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When the current changes from +2 A to - 2 A in 0.05s, and EMF of 8V is inducted in a coil. The coefficient of self-induction of the coil is

Question

When the current changes from +2 A to -2 A in 0.05s, and EMF of 8V is inducted in a coil. The coefficient of self-induction of the coil is

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Solution

The coefficient of self-induction, also known as inductance (L), can be calculated using Faraday's law of electromagnetic induction. The formula is:

L = - (ΔV / ΔI) * Δt

where:

  • ΔV is the change in voltage (EMF),
  • ΔI is the change in current, and
  • Δt is the change in time.

Given:

  • ΔV = 8 V,
  • ΔI = -2 A - (+2 A) = -4 A (because the current changes from +2 A to -2 A), and
  • Δt = 0.05 s.

Substitute these values into the formula:

L = - (8 V / -4 A) * 0.05 s L = 2 H

So, the coefficient of self-induction of the coil is 2 Henry (H).

This problem has been solved

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