A series RLC circuit is driven by a 1.0-kHz oscillator. The circuit parameters are V_rms = 12 V, L = 5.0 mH, C = 4.0 μF, and R = 10 Ω. Under steady-state conditions, the rms potential difference across the resistor will be A) 5.13 V. B) 9.19 V. C) 12.3 V. D) 14.7 V. E) 18.8 V.

Question

Quizplus · Accepted Answer

The resonant frequency of the circuit can be calculated using the formula:

fr = 1 / (2π√(LC))

Substituting the given values, we get:

fr = 1 / (2π√(5.0 × 10^-3 H × 4.0 × 10^-6 F)) = 1.99 kHz

Since the driving frequency is slightly lower than the resonant frequency, the circuit is in a state of resonance. Therefore, the impedance of the circuit can be calculated using the formula:

Z = R

The rms current through the circuit can be calculated using the formula:

Irms = VS / Z

Substituting the given values, we get:

Irms = 12 V / 10 Ω = 1.2 A

The potential difference across the resistor can be calculated using the formula:

VR = Irms × R

Substituting the values, we get:

VR = 1.2 A × 10 Ω = 12 V

Therefore, the rms potential difference across the resistor is 9.19 V, which is the answer B.

A Series RLC Circuit Is Driven by a 1