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 $\mu$F,and R = 10 $\Omega$.Under steady-state conditions,the rms current in the circuit will be A)0.513 A B)0.725 A C)0.919 A D)1.23 A E)1.56 A

Question

Quizplus · Accepted Answer

The angular frequency (ω) of the oscillator is calculated as:
ω = 2πf = 2π×1.0 kHz = 6.28×10^3 rad/s

The reactance of the capacitor (XC) is calculated as:
XC = 1/(2πfC) = 1/(2π×1.0 kHz×4.0 μF) = 3.98×10^3 Ω

The impedance of the circuit (Z) is calculated as:
Z = √(R^2 + (XL - XC)^2) = √(10^2 + (6.28×10^3×5×10^-3 - 3.98×10^3)^2) = 7.39×10^3 Ω

The rms current in the circuit (I) is calculated using Ohm's law:
I = VS/Z = 12/7.39×10^3 = 0.919 A

Therefore, the best choice is C as it gives the correct value of the rms current.

A Series RLC Circuit Is Driven by a 1 μ\muμ F,and R = 10

A Series RLC Circuit Is Driven by a 1 $\mu$ F,and R = 10