Basic_Electrical_Engineering_4th_edition
.pdf
TRANSFORMER |
205 |
5.3EMF EQUATION OF A TRANSFORMER
Referto Fig. 5.1. Let N1 be the no. ofturns in the primary and N2 turns in the secondary and assume that the secondary is open as shown in Fig. 5.1. Hence no current flows through the secondary of the transformer and under this condition the transformer acts as a reactance coil and the current 10 flowing through the primary will depend upon the supply voltage V1 and the impedance of the primary winding. Assume that the voltage V1 varies harmonically with fre quency[Hz and the core ofthe transformer is unsaturated i.e. flux qi increases linearly with the current. Note that qi is a function of mmf which is the product ofcurrent and no. of turns ofthe coil through which current flows and the resistance ofthe coil is negligible. The current is har monicallyvarying since the impressed voltage is harmonicallyvarying and the transformer acts as the inductive reactance. As the current increases from its initial value of zero, the flux qi increases in direct proportion and since the flux is also varying harmonically, it will induce an emfin the primary winding according to Faraday's law ofinduction and the polarity ofthe emf will be as per Lenz's law in opposition of the impressed voltage V1 . Since by hypothesis the primary has no resistance, the resistance ohmic drop is zero and hence the primary winding is the seat of only two opposing emfs namely the impressed voltage V1 and the counter emf E1 induced by the alternating flux and in accordance with Kirchhoffs voltage law, these two emfs must at every instant be equal and opposite.
Since the impressed voltage is varyingharmonically the counter emfalso must vary har
monically and hence the flux. Suppose flux at any instant is given by |
|
||||||||||
|
|
qi = <l>m sin mt |
|
|
|
|
.. (5. 1) |
||||
where <l>m is the maximum value offluxinwebers (Wb), it follows thatthe instantaneous value of |
|||||||||||
the induced emf. in N1 turns ofthe primary is |
|
|
|
|
|||||||
|
|
e |
1 |
= N |
d |
= - |
N1 |
w |
qim cos wt |
|
|
|
|
|
|
1 |
dt<jl |
|
N1 |
|
|
||
|
|
|
|
= 2rrfN1 <l>m sin |
(mt - rr12) |
...(5.2) |
|||||
This means the flux induces a sinusoidal voltage which lags behind the flux itself by. go0 |
|||||||||||
intimephase. |
|
|
|
|
|
|
|
|
|
|
|
The maximum valueofthe induced voltage from equation 5.2 is |
|
||||||||||
El |
m |
ax |
= |
2 |
|
|
|
|
|
...(5.3) |
|
|
|
-rtf<j>mNl |
|
|
|
|
|
||||
and hence its effective or rms value is
...(5.4) and this also equals the impressed voltage in magnitude. Fig. 5.3 shows the phasor diagram of the various quantities studied sofar (idealtransformer orlossless transformer).
Since 10 lags the supply voltage V1 by goo, the power supplied from the source is Vi10 cos go0 whichis zero. This result is a natural consequence ofour assumptionthat the transformer is lossless. Therefore, underthese conditionsJ0 = V'!X0 whereX0 is thereactanceofthe transformer.
|
Similarly emfinduced in the secondary winding is |
|
|
e2 = - N2 = 2rrfN2<j>m sin (mt- rr/2) |
...(5.5) |
or |
E2 = 4.44 fN2<1>rn |
...(5.6) |
2 1 2 |
ELECTRICAL ENGINEERING |
have these quantities more or less of the same order as the primary quantities and hence after referringthe secondaryquantitiestoprimaryside, electricalconnectionofthetwowindingsispossible.
Inthe preceding analysis we have seen that VI ::: lE' the difference is very small due to the drop ofpotential through the primaryleakage impedance. Also the exciting current I0 is a small fraction ofthe primaryfullload current, so that/2' (or12 referredto primary) is practicallyequal to II. Therefore, no appreciable errorwillbe introduced by transferringthe shuntbranchin Fig. 5.9 to the one shown in Fig. 5.10whichis the approximate equivalent circuit ofthe transformer.
Fig. 5.1 0. Approximate equivalent circuit of the transformer.
Since 10 is negligibly small as compared toII or12'
II '.:'. 121
and the phasor diagram for the approximate equivalent circuit in Fig. 5.10is shown in Fig. 5.11.
I'2
Fig. 5.1 1 . Phasor diagram (simplified) of approximate equivalent circuit.
In this phasor diagram V2 is taken as the reference phasor 12' lags V2 hy an angle 82, the p.f. angleoftheload. To V2 is added dropJ2'R2' parallel toJ2' andJ2'X2' perpendiculartoJ2'R2'. The phasor sum ofV2, J2'R2' andJ2'X2' givesE2' as shown in Fig. 5.10. We further add dropsJI RI and II XI to E2' to get VI, the supply voltage. The current JI lags VI by an angle 8I as shown in Fig. 5.11. In triangle ABC in Fig. 5.11 has side MD parallel to CB and DKis parallel to AC, hence
Rz' X2'
i.e. the ratio ofprimary resistance to secondary resistance referred to primary equals the ratio of their corresponding reactances. This is very important and proves quite useful when the trans formers are connectedinparallelandshare a load. The secondarycurrentofthe twotransformers will be in phase with each others.
Example 5.3. A singlephase transformer has 180 and 90 turns respectively in its second ary andprimary windings. The respective resistances are 0.233 Qand 0.067ohm. Calculate the
