An electrical device draws $2 \, kW$ power from $AC$ mains $(V_{rms} = 223 \, V = \sqrt{50000} \, V)$. The current lags in phase by $\tan \phi = -\frac{3}{4}$ compared to the 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?

(A) Given: Power $P = 2 \, kW = 2000 \, W$,$V_{rms} = 223 \, V$,$\tan \phi = -\frac{3}{4}$.
Since the current lags,the circuit is inductive,meaning $X_L > X_C$. Thus,$\tan \phi = \frac{X_L - X_C}{R} = \frac{3}{4}$.
$Z = \frac{V^2}{P} = \frac{223^2}{2000} = \frac{50000}{2000} = 25 \, \Omega$.
Using $Z^2 = R^2 + (X_L - X_C)^2$,we have $25^2 = R^2 + (\frac{3}{4}R)^2 = R^2 + \frac{9}{16}R^2 = \frac{25}{16}R^2$.
$625 = \frac{25}{16}R^2 \Rightarrow R^2 = 400 \Rightarrow R = 20 \, \Omega$.
$X_L - X_C = \frac{3}{4} \times 20 = 15 \, \Omega$. Therefore,$X_C - X_L = -15 \, \Omega$.
$I_M = \frac{V_m}{Z} = \frac{\sqrt{2} \times 223}{25} \approx 12.6 \, A$.
If $R, X_L, X_C$ are doubled,$Z' = \sqrt{(2R)^2 + (2(X_L - X_C))^2} = 2Z = 50 \, \Omega$.
$I' = \frac{V}{Z'} = \frac{I}{2} = 6.3 \, A$. Power $P' = V I' \cos \phi = \frac{P}{2} = 1 \, kW$.