An inductor is connected to the terminals of a battery that has an emf of 12.0 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 12.0 V and negligible internal resistance. The current is 4.86 mA at 0.700 ms after the connection is completed. After a long time the current is 6.80 mA.
What are
(a) the resistance R of the inductor and
(b) the inductance L of the inductor?

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Thu Thủy 5 years 2021-08-13T12:57:21+00:00 1 Answers 73 views 1

Answers ( )

  1. RuslanHeatt
    1
    2021-08-13T12:59:02+00:00
    Reply

    Answer:

    a) 1764.71 ohms

    b) 1.73 H

    Explanation:

    From the question, we can identify the following parameters;

    Vo =12 V , i = 4.86 mA, t =0.700 ms, io =6.80 mA

    (a) Indcued emf V = L di/dt =0

    From ohms law Vo = ioR

    R = 12/6.80*0.001

    R=1764.71 ohms

    (b) For LR circuit

    i =io (1-e^-t/T)

    Time constant T = L/R

    4.86 = 6.80 (1-e^-0.7*10^-3/T)

    divide both side by 6.8

    0.715 = 0.0007/T

    L/R = 0.0007/0.715

    L/R = 0.000979020979

    Substitute R from above

    L = 0.000979020979 * 1764.71

    L =1.73 H

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