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A series circuit consisting of an inductance free resistance R =0.16k Omega and a coil with active resistance is connected to the mains effective voltage V =220V Find the heat power generated in the coil if the effective voltage values across the resistance R an the coil equal to V_(1) =80 V and V_(2) =180V respectively . |
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Answer» Solution :Current in the circuit `V_(R) = i_(RMS)R` `80 = I_(rms) xx 160 implies i_(rms) = (1)/(2)A` `V_(Ro) = i_(rms) R_(0) = (R_(0))/(2)` `V_(L) =i_(rms) X_(L) = (X_(L))/(2)` `(V_(R) + V_(R_(0)))^(2) + V_(L)^(2) = (200)^(2)` `(80 + (R_(0))/(2))^(2) + ((X_(L))/(2))^(2) =(200)^(2) - (1800)^(2)` `(80 + R_(0)) (80) = (400) (40)` POWER consumed in coil ` P = i_(rms)^(2) R_(0) = ((1)/(2))^(2) xx 120 = 30 W`  .
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