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. `10` amp
B. `12.5` amp
C. `20` amp
D. `25` amp

Answer

A

Explanation

`R=125/12.5 =10 Omega`
`X_(L)=omegaL=2pinL=V/I=125/10=12.5`
`:. 2pinL=12.5`
or `2piL=12.5/50=0.25`
`:. X_(L)=2piLxxn=0.25xx40=10 Omega`
Impedence of the circuit
`Z=sqrt(R^(2)+X_(L)^(2))=10sqrt(2)` ohm
`:.` Current `=(100sqrt(2))/(10sqrt(2))=10` amp