An RLC circuit is composed of a 4.5 resistor, a 250 mF capacitor and a 2.50 mH i

Question

An RLC circuit is composed of a 4.5 resistor, a 250 mF capacitor and a 2.50 mH inductor. If the RLC circuit is connected to an alternating current (AC) with a peak voltage of 9 V, calculate the peak current at the resonance angular frequency (omega) of 40.0 Hz.

An RLC circuit is composed of a 4.5 resistor, a 250 mF capacitor and a 2.50 mH inductor. If the RLC circuit is connected to an alternating current (AC) with a peak voltage of 9 V, calculate the peak current at the angular frequency (omega) of 4000 Hz. Why is this problem different significantly from the earlier problem?

Explanation / Answer

impedance Z = sqrt(R^2 + (Xl-Xc)^2)

XL = inductive reactance = wL = (2*pi*f*L)

XL = 2*pi*40*2.50*10^-3 = 0.6283 ohms

Xc = capcitive reactance = 1/wc = 1/(2*pi*f*C)

Xc = 1/(2*pi*40*250*10^-3) = 0.0159 ohms

Z = sqrt(4.5^2+(0.6283-0.0159)^2) = 4.54 ohms

peak current Imax = Emax/z = 9/4.54 = 1.98 A <<-----answer

_________________

impedance Z = sqrt(R^2 + (Xl-Xc)^2)

XL = inductive reactance = wL = (2*pi*f*L)

XL = 2*pi*4000*2.50*10^-3 = 62.83 ohms

Xc = capcitive reactance = 1/wc = 1/(2*pi*f*C)

Xc = 1/(2*pi*4000*250*10^-3) = 0.000159 ohms

Z = sqrt(4.5^2+(062.83-0.000159)^2) = 63 ohms

peak current Imax = Emax/z = 9/63 = 0.143 A <<-----answer

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