When an R L C series circuit is connectedto a 120 V rms, 61 Hz line, the rmscurr

Question

When an RLC series circuit is connectedto a 120 V rms, 61 Hz line, the rmscurrent in the circuit is 10.8 A and thecurrent leads the line voltage by 45°. (a) Find the average power supplied to thecircuit.
1 kW

(b) What is the resistance in the circuit?
2

(c) If the inductance in the circuit is 46mH, find the capacitance in the circuit.
3 µF

(d) Without changing the inductance, by how much should you changethe capacitance to make the power factor equal to 1?
4 µF

(e) Without changing the capacitance, by how much should you changethe inductance to make the power factor equal to 1?
5 mH (a) Find the average power supplied to thecircuit.
1 kW

(b) What is the resistance in the circuit?
2

(c) If the inductance in the circuit is 46mH, find the capacitance in the circuit.
3 µF

(d) Without changing the inductance, by how much should you changethe capacitance to make the power factor equal to 1?
4 µF

(e) Without changing the capacitance, by how much should you changethe inductance to make the power factor equal to 1?
5 mH

Explanation / Answer

Given rms voltage   Vrms = 120V frequency f = 61 Hz rms current I rms = 10.8A current leads the voltage by    = 45 o 1) Average power supplied to the circuit            P= I rms Vrms   cos                 = 10.8* 120 * cos 45               = 916.4 W impedance in the circuit            Z= V rms   / I rms               = 120 / 10.8               = 11.11 2) Resistance    R =   Z cos                           = 11.11 * cos 45                            = 7.8559 3) impedance Z = R2 + (XL - XC )2    11.11 = R2 + (XL - XC )2         R2 + ( XL - XC)2 = 123.4321                            ( XL - XC )2 =  123.4321 - 7.85592                             ( XL - XC )   = 7.8520                                  XL - 7.8520 = X C                              (2 f L ) - 7.8520 = X C                                   17.621 - 7.8520 = X C                                      9.769   = 1 / 2 f C Capacitance of the capacitor in thecircuit                          C = 1 / ( 2*3.14**61 * 9.769 )                               = 2.672*10-4 F d) power factor = 1          Cos =1             R / Z = 1                R = Z           That means   X L = XC                                   2 f L = 1 / 2 f C                  Capacitance   C = 1 / 42 f2 L                                             = 1 / ( 4 * 3.142 * 612 *46*10-3)                                              = 1.481*10-4 F e)       L   = 1 /42 f 2 C                   = 1 / ( 4 * 3.142 * 612*2.672*10-4 )                    = 25 m H

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