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A steady current of 3 A produces a flux of 130 uWb in a coil of 240 turns. What is the inductance (in mH) of the coil? A. 15.

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Answer #1

inductance = N\phi / I

where N = number of turns

\phi = magnetic flux , I = current

L = 240*130*10^-6 / 3

= 0.0104 H

= 10.4 mH

so the inductance is 10.4 mH

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