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Question
An electrical device draws 2kW power from AC mains (voltage 223V (rms) = `sqrt(50,000)` V). The current differs (lags) in phase by `phi(tan phi = (-3)/4)` as compared to voltage. Find (i) R, (ii) XC – XL, and (iii) IM. Another device has twice the values for R, XC and XL. How are the answers affected?
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Solution
Current lags the voltage so
P = 2000W
V2 = 50,000V XC >XL
`tan phi = (-3)/4`
Im = I0
P = `V^2/Z`
2000 = `(50,000)/Z`
Z = `(50,000)/2000` = 25 Ω
| Z = `sqrt(R^2 + (X_C - X_L)^2)` |
`R^2 + (X_C - X_L)^2` = 252 .....(I)
`R^2 + (X_C - X_L)^2` = 625
`tan phi = (-3)/4`
`(X_C - X_L)/R = (-3)/4`
`X_C - X_L = (-3)/4 R`
`(X_C - X_L)^2 = 9/16 R^2` .....(II)
Put the value of `(X_C - X_L)^2` in (I)
`R^2 + 9/16 R^2` = 625
`25/16 R^2` = 625
R2 = 16 × 25
R = 4 × 5 = 20 Ω
`(X_C - X_L)/R = (-3)/4`
`X_C - X_L = (-3)/4 xx 20`
`X_C - X_L` = – 15 Ω
`I_(rms) sqrt(2)I_(rms) = sqrt(2) xx 9`
I0 = 12.6 A
As (i) R, XC, XL all are double tan `phi` = XL – XC R will remain in same. (ii) Z will become double then I = VZI = VZ become half as value of V does not change. (iii) As I become half P = VI will become again half as voltage remains same.
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