When a resistor is connected across the terminals of an ac generator (112V) that

Question

When a resistor is connected across the terminals of an ac generator (112V) that has a fixed frequency, there is a current of .500 A in the resistor. When an inductor is connected across the terminals of the same generator, there is a current of of 0.400 A in the inductor. When both the resistor and inductor are connected in series between the terminals of this generator, what are the (a) impedance of the series combination and (b) the phase angle between the current and the voltage of the generator?

Explanation / Answer

a) When the resistor is connected across the terminals ofthe battery then the current is                          i = /R                         0.500A=112V/R ==============> R = 224 When the inductance of the inductor is connected across theterminals of the battery then the current is                         0.400 =112V/XL ==============>XL =280 Then the resistor and inductor are conncected in series thenthe impedance of the series combination is                        Z =R2 +XL2                           = 358.6 b) The phase angle between the current and the voltage ofthe generator is                       Tan = (XL/R)                              =Tan-1(280/224)                                 =51.340                           = 358.6 b) The phase angle between the current and the voltage ofthe generator is                       Tan = (XL/R)                              =Tan-1(280/224)                                 =51.340

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