Consider a circuit with a 102.0 resistor, a 0.098 F capacitor, and a 315.0 mH in

Question

Consider a circuit with a 102.0 resistor, a 0.098 F capacitor, and a 315.0 mH inductor connected in series to a voltage source with amplitude 270 V , with the source operated at an angular frequency that is twice the resonant frequency. Part A: Calculate the ratio of the circuit's impedance at 20 to its impedance at resonance, Part B: Calculate the ratio of the maximum current at 20 to the maximum current at resonance. Part C: Calculate the ratio of the maximum voltage across the resistor at 20 to the maximum voltage across the resistor at resonance.

Explanation / Answer

AT resonance Wo^2 = 1/LC

Wo^2 = 1/(0.315 * 0.098e -6)

Wo = 5691.56 rad/s

also at Resonance Impedence Z^2 = R^2

Zo = R

as XL = XC

When angular frequency = 2WO

XL = 2woL = 2* 5691.56 * 0.315 = 3585.68 ohms

XC = 1/(2woC ) = 1/(2 * 5691.56 * 0.098e -6)

Xc = 896.42 ohms

so

Z^2 = (R^2) + (XL-Xc)^2

Z^2 = (102^2) + (3585.68 - 896.42)^2

impedence at 2WO = Z2 = 2691.19 ohms

so ratio of imedence Z/Zo = 2691.19/102 = 26.38

-------------------------------

Current at 2WO = Vmax/Z = 270*1.414/(2691.19) = 0.141 Amps

Curuent at Wo = Vmax/R = 270*1.414/(102 = 3.74 Amps

so ratio of Current I at 2Wo to Wo = 0.141/3.74 = 0.0376

--------------------------------------

V max at 2WO = IZ = 0.141 * 2691.19 = 379.5 Volts

Vmax at W0 = iR = 3.74 * 102 = 381.48

ratio = 379.5/381.48= 1 nearly

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