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JEE Main 2015 · 10 Apr · Shift 1, Physics question 21

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

Choose one option
Answer and worked solution

Official answer: B

L=εΔI/Δt=503/0.1=1.67L=\dfrac{\varepsilon}{\Delta I/\Delta t}=\dfrac{50}{3/0.1}=1.67 H.

Topics:Self inductance
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