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An electrical device draws 2 kW power from AC mains (voltage 223V (rms) V_(rms) = sqrt(50000) V ). The current differs (lags) in phase by (tan phi=-3/4) As compared to voltage. Find (i) R, (ii) X_C - X_L , and (iii) I_M. Another device has twice the values for R, X_C and X_L. How are the answers affected ? |
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Answer» SOLUTION :Given , power drawn P=2 kW=2000 W `tan phi =-3/4 , V_"RMS" =V=223` V `I_M`= maximum current `I_m`=? , R=? , `X_C-X_L` = ? Power `P=V^2/Z RARR Z=V^2/P=(223xx223)/(2xx10^3)` `therefore P=24.86 W approx` 25 W `Z=sqrt(R^2+(X_L-X_C)^2)` `25=sqrt(R^2+(X_L-X_C)^2)` `therefore 625=R^2+(X_L-X_C)^2` but tan `phi=(X_L-X_C)/R` `therefore 3/4 =(X_L-X_C)/R` `therefore X_L-X_C=(3R)/4`...(1) `therefore 625=R^2+((3R)/4)^2` `therefore 625=R^2+(9R^2)/16=(25R^2)/16` `therefore R^2=(625xx16)/25` `therefore R^2=25xx16` (a) `therefore R=20 Omega` (b)`X_L-X_C=(3R)/4 = (3xx20)/4 = 15 Omega` [ `because` equ. (1)] (c) Main current , `I_M=I_m=V_m/Z=(sqrt2V_(rms))/Z=(sqrt2xx223)/25` `therefore I_M`=12.612 A `approx` 12.6 A If R, `X_L, X_C` are all doubled , tan `phi` CHANGE from `tan phi =(X_L-X_C)/R` In `Z=sqrt(R^2+(X_L-X_C)^2), R, X_L, X_C` are doubled , then Z is doubled and `I=V/Z` as Z is doubled current I is halved and in power P = VI, V is constant so power is also halved when Z is doubled. |
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