If an AC voltage 100 V is applied between points A and B, then current of 1 A and phase difference between current and voltage will be `Deltaphi=37^(@)` . IF the same voltage is applied between points B and C, then current of 5 A and `Deltaphi =53^(@)` What will be the impedance `(in Omega)` in the chain, if the same voltage is applied between points A and C? Round off to nearest integer.

Answer

`0119`

Explanation

`tanphi=(wL_(1))/R_(1)=3/4`
`wL_(1)=3/4R_(1)`
`1=100/sqrt(9/16R_(1)^(2)+R_(1)^(2))`
`R_(1)=80Omega omegaL_(1)=60Omega`
`(wL_(2))/R_(2)=4/3 Rightarrow wL_(2)=4/3R_(2)`
`5=100/sqrt(16/9 R_(2)^(2)+R_(2)^(2)) Rightarrow R_(2)=12 Omega, WL_(2)=16Omega`
`z=sqrt((wL_(1)+wL_(2))^(2)+(R_(1)+R_(2))^(2)) =119 Omega`