Current in a circuit falls from 5.0 A to 0.1 A in 0.1 s. If an average e.m.f. of 200 V is induced, give an estimate of the self-inductance of the circuit. (Answer: 4 H) 



Data Given:

Initial current = I₁ = 5.0 A

Final current = I₂ = 0 A

Change in time interval = Δt  = 0.1 s

Average e.m.f = Ɛ = 200 V


To Find:

Self-inductance = L = ?


Solution:

The formula for self-inductance is given by:

L = Ɛ `\frac {Δt}{ΔI}`

where the change in current = ΔI = I₁ - I₂ = 5 A - 0 A = 5 A

by putting the corresponding values

L = 200 V ｘ`\frac {0.1 s}{5 A}`

L = 200 V ｘ0.02 s A⁻¹

L = 4 H -----------------Ans.



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