Question

Only need g and h answered. I put the other questions/answers in case any of them are needed to answer the last two parts. Thank you for any help in advance!

The output voltage of an AC generator is given by Δv = (1.14 ✕ 10² V)sin(41πt). The generator is connected across a 0.500 H inductor. Find the following.

(a) frequency of the generator
20.5 Hz

(b) rms voltage across the inductor
82.22 V

(c) inductive reactance
64.4 Ω

(d) rms current in the inductor
1.27 A

(e) maximum current in the inductor
1.79 A

(f) average power delivered to the inductor
0 W

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

i

=

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

Answer 1

    The output voltage of an AC generator is
                              ΔV = (116.28 V)sin (41πt)       ...... (1)     
      The inductance of the inductor, L = 0.500 H   
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a)
      The general equation for the output voltage of an AC generator is
                              ΔV = V0sin (ωt)             ...... (2)
      Here, the maximum voltage of an AC generator is V0  
                The angular frequency, ω = 2πf
      Comparing the equations (1) and (2), we get
      The angular frequency, ω = 41π
      Therefore, the frequency of the generator is
                       f = ω/2π
                            = 41π/2π
                            = 20.5 Hz
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b)
      The maximum voltage, V0 = 116.28 V
       The rms voltage across the inductor is
                    Vrms = V0/√2
                            = (116.28 V)/√2
                            = 82.23 V
                            = 82.2 V
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c)
      The inductive reactance is
                  XL = ωL
                       = (2πf)L
                       = (2π)(20.5 Hz)(0.500 H)
                       = 64.4 Ω
                     
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d)
    
      The rms current in the inductor is
                   Irms = Vrms/XL
                          = ( 82.2 V)/(64.4 Ω)
                         = 1.27 A
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e)
      The maximum current in the inductor is
                  Imax = Irms(√2)
                          = (1.27 A)√2
                          
                          = 1.79 A       (nearly)
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f)
      The average power delivered to the inductor is
                  Pavg = IrmsVrms
                         = (1.27 A)(82.2 V)
                         = 104.4 W
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g)
      The instantaneous current is
                   i = imaxsin(ωt-π/2)
                      = imaxsin(2πft-π/2)
                      = (1.79 A)sin(41πt-π/2)
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h)
      The instantaneous current, i = 1.00 A
                1 A = (1.79 A)sin(41πt-π/2)
                sin(41πt-π/2) = 1/(1.79 A)
                (41πt-π/2) = sin-1[1/(1.79 A)]
                         41πt = 33.9+π/2
      Then the required time is
              t = (0.2633 + 0.012) s
                = 0.2753 s