Self-inductance 46011

Calculate the self-inductance of the coil, which has 1,300 turns, if the inductive current through the coil changes by 6 * 10 ^ -5 Wb by a constant current change of 2A.

Correct answer:

L =  0.03 mH

Step-by-step explanation:

N=1300 I=2 A Φ=6 102=0.06 mWb  Φ=L I  L=Φ/I=0.06/2=0.03 mH



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