When a resistor is connected across the terminals of an acgenerator (114 V) that

Question

When a resistor is connected across the terminals of an acgenerator (114 V) that has a fixed frequency, there is a current of0.440 A in the resistor. When an inductor is connected across theterminals of the same generator, there is a current of 0.691 A inthe inductor. When both the resistor and the inductor are connectedin series between the terminals of this generator, what is(a) the impedance of the series combination and(b) the phase angle between the current and thevoltage of the generator? Note: The ac current and voltage arerms values and power is an average value unless indicatedotherwise.

Explanation / Answer

   Resistance   R   =   V/ IR                            =   114/ 0.440                            =   259.09       Inductivereactance   XL   =   V/ IL                                           =   114/ 0.691                                           =   164.98       a.   Impedance   Z   =   (R2   +   XL2)                                     =   (259.092   +   164.982)                                     =   307.16                                              =   114/ 0.691                                           =   164.98       a.   Impedance   Z   =   (R2   +   XL2)                                     =   (259.092   +   164.982)                                     =   307.16       b.   phaseangle       =  cos-1(R / Z)             =   cos-1(259.09 / 307.16)                =   32.490          i.e. thevoltage leads the current by 32.490.          i.e. thevoltage leads the current by 32.490.

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