A resistor is connected across the terminals of an AC generator,with a voltage o

Question

A resistor is connected across the terminals of an AC generator,with a voltage of 112, that has a fixed frequency, and a current of0.5A in the resistor. When an inductor is connected across theterminals of the generator, there is a current of 0.4A in theinductor . When the resistor and the inductor are connected inseries between the terminals of the generator what is the impedenceor the series combination and what is the phase angle between thecurrent and voltage of the generator.
I don't even know which equation to use, or even where to begin.Please help!

Thanks!

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

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---