In a pure inductive circuit current always lags behind the emf by 2π. If v(t)=v0sinωt then I=I0sin(ωt−2π) Now, given v(t)=100sin(500t) and I0=ωLE0=500×0.02100[∵L=0.02H] I0=10sin(500t−2π) I0=−10cos(500t)
JEE Main 2014 — Physics Electromagnetism
A sinusoidal voltage V(t)=100sin(500t) is applied across a pure inductance of L=0.02H. The current through the coil is:
Held on 12 Apr 2014 · Verified 6 Jul 2026.
10cos(500t)
−10cos(500t)
10sin(500t)
−10sin(500t)
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