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An electrical device draws 2 kW power fr...

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 ?

Text Solution

Verified by Experts

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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