An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current is 4.86 mA

Question

An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current is 4.86 mA at 0.940 ms after the connection is completed. After a long time, the current is 6.45 mA. What are (a) the resistance R of the inductor and (b) the inductance L of the inductor?

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Eirian 3 years 2021-08-14T19:47:32+00:00 1 Answers 86 views 0

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    2021-08-14T19:48:37+00:00

    Answer:

    (a) The resistance R of the inductor is 2480.62 Ω

    (b) The inductance L of the inductor is 1.67 H

    Explanation:

    Given;

    emf of the battery, V = 16.0 V

    current at 0.940 ms = 4.86 mA

    after a long time, the current becomes 6.45 mA = maximum current

    Part (a) The resistance R of the inductor

    R = \frac{V}{I_{max}} = \frac{16}{6.45*10^{-3}} = 2480.62 \ ohms

    Part (b)  the inductance L of the inductor

    \frac{Rt}{L} = -ln(1-\frac{I}I_{max}})\\\\L = \frac{Rt}{-ln(1-\frac{I}I_{max})}}

    where;

    L is the inductance

    R is the resistance of the inductor

    t is time

    L = \frac{Rt}{-ln(1-\frac{I}I_{max})}} = \frac{2480.62*0.94*10^{-3}}{-ln(1-\frac{4.86}{6.45})} \\\\L =\frac{2.3318}{1.4004} = 1.67 \ H

    Therefore, the inductance is 1.67 H

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