An ideal choke takes a current of 8A when connectd to an AC source of 100V and 50 Hz. A pure resistor under the same condition strikes a current of 10A. If two are connected in series to an AC supply of 100V and 40Hz, then the current in the series combination of above resistor and inductor `sqrt(10x)`A. Find value of x

Options

A. `3sqrt(2)A`
B. `sqrt(2)A`
C. `5sqrt(2)A`
D. `2sqrt(2)A`

Answer

C

Explanation

`X=(e)/(I)=(100)/(8)=12.5Omega`
Now `Xpropf`
`therefore (X^(2))/(X_(1))=(f_(2))/(f_(1))`
`X_(2)=(40xx12.5)/(50)=10Omega`
Now `Z=sqrt(R^(2)xxX_(2)^(2))=sqrt((10)^(2)+(10)^(2))`
`=10sqrt(2)Omega`
Now `I=(e)/(Z)=(100)/(10sqrt(2))=(10)/(sqrt(2))=(5xx2)/(sqrt(2))`
`=5sqrt(2)A`