When a resistor is connected across the terminals of an acgenerator (104 V) that

Question

When a resistor is connected across the terminals of an acgenerator (104 V) that has a fixedfrequency, there is a current of 0.505 Ain the resistor. When an inductor is connected across the terminalsof this same generator, there is a current of 0.384 A in the inductor. When both the resistor andthe inductor are connected in series between the terminals of thisgenerator, calculate the following. (a) the impedance of the series combination
1

(b) the phase angle between the current and the voltage of thegenerator
2°

thanks for your help! :)
(a) the impedance of the series combination
1

(b) the phase angle between the current and the voltage of thegenerator
2°

thanks for your help! :)

Explanation / Answer

   a.   Resistance   R   =   V/ IR                                  =   104/ 0.505                                  =   205.94             Inductivereactance   XC   =   V/ IC                                                 =   104/ 0.384                                                 =   270.83          When both inductor andresistor are connected net impedance       Z   =   (R2   +   XL2)             =   (205.942   +   270.832)             =   340.24       b.   Phaseangle         =   tan-1(XL / R)                                       =   tan-1(270.83 / 205.94)                                        =   52.750                                                     =   104/ 0.384                                                 =   270.83          When both inductor andresistor are connected net impedance       Z   =   (R2   +   XL2)             =   (205.942   +   270.832)             =   340.24       b.   Phaseangle         =   tan-1(XL / R)                                       =   tan-1(270.83 / 205.94)                                        =   52.750    

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