A constant potential difference = 10V is applied to the coil with an inductance of 0.25 Hn.

A constant potential difference = 10V is applied to the coil with an inductance of 0.25 Hn. How much the current in the coil will increase in t = 1s. Ignore the coil resistance.

The potential difference will be equal to the voltage and EMF of self-induction and is calculated by the formula:

φ1 – φ2 = U = Еі = L * ΔI / Δt, where φ1 – φ2 is the given potential difference (φ1 – φ2 = 10 V), L is the inductance of the coil (L = 0.25 H), ΔI is the required increase in current (A), Δt – time interval (Δt = 1 s).

Let us express from the formula the increase in the current in the coil:

ΔI = φ1 – φ2 * Δt / L.

Let’s do the calculation:

ΔI = φ1 – φ2 * Δt / L = 10 * 1 / 0.25 = 40 A.

Answer: The current in the coil will increase by 40 A.