# When the current strength in the coil changed from 5A to 20A in 0.1 s, an EMF of self-induction

When the current strength in the coil changed from 5A to 20A in 0.1 s, an EMF of self-induction appeared in it, equal to 20V. What is the inductance of the coil?

Task data: In (initial current strength) = 5 A; Ik (final current in the taken coil) = 20 A; Δt (duration of current change) = 0.1 s; ε (emerging EMF of self-induction) = 20 V.
To determine the desired inductance of the coil taken, we use the formula: ε = | L * ΔI / Δt |, whence L = ε * Δt / ΔI = ε * Δt / (Iк – In).
Let’s perform the calculation: L = 20 * 0.1 / (20 – 5) = 0.133 G.
Answer: The required inductance of the taken coil is 0.133 H.

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