Problem

For a series RLC circuit with a $40.0 \, \Omega$ resistor, a $3.00 \, \mathrm{mH}$ inductor, and a $5.00 \, \mu\mathrm{F}$ capacitor: (a) Calculate the resonant frequency.
(b) Determine the RMS current at resonance if $V_{\mathrm{rms}} = 120 \, \mathrm{V}$.

Data

• Resistance: $R = 40.0 \, \Omega$
• Inductance: $L = 3.00 \, \mathrm{mH} = 3.00 \times 10^{-3} \, \mathrm{H}$
• Capacitance: $C = 5.00 \, \mu\mathrm{F} = 5.00 \times 10^{-6} \, \mathrm{F}$
• RMS Voltage: $V_{\mathrm{rms}} = 120 \, \mathrm{V}$

Prerequisite Concepts

1. Resonant frequency for a series RLC circuit:

$$f_0 = \frac{1}{2\pi \sqrt{LC}}$$

2. At resonance, the circuit behaves as purely resistive, and the impedance $Z = R$. The RMS current is given by:

$$I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{R}$$

Solution

Part (a): Calculate the Resonant Frequency

$$f_0 = \frac{1}{2\pi \sqrt{LC}}$$

Substitute the given values:

$$f_0 = \frac{1}{2 \times 3.14 \times \sqrt{(3.00 \times 10^{-3} \, \mathrm{H}) \times (5.00 \times 10^{-6} \, \mathrm{F})}}$$ $$f_0 = \frac{1}{6.28 \times \sqrt{1.50 \times 10^{-8}}}$$ $$f_0 = \frac{1}{6.28 \times 1.225 \times 10^{-4}}$$ $$f_0 = \frac{1}{7.691 \times 10^{-4}}$$ $$f_0 = 1.30 \times 10^{3} \, \mathrm{Hz}$$

Part (b): Calculate the RMS Current at Resonance

At resonance, the circuit impedance $Z = R$. The RMS current is:

$$I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{R}$$

Substitute the values:

$$I_{\mathrm{rms}} = \frac{120 \, \mathrm{V}}{40.0 \, \Omega}$$ $$I_{\mathrm{rms}} = 3.00 \, \mathrm{A}$$

Answer

• Resonant Frequency: $f_0 = 1.30 \, \mathrm{kHz}$
• RMS Current at Resonance: $I_{\mathrm{rms}} = 3.00 \, \mathrm{A}$

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