Question 3.14: A transformer has 900 turns on its primary winding and 300 t...

Given V_{1} = 230  V.
The induced EMF in the primary windings is E_{1}.  E_{1} is slightly less than V_{1} because there will be some voltage drop in the winding.
V_{1} > E_{1}, V_{1} − E_{1} = Voltage drop in the primary winding due to current, I_{0} flowing through it.
The no-load current, I_{0} is very small as compared to the current that would flow when some electrical load is connected across the secondary winding. Here, the transformer is on no-load, i.e., no load has been connected to its secondary winding.
If we neglect the no-load voltage drop in the winding, we can write V_{1} = E_{1}.

E_{1}= 4.44  Φ_{m}  f  N_{1}

and                                             E_{2}= 4.44  Φ_{m}  f  N_{2}

therefore,                                   \frac{E_{1}}{E_{2}}=\frac{N_{1}}{N_{2}}

or,                                                E_{2} = E_{1}\left( \frac{N_{2}}{N_{1}}\right) =230 \left(\frac{300}{900}\right) = 76.7  V

Again,                                        E_{1}= 4.44  Φ_{m}  f  N_{1}

= 4.44  B_{m}  A  f  N_{1}

when B_{m} is the maximum flux density and A is the cross-sectional area of the core.

Substituting values

230 = 4.44  B_{m}  ×  \frac{64}{10^{4}}  ×  50  ×  900

or,                                               B_{m} = \frac{230  × 10^{4}}{4.44  ×  64  ×  45000}  = \frac{2300}{4.44  ×  64  ×  45}

or,                                               B_{m} = 0.18  Wb/m^{2}