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

Options

A. (A) `10//sqrt(2)A`
B. (B) 12.5 A
C. (C) 20A
D. (D) 10A

Answer

A

Explanation

`R=(125)/(12.5) = 10 Omega`
`(X_L)= omega L = 2 pi f L =(V)/(I) = (125)/(10)=12.5`
`:. X'_(L)=2 pi L xx f' = 0.25 xx 40 = 10 Omega`
Impedance of the circuit
`Z=sqrt(R^(2)+X_(L)^(2))=10 sqrt(2) Omega `
`:. Current =(100)/(10 sqrt(2)) = 10 sqrt(2) A`.