An ideal choke coil takes a current fo 8 ampere when connected to an `AC` supply of `100` volt and `50 Hz`. A pure resistor under the same conditions takes a current of 10 ampere. If the two are connected to an `AC` supply of `150` volts and `40 Hz` then the current in a series combination of the above resistor and inductor is

Options

A. `(15)/(sqrt(2))A`
B. 8A
C. 18A
D. 10A

Answer

A

Explanation

`{:(" "i = (V)/(X_(L))," "i = (V)/(R)),(rArr 8 = (100)/(omega L),rArr 19 = (100)/(R)),(rArr L = (100)/( 2 pi xx 50 xx 8),rArr R = 10 Omega ):}`
`rArr L = (1)/( 8 pi ) H`
`X_(L) = 2 pi f. L`
`= 2 pi xx 40 xx (1)/( 8 pi) = 10 Omega `

`Z = sqrt(R^(2) + X_(L)^(2))= sqrt(10^(2) + (10)^(2)) = 10 sqrt(2) Omega `
Current in the circuit ` I = (V)/(Z) =(150)/(10sqrt(2)) = (15)/(sqrt(2)) A `