An ac generator has a frequency of 5.0 kHz and a voltage of 45 V. When an inductor is connected between the terminals of this generator, the

Question

An ac generator has a frequency of 5.0 kHz and a voltage of 45 V. When an inductor is connected between the terminals of this generator, the current in the inductor is 65 mA. What is the inductance of the inductor?

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Khánh Gia 3 months 2021-07-21T09:50:58+00:00 2 Answers 1 views 0

Answers ( )

    0
    2021-07-21T09:52:08+00:00

    Answer:

    Explanation:

    Given that,

    Frequency f = 5 kHz = 5000Hz

    Voltage V=45V

    Current In inductor I = 65mA

    I = 65 × 10^-3 = 0.065A

    We want to find inductance L

    We know that the reactive inductance cam be given as

    XL = 2πFL

    Where

    XL is reactive inductance

    F is frequency

    L is inductance

    Then,

    L = XL/2πF

    From ohms law

    V = IR

    We can calculate the receive reactance of the inductor

    V = I•XL

    Then, XL = V/I

    XL = 45/0.065

    XL = 692.31 ohms

    Then,

    L = XL/2πF

    L = 692.31/(2π×5000)

    L = 0.02204 H

    Then, L = 22.04 mH

    The inductance of the inductor is 22.04mH

    0
    2021-07-21T09:52:35+00:00

    Answer:

    0.022 H

    Explanation:

    From Alternating current,

    V = (XL)I………………… Equation 1

    Where V = Voltage of the generator, I = current in the inductor, XL = Inductive reactance of the generator.

    Make XL the subject of the equation

    XL = V/I…………….. Equation 2

    Given: V = 45 V, I = 65 mA = 0.065 A

    Substitute into equation 2

    XL = 45/0.065

    XL = 692.31 Ω

    But,

    XL = 2πFL…………………… Equation 3

    Where F = frequency of the generator, L = Inductance of the inductor

    Make L the subject of the equation

    L = XL/(2πF)……………… Equation 4

    Given: F = 5.0 kHz = 5000 Hz, π = 3.14, XL =  692.31 Ω

    L = 692.31/(2×3.14×5000)

    L = 0.022 H

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