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Q.

An ideal choke takes a current of 10 A when connected to an ac supply of 125 V and 50 Hz. A pure resistor under the same conditions takes a current of 12.5 A. If the two are connected to an ac supply of 100 V and 40 Hz, then the current in series combination of above resistor and inductor is

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a

10/2A

b

12.5 A

c

20 A

d

10 A

answer is A.

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Detailed Solution

R=12512.5=10ΩXL=ωL=2πfL=VI=12510=12.5∴XL'=2πL×f'=0.25×40=10ΩImpedance of the circuitZ=R2+XL2=102Ω∴ current =100102=10/2A

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