1.

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.

Answer»

SOLUTION :Here `I_(1), = 5.0 A, I_(2) = 0.0A, 1 = 0.1 s and VAREPSILON = 200 V`
as `varepsilon=L(I_(1)-I_(2))/timplies L=(varepsilont)/(I_(1)-I_(2))=(200Vxx0.1s)/(5.0A-0.0A)=4.0H.`


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