A coil and an inductance `-` free resistance `R=25 Omega` are connected in parallel to the `ac1` mains. Find the heat power generated in the coil provided a current `I=0.90 A` is drawn from the mains. The coil and the resistance `R` carry currents `I_(1)=0.50 A` and `I_(2)=0.60 A` respectively.

Explanation

Here, `I_(2)=(V)/(R), V=` effective voltage
`I_(1)=(V)/( sqrt(R^(2)+X_(L)^(2))) `
and `I=(Vsqrt((R+R_(1))^(2)=X_(1)^(2)))/(Rsqrt(R_(1)^(2)+X_(L)^(2)))=(V)/(R_(eff))`
`R_(eff)` is the impedance of the coil `&` the resistance in parallel .
Now `(I^(2)-I_(2)^(2))/(I_(2)^(2))=(R^(2)+2R R_(1))/(R_(1)^(2)+X_(1)^(2))=((I_(1))/( I_(2)))^(2)+(2R R_(1))/( R^(2)+X_(L)^(2))`
Now mean power consumed in the coil
`=I_(1)^(2)R_(1)=(V^(2)R_(1))/(R_(2)+X_(L)^(2))=I_(2)^(2)R.(I^(2)-I_(1)^(2)-I_(2)^(2))/(2I_(2)^(2))=(1)/(2)R(I^(2)-I_(1)^(2)-I_(2)^(2))=2.5W.` ltbr.