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In an L-R-C circuit, resonance frequency is a critical concept. At this point, the circuit oscillates most naturally. It means the reactive effects of the inductor and capacitor perfectly cancel out, minimizing their impedance influences. The resonance frequency \( \omega_0 \) is crucial for various applications, including filters and tuners.To find it, we use \( \omega_0 = \frac{1}{\sqrt{LC}} \). This formula highlights how both inductance \( L \) and capacitance \( C \) influence the frequency. For this exercise, substituting \( L = 0.400 \) H and \( C = 6.00 \times 10^{-8} \) F, we calculate:

• \( \omega_0 \approx 2\times 10^4 \) radians/sec

At this frequency, the circuit naturally resonates, causing interesting and useful behaviors like zero net reactance.

Impedance is the opposition a circuit presents to the flow of alternating current. It's a combination of resistance and the frequency-dependent reactance from capacitors and inductors. Represented as \( Z \), impedance impacts how circuits respond to AC signals, and changes with frequency.In resonance, impedance \( Z \) simplifies to just the resistance \( R \) because inductive and capacitive reactances cancel out. Thus, in our exercise, \( Z = R = 300 \) Ω. This property makes resonant circuits so valuable for minimizing impedance, allowing maximum current flow.

Ohm's Law is fundamental in electrical circuits, connecting voltage, current, and resistance (or impedance for AC circuits). The law states \( V = IR \), where \( V \) is voltage, \( I \) is current, and \( R \) is resistance (or \( Z \) for impedance).In our scenario, at resonance, the circuit’s total impedance equals its resistance. So, Ohm's Law helps determine voltage from current in a straightforward manner:

• Voltage amplitude of the source: \( V_m = I_m \times R = 0.500 \times 300 = 150 \), yielding a voltage amplitude of 150 V.
• Voltage across each component can also be found using Ohm's Law, adapting to reactance for inductor and capacitor.

Average power in AC circuits is essential because it represents the real power consumed over time. Average power \( P_{\text{avg}} \) for an L-R-C circuit is given by \( P_{\text{avg}} = \frac{1}{2}I_m^2R \), where \( I_m \) is the current amplitude and \( R \) is resistance.For the problem at hand, we calculate:

• \( P_{\text{avg}} = \frac{1}{2} \times (0.500)^2 \times 300 \approx 37.5 \text{ W} \)

This value reflects the effective power, crucial for energy efficiency in electrical applications. It conveys how much of the circuit's power contributes to performing useful work.