The current in an inductive coil increases from zero to 6 A in 0.2 s, due to which an induced emf of 2.4v is generated .calculate the self -inductance of the coil
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E = -Ldi/dt
2.4 = -L ( 0-6)/0.2
L = 2.4*0.2/6
L = 0.08 H
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