Question 3.2: The BH curve data for the core of the transformer shown in F...

Substituting values in Eq.(3.5)
E_{1} =\sqrt{2} \pi fN_{1} \phi _{max} =4.44fN_{1} \phi _{max}
200=4.44\times 50\times 150\times \phi _{max}
or   \phi _{max}=6.06 mWb
B_{max} =\frac{6.06\times 10^{-3} }{10\times 5\times 10^{-4}}  =1.212 T
From the data of BH curve of Problem 2.10, we get
H_{max} =250 AT/ m
AT_{max} =250l_{c} =250\times 2\left(30+35\right) \times 10^{-2}=325
I_{m\left(max\right) } =\frac{325}{150} =2.17 A
I_{m\left(rms\right) } =\frac{2.17}{\sqrt{2} } =1.53 A
Core volume =2 \left(20\times 10\times 5\right) + 2 \left(45\times 10\times 5\right) =6500 cm^{3}
Weight of core = 6500 \times 7.9\times 10^{-3} =51.35 kg
Core loss = 51.35 \times 3 = 154.7 W
I_{i} =\frac{154.7}{200} = 0.77 A
Referring to the phasor diagram of Fig.3.7
\bar{I_{0} } =0.77-j 2.17 = 2.3∠-70.5°
I_{0}= 2.3 A (no-load current)
No-load          pf = \cos71.5° = 0.334 lagging