Question

In an \(L-R-C\) series circuit, \(L=0.280 \mathrm{H}\) and \(C=\) \(4.00 \mu \mathrm{F}\). The voltage amplitude of the source is \(120 \mathrm{~V}\). (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 1.70 A. What is the resistance \(R\) of the resistor? (c) At the resonance angular frequency, what are the peak voltages across the inductor, the capacitor, and the resistor?

Short answer

(a) \(2998 \; \mathrm{rad/s}\); (b) \(70.59 \; \Omega\); (c) \(V_L = 1427 \; \mathrm{V}, V_C = 1427 \; \mathrm{V}, V_R = 120 \; \mathrm{V}\).

Step-by-step solution

Identify Known Variables

We are given that the inductance \( L = 0.280 \; \mathrm{H} \), the capacitance \( C = 4.00 \; \mu \mathrm{F} = 4.00 \times 10^{-6} \; \mathrm{F} \), and the voltage amplitude \( V_0 = 120 \; \mathrm{V} \). Furthermore, the current amplitude at resonance is \( I_0 = 1.70 \; \mathrm{A} \).

Calculate Resonance Angular Frequency

Use the formula for the resonance angular frequency: \( \omega_0 = \frac{1}{\sqrt{LC}} \).Substitute \( L = 0.280 \; \mathrm{H} \) and \( C = 4.00 \times 10^{-6} \; \mathrm{F} \) into the formula:\[\omega_0 = \frac{1}{\sqrt{0.280 \times 4.00 \times 10^{-6}}} = \frac{1}{\sqrt{1.12 \times 10^{-6}}} \approx 2998 \; \mathrm{rad/s}.\]

Determine Resistance at Resonance

At resonance, the impedance is equal to the resistance, \( R \). Use the relation \( V_0 = I_0 R \) to find \( R \):\[R = \frac{V_0}{I_0} = \frac{120}{1.70} \approx 70.59 \; \Omega.\]

Calculate Peak Voltage Across Inductor and Capacitor

At resonance, the peak voltage across the inductor \( V_L \) is given by \( V_L = I_0 \cdot \omega_0 \cdot L \) and across capacitor \( V_C \) by \( V_C = \frac{I_0}{\omega_0 \cdot C} \).For inductor:\[V_L = 1.70 \times 2998 \times 0.280 \approx 1427 \; \mathrm{V}.\]For capacitor:\[V_C = \frac{1.70}{2998 \times 4.00 \times 10^{-6}} \approx 1427 \; \mathrm{V}.\]

Calculate Peak Voltage Across Resistor

At resonance, the peak voltage across the resistor \( V_R \) is simply \( V_R = I_0 \cdot R \):\[V_R = 1.70 \times 70.59 \approx 120 \; \mathrm{V}.\]This matches the source voltage, confirming our calculations.

Key concepts

Resonance Angular Frequency

In an LRC circuit, the resonance angular frequency, represented as \( \omega_0 \), is a crucial concept. It determines the frequency at which the circuit naturally prefers to oscillate. Understanding this enables us to predict the circuit's behavior under various conditions.
At resonance, the impedance of the circuit is minimized, typically resulting in the maximum possible current flow. The formula to calculate the resonance angular frequency is:

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

where \( L \) is the inductance in henries (H) and \( C \) is the capacitance in farads (F).
Using our exercise values, \( L = 0.280 \, \mathrm{H} \) and \( C = 4.00 \times 10^{-6} \, \mathrm{F} \), we find \( \omega_0 \) by substituting the known values into the formula. After calculation, \( \omega_0 \approx 2998 \, \mathrm{rad/s} \).
This frequency is where the inductor and capacitor store and release energy maximally, leading to a significant current with minor resistance influence.

Impedance

Impedance in an LRC circuit, denoted as \( Z \), is a measure of the circuit's opposition to the current when voltage is applied. It is more than simple resistance as it considers the effects of inductance and capacitance.
At resonance, however, the impedance simplifies greatly.

• In a series LRC circuit at resonance, the inductive reactance \( \omega L \) and capacitive reactance \( \frac{1}{\omega C} \) cancel each other out.
• Thus, the impedance \( Z \) equals the resistance \( R \).

This simplification allows us to calculate \( R \) directly from the peak current and voltage values using the equation:
\[ R = \frac{V_0}{I_0} \]
where \( V_0 \) is the source's voltage amplitude and \( I_0 \) is the current amplitude. From the given exercise, using \( V_0 = 120 \, \mathrm{V} \) and \( I_0 = 1.70 \, \mathrm{A} \), we find that \( R \approx 70.59 \, \Omega \). This resistance is crucial as it dictates the circuit's energy loss at resonance.

Peak Voltage Calculation

In an LRC circuit at resonance, understanding the distribution of voltage across each component is vital. The peak voltage across the inductor \( V_L \), the capacitor \( V_C \), and the resistor \( V_R \) can be calculated as follows:
For the inductor:

• \( V_L = I_0 \cdot \omega_0 \cdot L \)

Substituting the known values leads to \( V_L \approx 1427 \, \mathrm{V} \).
For the capacitor:

• \( V_C = \frac{I_0}{\omega_0 \cdot C} \)

This also results in \( V_C \approx 1427 \, \mathrm{V} \).
The equality of \( V_L \) and \( V_C \) at resonance implies energy is perfectly exchanged between the inductor and capacitor.
Finally, for the resistor:

• \( V_R = I_0 \cdot R \)

From calculations, \( V_R \approx 120 \, \mathrm{V} \), matching the source, which confirms our understanding of the voltage distribution at resonance. This helps recognize the separate roles each component plays in the circuit.