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Question:

When current in a coil changes from 5 A to 2 A in 0.1 s, an average of 50 V is produced. The self-inductance of the coil is?

1.67 H

3 H

6 H

0.67 H

Solution:

Inductor opposes change in current, as the current is decreased, the inductor will convert magnetic energy into electrical energy by creating a potential difference. The voltage drop across the inductor is given by

V = -L(ΔI/Δt)

where:

  • V is the induced voltage
  • L is the self-inductance
  • ΔI is the change in current
  • Δt is the change in time

Given:

  • V = 50 V
  • ΔI = 5 A - 2 A = 3 A
  • Δt = 0.1 s

We can rearrange the formula to solve for L:

L = -V(Δt/ΔI)

Substituting the given values:

L = -(50 V)(0.1 s) / (3 A)

L = -5 V⋅s / 3 A

L ≈ -1.67 H

Since inductance is always positive, we take the absolute value:

L ≈ 1.67 H

Therefore, the self-inductance of the coil is approximately 1.67 H.