A coil with inductance `L=0.70 H` and active resistance `r=20 Omega` is connected in series with an inductance `-` free resistance `R`. An alternating voltage with effective value `V=220 V` and frequency `omega=314 s^(-1)` is applied across the terminals of this circuit. At what value of the resistance `R` will the maximum heat power be generated in the circuit ? What is it equal to ?

Explanation

`i_(R) = (V)/(Z) = (V)/([(R + r)^(2) + (omegaL)^(2) ]^(1//2)`
Power in the circuit
` P = i_(rms)^(2) (R +r) = (V^(2) (R +r))/((R + r)^(2) + (omegaL)^(2)`
For P to be maximum
`(dP)/(dR) = ([ R + r)^(2) + (omegaL)^(2) xx 1- (R + r) xx 2(R + r))/([(R + r)^(2) + (omegaL)^(2) ]^(2)) V^(2) = 0`
`(R + r)^(2) + (omegaL)^(2) -2 (R + r)^(2) =0`
`(R + r)^(2) =(omegaL)^(2)`
`R + r = omegaL`
`R =omegaL -r = (22)/(7) xx 100 xx 0.7 -20`
`=220 - 20 =200 Omega`
`P_(max) = (V^(2) (R + r))/((R +r)^(2) + (R + r)^(2)) = (V^(2))/(2(R +r))`
`(220)^(2)/(2(200)) =110W`
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