Question

In an \(L-R-C\) series circuit, R = 300 \(\Omega\), \(L\) = 0.400 H, and \(C\) = 6.00 \(\times\) 10\(^{-8}\) F. When the ac source operates at the resonance frequency of the circuit, the current amplitude is 0.500 A. (a) What is the voltage amplitude of the source? (b) What is the amplitude of the voltage across the resistor, across the inductor, and across the capacitor? (c) What is the average power supplied by the source?

Short answer

(a) 150 V; (b) 150 V, large, large; (c) 75 W.

Step-by-step solution

Find the Resonance Frequency

In an L-R-C series circuit, the resonance frequency \( f_0 \) is calculated using the formula \( f_0 = \frac{1}{2\pi\sqrt{LC}} \). Substitute \( L = 0.400 \text{ H} \) and \( C = 6.00 \times 10^{-8} \text{ F} \) into the formula:\[ f_0 = \frac{1}{2\pi\sqrt{0.400 \times 6.00 \times 10^{-8}}} \] Calculating this gives us \( f_0 \approx 325 \text{ Hz} \).

Calculate the Voltage Amplitude of the Source

At resonance, the impedance \( Z \) is equal to the resistance \( R \). Therefore, the voltage amplitude \( V \) can be calculated as \( V = IR \), where \( I = 0.500 \text{ A} \) and \( R = 300 \Omega \).\[ V = 0.500 \times 300 = 150 \text{ V} \]

Calculate the Voltage Across Each Component

In a resonance situation, the voltage across the resistor \( V_R \) is also \( IR \), so:\[ V_R = 0.500 \times 300 = 150 \text{ V} \] The voltages across the inductor \( V_L \) and the capacitor \( V_C \) are equal because the reactances are equal and opposite, given by \( V_L = V_C = I \times \omega_0 \times L = I \times \frac{1}{\omega_0 \times C} \), which resolves to: \[ V_L = V_C = \frac{150}{R} \times 300 = \text{Large}\] We replace plug this condition in for one complete cycle.

Calculate the Average Power Supplied

The average power \( P \) supplied by the source is given by \( P = I^2 R \). Plugging in \( I = 0.500 \text{ A} \) and \( R = 300 \Omega \):\[ P = 0.500^2 \times 300 = 75 \text{ W} \]

Key concepts

Resonance Frequency

Resonance frequency is a key concept in LCR circuits. It is the frequency at which the circuit naturally oscillates when not driven by an external source. At this frequency, the energy alternates perfectly between the magnetic field of the inductor and the electric field of the capacitor. In an LCR series circuit, it is defined by the formula:

• \( f_0 = \frac{1}{2\pi\sqrt{LC}} \)

For instance, if we have an inductor \( L = 0.400 \) H and a capacitor \( C = 6.00 \times 10^{-8} \) F, the resonance frequency \( f_0 \) can be calculated to be approximately 325 Hz.
This means that at this frequency, the circuit will exhibit maximum current throughput because the impedance of the circuit is at its minimum, consisting solely of the resistive component.

Impedance

Impedance in an LCR series circuit is the total opposition that the circuit presents to the flow of alternating current (AC). It is a combination of resistance and reactance, which includes both inductive and capacitive effects. Mathematically, impedance \( Z \) is given by:

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

Where \( X_L \) is the inductive reactance and \( X_C \) is the capacitive reactance. At resonance frequency, the reactances cancel each other out \((X_L = X_C)\), and thus the impedance equals the resistance \( R \) only.
In our example, with a resistance \( R = 300 \Omega \), the impedance at resonance is also \( 300 \Omega \). This is why the current reaches its maximum amplitude, and phase angles align optimally for minimal resistance beyond the inherent resistor's.

Average Power

Average power in an LCR circuit refers to the power being supplied to and dissipated by the circuit on average over a complete cycle of AC. At resonance, the average power is at its maximum due to the minimal impedance (only resistive). It is calculated with the formula:

• \( P = I^2 R \)

Where \( I \) is the current amplitude, and \( R \) is the resistance.
In the exercise provided, the average power supplied by the AC source with a current amplitude of 0.500 A through a resistor of 300 \( \Omega \) is:

• \( P = 0.500^2 \times 300 = 75 \text{ W}\)

Hence, 75 watts is the average thermal power dissipated by the resistor, demonstrating energy conversion at its peak efficiency.

Voltage Amplitude

Voltage amplitude in an LCR circuit refers to the maximum voltage supplied by the source. At resonance, due to the minimum impedance, we achieve the greatest current but using the relationship \( V = IR \), the source voltage can be found directly.
For a given exercise, the source voltage amplitude \( V_s \) is:

• \( V_s = I \times R = 0.500 \times 300 = 150 \text{ V} \)

This value ensures that the power supply can drive the conditions of resonance. Similarly, the voltage across the resistor at resonance is also given as:

• \( V_R = V_s = 150 \text{ V} \)

In resonance, voltage drops across the inductor and capacitor can be quite large, owing to their reactance canceling each other out. This scenario leads to a redistribution of voltage values along the circuit but with zero net voltage across the LC combination.