A 100 Vrms AC source is connected to a series RLC circuit consisting of a 18 O r

Question

A 100 Vrms AC source is connected to a series RLC circuit consisting of a 18 O resistor, a 10 mH inductor, and a 60 µF capacitor, The voltage source frequency is 158 Hz. The phase angle of the source voltage is 0°.



1) What is the impedance phasor for the resistor?
Magnitude ____ O
Phase angle ____ degrees

2) What is the impedance phasor for the inductor?
Magnitude ____ O
Phase angle ____ degrees

3) What is the impedance phasor for the capacitor?
Magnitude ____ O
Phase angle ____ degrees

4) What is the total impedance phasor?
Magnitude ____ O
Phase angle ____ degrees

5) What is the current phasor?
Magnitude ____ A
Phase angle ____ degrees

6) What is the resistor voltage phasor?
Magnitude ____ V
Phase angle ____ degrees

7) What is the inductor voltage phasor?
Magnitude ____ V
Phase angle ____ degrees

8) What is the capacitor voltage phasor?
Magnitude ____ V
Phase angle ____ degrees

9) What is the power delivered by the voltage source?
____ W

10) What is the power dissipated by the resistor?
____ W

11) What is the resonant frequency of this circuit?
____ Hz

12) What will be the power delivered by the voltage source when the voltage source is operated at the resonant frequency?
____ W

Explanation / Answer

Given capacitance of the capacitor C = 60 F resistance of the resistor   R = 18 inductance of the inductor    L = 1 0mH     frequency of soource f =    158 Hz phase angle of source voltage is   = 0 o   1) impedance phasor for the resistor is   Z = R                                                                     = 18 phase angle = tan -1 ( XL - XC )/ R                           =      0 0    2) impedance phasor for the inductor is                      Z = XL                                = 2 f L                           = 2 ( 158 Hz ) ( 10 *1 0^ -3 H )                            = 9.927    phase angle = tan -1 ( XL - XC )/ R                           =   90 degrees 3)    impedance phasor for the Capaciotr is                      Z = XC                                =1 / 2 f C                           = 1 / 2 ( 158 Hz ) ( 60 *1 0^ -6F )                            = 16.788 phase angle = tan -1 ( XL - XC )/ R                           =  - 90 degrees 4) What is the total impedance phasor                  Z = ( R 2 + ( XL -XC ) 2                        = .............        phase angle = tan -1 ( XL - XC )/ R                                   = tan -1 (    9.927 - 16.788 / 18 )                                  = -20.86 degrees   please post remaining question as other post                      Z = XC                                =1 / 2 f C                           = 1 / 2 ( 158 Hz ) ( 60 *1 0^ -6F )                            = 16.788 phase angle = tan -1 ( XL - XC )/ R                           =  - 90 degrees 4) What is the total impedance phasor                  Z = ( R 2 + ( XL -XC ) 2                        = .............        phase angle = tan -1 ( XL - XC )/ R                                   = tan -1 (    9.927 - 16.788 / 18 )                                  = -20.86 degrees   please post remaining question as other post                       

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