When an ideal choke is connected to an ac source of 100V and 50Hz, a current of 8 A flows through the circuit, A current of 10 A flows throught the circuit when a pure resistor is connected instead of the choke coil. IF the two are connected in series with an ac supply of 100 V and 40 Hz, then the current in the circuit is

Options

A. `10A`
B. `8A`
C. `5sqrt2A`
D. `10sqrt2A`

Answer

C

Explanation

`X_L=omegaL=100/8`
`thereforeL=100/(8omega)=1/(8pi)``R=100/10=10Omega`
Where R and L are connected in series,
`Z=sqrt((1/(8pi)times2pitimes40)^2+10^2)=10sqrt2`
`thereforeI=V/Z=100/(10sqrt2)=5sqrt2A`