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A circuit with current increasing at a rate of 4 A/s contains an inductor, L. If the induced emf is -2 V, what is the inductance of the inductor?

Respuesta :

Answer:

L=500 mH

Explanation:

Here di/dt = 4A/s,    ε= -2V

Inductance of inductor, induced emf and rate of change of current have the following relation.

ε= [tex] - L\frac{di}{dt}[/tex]

⇒L= - ε/[tex]\frac{di}{dt}[/tex]

⇒L= -(-2)/ 4

⇒ L= 0.5 H   or

⇒ L= 500 mH

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