A series circuit consisting of an inductance `-` free resistance `R=0.16 k Omega` and coil with active resistance is connected to the mains with effective voltabe `V=220 V`. Find the heat power generated in the coil if the effective voltage values across the resistance `R` and the coil are equal to `V_(1)=80V` and `V_(2)=180V` respectively.

Explanation

We have
`V_(1)=( VR)/( sqrt((R+R_(1))^(1)=X_(L)^(2))), V_(2)=(Vsqrt(R_(1)^(2)+X_(L)^(2)))/(sqrt((R+R_(1))^(2)+X_(L)^(2)))`
so `(R+R_(1))^(2)+X_(L)^(2)=((VR)/( V_(1)))^(2), R_(1)^(2)+X_(l)^(2)=((V_(2)R)/(V_(1)))^(2)`
Hence ` R^(2)+2R R_(1)=(R^(2))/(V_(1)^(2))(V^(2)-V_(2)^(2))`
`R_(1)=(R)/(2V_(1)^(2))(V^(2)-V_(2)^(2)-V_(1)^(2))`
Heat generated in the coil `=(V^(2)R_(1))/( (R_(1)+R_(2))^(2)=X_(L)^(2))=(V_(1)^(2))/(R^(2))xxR_(1)=(V_(1)^(2))/(R^(2))xx(R^(2))/(2V_(1)^(2))(V^(2)-V_(1)^(2)-V_(2)^(2))`
`=(V-V_(1)^(2)-V_(2)^(2))/( 2R)=30 W`