In a coil with an inductance of 5 mg, the current intensity increases uniformly from 0 to a certain final

| July 23, 2021 | education

In a coil with an inductance of 5 mg, the current intensity increases uniformly from 0 to a certain final value within 0.5 s. In this case, an EMF of self-induction of 0.2 V appears in it. Determine the final value of the current in the coil.

To calculate the final value of the current in the specified coil, we use the formula: εi = L * (Ik – In) / Δt, whence we express: Ik – In = εi * Δt / L and Ik = In + εi * Δt / L.

Variables: In – initial current (In = 0 A); εi – emerged EMF of self-induction (εi = 0.2 V); Δt – elapsed time (Δt = 0.5 s); L is the inductance of the indicated coil (L = 5 mH = 5 * 10-3 H).

Let’s perform the calculation: Iк = Iн + εi * Δt / L = 0 + 0.2 * 0.5 / (5 * 10-3) = 20 A.

Answer: The final value of the current in the indicated coil is 20 A.



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