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) 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?

Answer 1

Current lags the voltage so 

P = 2000W 

V² = 50,000V  X_C >X_L

`tan phi = (-3)/4`

I_m = 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, X_C, X_L all are double tan `phi` = X_L – X_C 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.