An average induced e.m.f of 0.4V appears in a coil when the current in it is changed from 10 A in one direction to 10 A in opposite direction in 0.4 second.find the coefficient of self induction of the coil.
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The coefficient of self induction of the coil is 8×10^−3 H.
Explanation:
Given data:
e = 0.4 V
dI = 10 − (−10) = 20 A
dt = 0.40 s
L=?
Solution:
e = LdI / dt
L = edt / dt = 0.4×0.4 / 20
L = 8×10^−3 H
Hence the coefficient of self induction of the coil is 8×10^−3 H.
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