When the transformer is loaded, the current I

_{2}flows through the secondary winding. The magnetic and phase of I_{2}is determined by the load. If load is inductive, I_{2}lags V_{2}. If load is capacitive, I_{2}leads V_{2}while for resistive load, I_{2}is in phase withV_{2}. There exists a secondary m.m.f. N

_{2}I_{2}due to which secondary current sets up its own flux Φ_{2}. This flux opposes the main flux Φ which is produced in the core due to magnetising component of no load current. Hence the m.m.f. is N_{2}I_{2}called demagnetising ampere-turns. This is shown in the Fig.1(a). The flux Φ

This additional current drawn by primary is due to the load hence called load component of primary current denoted as I

_{2 }momentarily reduces the main flux Φ, due to which the primary induced e.m.f. also E_{1}reduces.This additional current drawn by primary is due to the load hence called load component of primary current denoted as I

_{2}**'**as shown in the Fig.1(b).Fig. 1 Transformer on load |

This current I

_{2}**'**is in antiphase with I_{2}. The current sets up its own flux Φ_{2}**'**which opposes the flux Φ_{2}and helps the main fluxΦ. This flux Φ_{2}**'**neutralises the flux Φ_{2}**produced by I**_{2}. The m.m.f. i.e. ampere turns N_{2}I_{2}**'**balances the ampere turns N_{2}I_{2}. Hence the net flux in the core is again maintained at constant level.**Key**

**point**: Thus for any load condition, no load to full load the flux in the core is practically constant.

The load component current I

_{2}**'**always neutralises the changes in the loads. Hence the transformer is called constant flux machine. As the ampere turns are balanced we can write,

N

N

_{2}I_{2}=N_{2 }I_{2}**'****.**I^{.}._{2}**'**=(N_{2}/N_{1}) = K I_{2 }..................(1) Thus when transformer is loaded, the primary current I

_{1 }has two components :1. The no load current I

_{o }which lags V_{1 }by angle Φ_{o}. It has two components I_{m }and I_{c}.2. The load component I

_{2}**'**which in antiphase with I_{2}. And phase of I_{2}is decided by the load. Hence primary current I

_{1 }is vector sum of I_{o }and I_{2}**'**.**.**Ī^{.}._{1}= Ī_{o}+ Ī_{2}...............(2) Assume inductive load, I

_{2}lags E_{2}by Φ_{2}, the phasor diagram is shown in the Fig. 2(a). Assume purely resistive load, I

_{2}in phase with E_{2}, the phasor diagram is shown in the Fig.2(b). Assume capacitive load, I

_{2}leads E_{2}by Φ_{2}, the phasor diagram is shown in the Fig. 2(c). Actually the phase of I

_{2 }is with respect to V_{2 }i.e. angle Φ_{2}is angle between I_{2 }and V_{2}. For the ideal case, E_{2}is assumed equal to V_{2 }neglecting various drops. The current ratio can be verified from this discussion. As the no load current I

I_{o }is very small, neglecting I_{o }we can write,_{1 }

__~__I

_{2}

**'**

_{}

Balancing the ampere turns,

N

N

_{1 }I_{1 }= N_{1 }I_{1 }= N_{2 }I_{2 }**.**N^{.}._{2 }/N_{1 }= I_{1 }/I_{2 }= K Under full load conditions when I

_{o }is very small compared to full load currents, the ratio of primary and secondary current is constant.**Example**: A 400/200 V transformer takes 1 A at a power factor of 0.4 on no load. If the secondary supplies a load current of 50 A at 0.8 lagging power factor, calculate the primary current.

**Solution**: The given values are

I

K = E

The angle of I

Now cos Φ

I

Consider the phasor diagram shown in the Fig. 3. The flux Φ is the reference.

Now cos Φ

Ī

Resolve I

x component of I

y component I

x component of I

y component of I

Ī

Thus the two components of I

This is the primary current magnitude.

While tan Φ

Hence the primary power factor is,

cos Φ

_{o }= 1 A, cos Φ_{o}= 0.4, I_{2}= 50 A and cos Φ_{2}= .08K = E

_{2}/E_{1}= 200/400 = 0.5**.**I^{.}._{2}**'**= K I_{2}= 0.5 x 50 = 25 AThe angle of I

_{2}**'**is to be decided from cos Φ_{2}= 0.8Now cos Φ

_{2}= 0.8**.**Φ^{.}._{2}= 36.86^{o }I

_{2}**'**is antiphase with I_{2}which lags E_{2}**by 36.86**^{o }Consider the phasor diagram shown in the Fig. 3. The flux Φ is the reference.

Now cos Φ

_{o}= 0.4**.**Φ^{.}._{o}= 66.42^{o }Ī

_{1}= Ī_{2}**'**+ Ī_{o }............ vector sumResolve I

_{o }and I_{2}**'**into two components, along reference Φ and in quadrature with Φ in phase with V_{1}.x component of I

_{o }= I_{o }sin Φ_{o}= 0.9165 Ay component I

_{o }= I_{o }cos Φ_{o}= 0.4 A**.**Ī^{.}._{o}= 0.9165 + j 0.4 Ax component of I

_{2}**'**= I_{2}**'**sin Φ_{2}= 25 sin (36.86^{o }) = 15 Ay component of I

_{2}**'**= I_{2}**'**cos Φ_{2}= 25 x 0.8 = 20 A**.**I^{.}._{2}**'**_{}= 15 + j 20 AĪ

_{1}= 0.9165 + j 0.4 + 15 + j 20 = 15.9165 + j 20.4 AThus the two components of I

_{1}are as shown in the Fig.3(c).**.**I^{.}._{1}= √**(**(15.9165)^{2}+ (20.4)^{2}**)**= 25.874 AThis is the primary current magnitude.

While tan Φ

_{1}= 15.9165/20.4**.**Φ^{.}._{1}= 37.96^{o }Hence the primary power factor is,

cos Φ

_{1}= cos (37.96^{o }) = 0.788 lagging**Key point**: Remember that Φ

_{1}is angle between V

_{1}and I

_{1}and as V

_{1}is vertical, Φ

_{1}is measured with respect to V

_{1}. So do not convert rectangular to polar as it gives angle with respect to x-axis and we want it with respect to y-axis.

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