Two coils A and B are connected in series across a $240V, 50H_Z$ supply. The resistance of A is 5 Ω and inductance of B is 0.015H. If the input from supply is 3kW and 2Kvar, find inductance of A and resistance of B. Calculate voltage across each coil.
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$P = 3kW \\ Q = 2KVAR \\ \tan^{-1}⁡∅= \dfrac QP= \dfrac 23=0.67 \\ ∅= \tan^{-1}⁡(0.67) \\ = 33.690 \\ P = V. I. \cos⁡∅ \\ 3000 = 240×I × \cos⁡(33.69) \\ I=15.02A \\ Z_T= \dfrac VI= \dfrac{240}{15.02}=15.98Ω \\ \bar{Z}_T ̅ = Z_T ⦟ ∅ \\ = 15.98 ⦟ 33.69 \\ =( 13.3+j 8.86 )Ω \\ V_T= V_A+ V_B=13.3Ω \\ r_B=13.3-5 \\ \boxed{r_B=8.3Ω} \\ V_B=2πfL_B \\ =2π×50×0.015 \\ \boxed{V_B=4.71Ω} \\ X_T= X_A+ X_B \\ X_A=8.86-4.71 \\ X_A=4.15Ω \\ X_A=2πfL_A \\ = 2π×50×L_A \\ \boxed{∴L_A=0.013H} \\ Z_A= \sqrt{r_A^2+X_A^2 }= \sqrt{5^2+4.15^2 }=6.5Ω \\ Z_B= \sqrt{r_B^2+X_B^2}= \sqrt{8.3^2+4.71^2 }=9.54Ω \\ \boxed{V_A= Z_A×I=6.5 ×15.02=97.63V \\ V_B= Z_B×I=9.54 ×15.02=143.29V}$