Question 2014.S3.39: The load shown in the figure absorbs 4 kW at a power factor ......
The load shown in the figure absorbs 4 kW at a power factor of 0.89 lagging.
Assuming the transformer to be ideal, the value of the reactance X to improve the input power factor to unity is ————
Given that:
Secondary side: V_{2}=110\ {\mathrm{V}},P_{2}=4\,{\mathrm{kW}},\cos{\phi_{2}}=0.89\,
Load power,
\begin{aligned}& P_2=V_2 I_2=\cos \phi_2 \\& I_2=\frac{4 \times 10^3}{110 \times 0.89}=40.86 A \\& I_1=K I_2 ; \text { where } K=\frac{N_2}{N_1}\end{aligned}
I_1=\frac{N_2}{N_1} \times I_2=\frac{1}{2} \times 40.6=20.436 A
To improve power factor to unity, the reactive power
Q={\frac{V_{1}^{2}}{X}}
Therefore,
\begin{aligned}X & =\frac{V_1^2}{Q}=\frac{V_1^2}{V I \sin \phi}=\frac{(220)^2}{220 \times 20.43 \times \sin (27)} \\& =23.72 \Omega \end{aligned}