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How does an inductor behave in a dc circuit?

Answer» An inductor, having a pure resistance R and a self-inductance L, has an impedance , `Z=sqrt(R^2+omega^2L^2)`. Here, `omega=2pif` where f=frequency of the electric source. Naturally, f=0 for dc circuits, so, `omega=0` and Z=R
This means that the inductor behaves as a pure resistance in dc-circuits, whereas its inductance L plays no role.


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