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20 July, 04:15

If two coils placed next to one another have a mutual inductance of 3.00 mH, what voltage (in V) is induced in one when the 2.50 A current in the other is switched off in 40.0 ms

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  1. 20 July, 04:40
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    -0.1875 V.

    Explanation:

    Using

    E₂ = MdI₁/dt ... Equation 1

    Where E₂ = Voltage induced in the second coil, M = mutual inductance of both coil, dI₁ = change in current in the first coil, dt = change in time.

    Given: M = 3.00 mH = 0.003 H, dI₁ = (0-2.50) = - 2.5 A, dt = 40 ms = 0.04 s.

    Substitute into equation 1

    E₂ = 0.003 (-2.5) / 0.04

    E₂ = - 0.1875 V.

    Hence the induced emf = - 0.1875 V.
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