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Q.

An ideal choke takes a current of 8A when connected to an ac supply of 100V and 50Hz. A pure resistor under the same conditions takes a current of 10A. if the two are connected to an ac supply of 150V and 40Hz, then the current in a series combination of the above resistor and inductor is

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a

10A

b

8A

c

18A

d

(15/2)A

answer is D.

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Detailed Solution

For an ideal choke, Iv=EvωL Or 8=100100π×L ∴L=18πH For a pure resistorlv=EvR or 10=100R ∴R=10Ω When connected in series:Iv=EvR2+(ωL)2 =150(10)2+(2π×40×18π)2 =150(10)2+(10)2=152A

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