Question

In an \(L-R-C\) series circuit, \(L\) = 0.280 H and C = 4.00 \(\mu\)F. The voltage amplitude of the source is 120 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) Approx. 2982 rad/s. (b) Approx. 70.6 ohms. (c) Peak voltages: 1420 V for inductor and capacitor; 120 V for resistor.

Step-by-step solution

Understanding Resonance Angular Frequency

In an L-R-C circuit, the resonance angular frequency is given by \( \omega_0 = \frac{1}{\sqrt{LC}} \). This formula is derived from the condition where the circuit's inductive reactance is equal to its capacitive reactance, minimizing impedance.

Calculate Resonance Angular Frequency

Using the given values, \( L = 0.280 \text{ H} \) and \( C = 4.00 \mu\text{F} = 4.00 \times 10^{-6} \text{ F} \), calculate \( \omega_0 \):\[\omega_0 = \frac{1}{\sqrt{0.280 \times 4.00 \times 10^{-6}}}\]Solving this gives:\[\omega_0 \approx 2982 \text{ rad/s}\]

Calculate Resistance at Resonance

At resonance, the current amplitude \( I \) is given by the voltage amplitude \( V \) divided by the resistance \( R \):\[I = \frac{V}{R} \Rightarrow R = \frac{V}{I}\]Given \( V = 120 \text{ V} \) and \( I = 1.70 \text{ A} \), calculate:\[R = \frac{120}{1.70} \approx 70.6 \text{ ohms}\]

Calculate Peak Voltage Across Components at Resonance

At resonance, the peak voltages across the inductor \( V_L \) and the capacitor \( V_C \) are given by \( V_L = V_C = \omega_0 L I \) where \( I \) is the current at resonance.Substituting the known values:\[V_L = \omega_0 L I = 2982 \times 0.280 \times 1.70 \approx 1420 \text{ V}\]And similarly, \( V_C = V_L \), hence \( V_C \approx 1420 \text{ V} \).The peak voltage across the resistor \( V_R \) is simply \( IR \):\[V_R = 1.70 \times 70.6 \approx 120 \text{ V}\]

Key concepts

Resonance Angular Frequency

In an L-R-C circuit, resonance occurs when the inductive reactance and capacitive reactance are equal, allowing the circuit to oscillate at its natural frequency with minimal impedance. This special frequency is known as the resonance angular frequency, denoted by \( \omega_0 \). The formula for calculating \( \omega_0 \) is \( \omega_0 = \frac{1}{\sqrt{LC}} \), where \( L \) is the inductance and \( C \) is the capacitance. At this frequency, the energy stored in the inductor and capacitor is perfectly exchanged with each cycle, resulting in maximum current flow.

• Minimizes impedance, allowing maximum current flow.
• Expression: \( \omega_0 = \frac{1}{\sqrt{LC}} \)

Using this concept, you can plug in the provided values for \( L = 0.280 \text{ H} \) and \( C = 4.00 \times 10^{-6} \text{ F} \) to calculate the resonance frequency of the circuit as approximately 2982 rad/s.

Impedance

Impedance in an L-R-C circuit refers to the total opposition to the current flow, combining both resistive and reactive elements. Unlike pure resistance, impedance depends on frequency. It reaches its minimum value at the resonance angular frequency, where the circuit's reactance components cancel each other out. At resonance, the impedance is purely resistive and equals the resistance of the circuit.

• Combines effects of resistance, inductance, and capacitance.
• At resonance, impedance equals the circuit's resistance \( R \).

For the given problem, when \( \omega_0 \approx 2982 \text{ rad/s} \), the impedance is minimized, allowing us to use the relation \( R = \frac{V}{I} \) to find the resistance, which is approximately 70.6 ohms.

Inductive Reactance

Inductive reactance, represented by \( X_L \), is the opposition to current flow due to the inductor in an AC circuit. It varies with frequency, proportional to both the inductance \( L \) and the angular frequency \( \omega \). The mathematical expression for inductive reactance is \( X_L = \omega L \).

• Increases with higher frequencies.
• Expression: \( X_L = \omega L \).

At resonance, \( X_L \) and the capacitive reactance \( X_C \) balance out, resulting in their effects canceling. For resonance frequency, \( \omega = \omega_0 \), calculated as approximately 2982 rad/s, the voltage across the inductor can be found using \( V_L = \omega_0 L I \), yielding about 1420 V.

Capacitive Reactance

Capacitive reactance \( X_C \) operates much like inductive reactance but is related to capacitors in AC circuits. It describes how much a capacitor opposes the flow of alternating current, decreasing as the frequency increases. The formula to compute \( X_C \) is \( X_C = \frac{1}{\omega C} \).

• Decreases with higher frequencies.
• Expression: \( X_C = \frac{1}{\omega C} \)

In our L-R-C circuit at resonance, \( X_C \) becomes equal to \( X_L \), nullifying their opposing influences, thereby reducing impedance. At resonance, the voltage across the capacitor is calculated similarly to the inductor using the formula \( V_C = \omega_0 L I \), which also results in approximately 1420 V.