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

Current in the circuit,
`V_(r) = I_(rms)R`
`rArr 80= I_(rms)xx160 rArr I_(rms)=1/2A`
Now, `V_(R0) = I_(rms)R_(R_(0)) = R_(0)/2`
and `V_(L) = I_(rms)R_(0)=R_(0)/2`
`rArr (V_(R)+V_(R0)^(2)+V_(L)^(2) = (220)^(2)`
`rArr (80+R_(0)/2)^(2) +(X_(L))/2^(2)= (220)^(2)`
Also, `V_(R0)^(2)+V_(L)^(2) = (180)^(2)`
`rArr (R_(0))/(2)^(2)+((X_(L))/(2))^(2)= (180)^(2)`
Subtracting Eqs. (i) and (ii), we get
`(80+R_(0))/(2))^(2)-((R_(0))/(2))^(2) = (220)^(2)-(180)^(2)`
`rArr (80+R_(0))(80) = (400)(40)`
Power consumed in coil,
`P=I_(rms)^(2)R_(0) = (1/2)^(2) xx 120=30W`