely proportional to the frequency of the AC supply and the capacitance itself.

Formula for capacitive reactance:

• \( X_C = \frac{1}{2\pi f C} \)

Where \(f\) is the frequency and \(C\) is the capacitance, measured in Farads.
In the context of our L-R-C circuit problem, knowing \( X_C \) helps us to determine the current flowing through the circuit, given the voltage across the capacitor. As capacitive reactance decreases with higher frequency, it allows more current to pass, emphasizing the importance of resonance in maximizing current flow.

Impedance Calculation

Impedance is the total opposition a circuit presents to the flow of AC. It is a combination of resistance and reactance and is measured in Ohms (\( \Omega \)). In an L-R-C circuit, the impedance diminishes at resonance because the inductive and capacitive reactances cancel each other out.
Impedance \( Z \) in a non-resonant series L-R-C circuit is calculated using:

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

However, at resonance, since \( X_L = X_C \), the formula simplifies to:

• \( Z = R \)

Understanding impedance is crucial, as it allows us to predict the behavior of the circuit in terms of current and voltage. At resonance, this understanding simplifies the calculation, allowing us to focus on the resistance, thus making it easier to calculate the voltage across the circuit.

Current in AC Circuits

In AC circuits, the current is the flow of electric charge that periodically reverses direction. It is essential to understand how current behaves in an L-R-C circuit, especially under resonance conditions where it reaches its maximum.Current in such a circuit can be determined by using Ohm’s law, where the voltage across a component is equal to the product of current and the component's impedance. In the case of an L-R-C series circuit at resonance, the current \( I \) can be calculated from:

• \( I = \frac{V_C}{X_C} \)

Where \( V_C \) is the voltage across the capacitor and \( X_C \) is the capacitive reactance.

Knowing the current is vital as it helps determine other parameters of the circuit, such as the overall voltage amplitude of the source. This understanding of current flow can also aid in diagnosing and optimizing circuit performance for practical applications.