4. An inductor of 10 mH is connected to an ac source of voltage 200 V and 50Hz. Calculate the inductive reactance and the current in the circuit.
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Answer:
The reactance of the inductor is
XL=ωL=(2π×50s-1)×(200×10-3H)
=62.8Ω
The peak current is
i0=
ε0
XL
=
210V
72.8Ω
=33.A.
As the current lags behind the voltage by
π
2
, the voltuage is zero when the current has it peak value.
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