An RLC series circuit is connected to a 240-V rms powersupply at a frequency of

Question

An RLC series circuit is connected to a 240-V rms powersupply at a frequency of 2.50 kHz. The elements in the circuit havethe following values: R = 38.0, C = 0.26 µF, andL = 39.0 mH. (a) What is the impedance of the circuit?
Enter anumber.

(b) What is the rms current?
Enter anumber. A

(c) What is the phase angle?
Enter anumber. °

(d) Does the current lead or lag the voltage?
---Select---lagsleadsneither lags nor leads

(e) What are the rms voltages across each circuit element?
VR = Enter anumber. V
VC = Enter anumber. V
VL = Enter anumber. V (a) What is the impedance of the circuit?
Enter anumber.

(b) What is the rms current?
Enter anumber. A

(c) What is the phase angle?
Enter anumber. °

(d) Does the current lead or lag the voltage?
---Select---lagsleadsneither lags nor leads

(e) What are the rms voltages across each circuit element?
VR = Enter anumber. V
VC = Enter anumber. V
VL = Enter anumber. V Enter anumber. Enter anumber. Enter anumber. Enter anumber. Enter anumber. Enter anumber.

Explanation / Answer

Given : .              Vrms = 240 V   ; frequency(f) = 2.50 k Hz ; .               R = 38.0 , C =0.26 µF, and L =39.0 mH. . (a)     Impedance is :   Z = R2 + ( XL -XC )2 -------------(1) .             XL   = 2 f L = 2 *3.14 * 2.50 * 103 * 39.0 * 10-3 = 612.3 .             XC   = 1 / 2 f C = 1 /( 2*3.14*2.50 * 103   * 0.26 * 10-6) .                                          =   244.97 .            Hence Z = (38)2 + ( 612.3 - 244.97)2    = 369.3 Ohms . (b) rms current is : .              Irms    = Vrms / Z = 240 / 369.3 =  0.649 A . (c)   Phase angle : .                     =  tan-1 [ (XL - XC) / R] = tan-1 [ (612.3 - 244.97) / 38 ] = --------- . (d)   lags . (e)   VR =  Irms R = 0.649 * 38 = ----------V . (f)    VC = Irms XC = 0.649 *244.97 = --------- V . (g)   VL   = Irms XL = 0.694 *612.3 = ----------V . Solve the above. . Hope this helps u!

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