In an L-R-C series circuit, the resistance is 380 Ohms, the inductance is 0.400 H, and the capacitance is 2.00×10−2 uF.

What is the resonant frequency of the circuit? (omega naught)

The capacitor can withstand a peak voltage of 570 volts. If the voltage source operates at the resonance frequency, what maximum voltage amplitude V_max can the source have if the maximum capacitor voltage is not exceeded? Three sig. fig.

Thanks!

To find the resonant frequency (ω₀) of an L-R-C series circuit, we can use the formula:

ω₀ = 1 / √(LC)

Given that the resistance (R) is 380 Ohms, inductance (L) is 0.400 H, and capacitance (C) is 2.00×10⁻² μF, we need to convert the capacitance to Farads (F) since the formula requires the capacitance in Farads.

1 μF = 10⁻⁶ F

So, 2.00×10⁻² μF = 2.00×10⁻² × 10⁻⁶ F = 2.00×10⁻⁸ F

Now, we can substitute the values into the formula for ω₀:

ω₀ = 1 / √(0.400 H × 2.00×10⁻⁸ F)

Now, let's calculate ω₀:

ω₀ = 1 / √(0.400 × 2.00×10⁻⁸)
= 1 / √(8.00×10⁻⁹)
= 1 / (2.83×10⁻⁵)
≈ 3.53×10⁴ rad/s

So, the resonant frequency (ω₀) of the circuit is approximately 3.53×10⁴ rad/s.

Now, let's move on to finding the maximum voltage amplitude (V_max) that the source can have without exceeding the maximum capacitor voltage.

The peak voltage (V_peak) across the capacitor and the voltage amplitude (V_max) are related by:

V_max = V_peak / √2

Given that the capacitor can withstand a peak voltage of 570 volts, we can calculate V_max:

V_max = 570 V / √2
≈ 402 V

Therefore, the maximum voltage amplitude (V_max) that the source can have without exceeding the maximum capacitor voltage is approximately 402 volts (rounded to three significant figures).

I hope this helps! Let me know if you have any further questions.