Question
Asked Aug 4, 2019
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A 45.1 mH inductor is connected as in the figure to an ac generator with εm = 30.0 V. What is the amplitude of the resulting alternating current if the frequency of the emf is (a)1.40 kHz and (b)11.2 kHz?

 

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Expert Answer

Step 1

(a) Write the expression for the amplitude of the resulting alternating current .

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I = current VL I = Vpotential difference across inductance X inductive reactance (1) Х, f frequency х, %3D 2л JL (2) |L inductance V1 (3) use equation (2) and (3) in (1) I = 27 fL (4)

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Step 2

Substitute the given values  in equation (4)  when the frequency of the emf is 1.40kHz

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E 30.0V (30.0V) 2T(1.40 x10 Hz) (45.1x103 H f =1.40kHz L 45.1mH =0.076 A

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Step 3

(b) substitute the given values in equation (4)w...

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(30.0V) I = f=11.2kHz] 27 (11.2x 10 Hz)(45.1x10 H =0.00946 A

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