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A coil has an inductance of 0.7 H and is joined in series with a resistance of . When an alternating emf of 220 V at 50 cps is applied to it, then the wattless component of the current in the circuit is (take )
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Correct option is B
Wattless component of ac=Ivsinϕ=EvZXLZ=EvXLZ2=220×ωLR2+ω2L2As ωL=0.7×2π×50=220ΩHence, wattless component of ac=200×2202202+2202=0.5 A
Similar Questions
In the circuit shown in figure, R is a pure resistor, L is an inductor of negligible resistance (as compared to R), S is a 100 V 50 Hz ac source of negligible resistance. With either key K1 alone or K2 alone closed, the current is I0. If the source is changed to 100 V, 100 Hz the current with K1 alone closed and with K2 alone closed will be, respectively,
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