A solenoidal inductor for a printed circuit board is being redesigned. To save weight, the number of turns is reduced by one-fifth, with the

Question

A solenoidal inductor for a printed circuit board is being redesigned. To save weight, the number of turns is reduced by one-fifth, with the geometric dimensions kept the same. By how much must the current change if the energy stored in the inductor is to remain the same?

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1 week 2021-07-22T20:38:07+00:00 1 Answers 2 views 0

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    2021-07-22T20:39:40+00:00

    Answer:

    The current must be 5 times larger

    Explanation:

    The equation for Inductance, is given by;

    L = (μ_o•N²•A)/l

    Where;

    μ_o is the permeability of the core material

    N is the number of turns

    A is the cross sectional Area

    l is the length of the coil in meters

    L is the inductance

    However, we also need to find an expression for the Energy Stored, and the equation is given by;

    U = (1/2)LI²

    Where;

    L is the inductance

    I is the current.

    We also know that U is proportional to the number of turns and the current.

    Thus, we can conclude that;

    N_2/N_1 = I_1/I_2

    From the question, we are told that the number of turns is reduces by 1/5. Thus,

    1/5 = I_1/I_2

    So, I_2 = 5I_1

    So the current must be 5 times larger

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