<p>A series of RLC circuit consists of a 4 &#937; resistor, a 4 mH inductor, and

Question

<p>A series of RLC circuit consists of a 4 &#937; resistor, a 4 mH inductor, and a 1 mF capacitor. It is connected to an ac voltage cource operating at an angular frequency &#969; = 1000 rad/sec with rms voltage &#916;v(rms) = 10V.</p>
<p>&#160;</p>
<p>(a) Calculate the resonance frequency of this circuit. Is the circuit operating above, below, or at resonance?</p>
<p>(b)Draw the impedance triangle and calculate the rms current flowing through the resistor.</p>
<p>(c) Calculate values for the average power dissipated in the resistor, the inductor, and the capacitor.</p>
<p>(d)Calcualte the peak voltages across the resistor, capacitor, and inductor and draw a corresponding phasor diagram.</p>

Explanation / Answer

A series RLC circuit consists of a resistance R = 4.0 Capacitance C = 1.00*10^-3 F , inductance L = 4.00*10^-3 H angular frequency is = 1000rad /s a) frequency of the circuit is f = 1/2 L C                                             = 1/ 2 4.00*10^-3 H   * 1.00*10^-3 F                                              = 79.6 Hz b )       c ) Total average power        P = V^2/ z     z = R^2 +( X_L - X_C)^2         = (R)^2 + ( L - 1/ C)^2         = (4.0)^2 + ( 1000rad /s * 4.00*10^-3 H   - 1/ 1000rad /s * 1.00*10^-3 F )         = 5.0 P = (10.0V)^2 / 5.0       = 20.0W c ) Total average power        P = V^2/ z     z = R^2 +( X_L - X_C)^2         = (R)^2 + ( L - 1/ C)^2         = (4.0)^2 + ( 1000rad /s * 4.00*10^-3 H   - 1/ 1000rad /s * 1.00*10^-3 F )         = 5.0 P = (10.0V)^2 / 5.0       = 20.0W

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