Practice portal › Electromagnetic Induction › Self-Inductance
Asked in GUJCET 2007 · Self-inductance and self-induced EMF
Given: ε = 5 V; the current changes from 3 A to 2 A, so |Δ I| = 1 A; Δ t = 1 ms = 10⁻³ s.
Idea: ε = L (|Δ I|)/(Δ t), so L = (ε Δ t)/(|Δ I|).
L = (5 × 10⁻³)/1 = 5×10⁻³ H.
So the self-inductance is 5 mH.
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