**EMF Equation of Transformer**

Let

N

_{1 }= Number of turns in primary windings.N

_{2}= Number of turns in second windings._{m }= Maximum flux in the core in Webbers._{m=}B_{m}.A,

f = Frequency of A.C input in H

_{z.}As shown in fig- flux increases from its zero value to maximum value

**Ø**in one quarter of the cycle i.e. in ¼ second._{m}Average rate of change of flux = (Ø

_{m /}¼f.)**= 4f Ø**

_{m}Wb/s or volt.Now rate of change of flux per turn means induced e.m.f in volts.

Average e.m.f /per turn = 4f

**Ø**volt._{m }If flux

**Ø**_{m }varies sinusoidally, then r.m.s value of induced .e.m.f is obtained by multiplying the average value with form factor.**Form factor =**

*r.m.s value / Average value*=1.11r.m.s value of e.m.f/turn = 1.11. 4 f

**Ø**volt_{m= }4.44f Ø_{m }now r.m.s value of the induced e.m.f in the whole primary winding.

=( induced e.m.f/turn) . number of primary turns

E

_{1}= 4.44fN_{1}**Ø**_{m………………………………………(1)}**E**

_{1}= 4.44fN_{1}B_{m}A. (Ø_{m}= B_{m}A)Similarly, r.m.s value of the e.m.f. induced in secondary is,

E

_{2}= 4.44fN_{2}**Ø**_{m}**E**

_{2}= 4.44fN_{2}B_{m}A. (Ø_{m}= B_{m}A)…………..(2)It’s seen from (1) and (2) that E

_{1}/N_{1}=E_{2}/N_{2}= = 4.44f**Ø**It means that e.m.f/ turn is the same is the same in both the primary and secondary windings._{m. }So in ideal Transformer on no-load, V

_{1}=E_{1}and E_{2}=V_{2}where E_{2}is the terminal voltage.By : Engr Wasim Khan

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It is really easy to understand the emf equation of a transformer. It is even easier to understand with the neat and clean diagram. Just like this post <a href="http://www.electricaleasy.com/2014/03/electrical-transformer-basic.html" rel="nofollow"> Electrical Transformer</a>. Thanks for the post. 🙂

How we find numerical question in this site.please tell me