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The output voltage of an AC generator is given by ? v = 119 V sin (27 ? t ). The

ID: 1286710 • Letter: T

Question

The output voltage of an AC generator is given by ?v = 119 V sin (27?t). The generator is connected across a 0.007 H inductor. Find the following.

(a) frequency of the generator
Hz

(b) rms voltage across the inductor
V

(c) inductive reactance
?

(d) rms current in the inductor
A

(e) maximum current in the inductor
A

(f) average power delivered to the inductor
W

(g) Find an expression for the instantaneous current. (Use the following as necessary: t.)


(h) At what time after t = 0 does the instantaneous current first reach 1.00 A? (Use the inverse sine function.)
s

i =

Explanation / Answer

Given data

The output voltage of an AC generator is

                              V = (119 V)sin (27t)     

The inductance of the inductor, L = 0.007 H    

a)

The equation for the output voltage of an AC generator is given by

                              V = V0sin (t)           

Here, V0   is the maximum voltage of an AC generator and = 2f is the angular frequency.

Comparing the equations V = (119 V)sin (27t)   andV = V0sin (t),

The angular frequency is given by,

             = 27

Therefore, the frequency of the generator is

                       f = /2

                            = 27/2

                            = 13.5 Hz

b)

The maximum voltage, V0 = 119 V

The rms voltage across the inductor is given by,

                    Vrms = V0/2
                            = (119 V)/2

                            = 84.14 V

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c)

The inductive reactance of the inductance is given by,

                  XL = L

                       = (2f)L

                       = (2)(13.5 Hz)(0.007 H)

                       = 0.5937

                       = 0.594

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d)

The rms current in the inductor is given by,

                   Irms = Vrms/XL

                          = ( 84.14 V)/(0.5937 )

                         = 141.7 A

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e)

The maximum current in the inductor is given by,

                  Imax = Irms(2)
                          = (141.7 A)2

                            = 200.39 A

                           = 200A     

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f)

  The average power delivered to the inductor is

                  Pavg = IrmsVrms

                         = (141.7 A)(84.14 V)

                         = 11922.6 W

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g)

The instantaneous current as function of t is given by,

                   i = imaxsin(t-/2)

                      = imaxsin(2ft-/2)

                      = (200.39 A)sin(27t-/2)

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h)

The instantaneous current is i = 1.00 A

                1.00 A = (200.39 A)sin(27t-/2)

                sin(27t-/2) = 1/(200.39 A)

                (27t-/2) = sin-1[1/(200.39 A)]

Solve for t,

                    t = 0.0219 s

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