In a coil, an increase in current from 5 A to 10 A in 100 ms induces an induced emf of 100 V. The self inductance of the coil is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 2 H
Using Faraday's law of induction,
$ε=-L\frac{ΔI}{Δt}$
$⇒L=\frac{εΔt}{ΔI}=\frac{100V×0.15}{5A}=\frac{10Vs}{5A}=2H$