# Equivalent Circuits of Transformer

# Equivalent Circuit of an Ideal Transformer at No Load

Under certain conditions, the transformer can be treated as an ideal transformer. The assumptions necessary to treat it as an ideal transformer are:

(a) Primary and secondary windings have zero resistance. This means that ohmic loss (*I*2 *R *loss), and resistive voltage drops in windings are zero.

(b) There is no leakage flux, i.e. the entire flux is mutual flux that links both the primary and secondary windings.

(c) Permeability of the core is infinite this means that the magnetizing current needed for establishing the flux is zero.

(d) Core loss (hysteresis as well as eddy current losses) are zero.

We know that:E2/E1=N2/N1=k

(*k *is a constant, known as voltage transformation ratio or turns ratio).

For an ideal transformer, *V*1 = *E*1 and *E*2 = *V*2.

Even at no load, a transformer draws some active power from the source to provide the following losses in the core :

(a) eddy-current loss, and

(b) hysteresis loss.

The current responsible for the active power is nearly in phase with *V*1 (applied voltage) and is known as core-loss current. A transformer when connected to supply, draws a current to produce the flux in the core. At no-load, this flux lags nearly by 90o behind the applied voltage *V*1. The magnetizing current, denoted by *Im *is in phase with the flux f and thus, lags behind the applied voltage by nearly 90o. The phasor sum of the core losscomponent of current *Ic *and the magnetizing current *Im *is equal to the no-loadcurrent *I*0 .

*I *= *I*0 cos f0 and *I *= *I*0 sin f0 *c m*

Core loss = *P*0 = *V*1 *I*0 (cos f0)

where f0 is the phase angle between *V*1 and *I*0, and, (cos f0) is the no load power factor.The phase relationship between applied voltage *V*1, no-load current *I*0, and its components*Ic*, *Im *is shown in Figure 4.3(a).

In the form of equivalent circuit, this can be represented as Figure 4.3(b), in which *Rc *is a resistance representing core loss and *Xm *is an inductive reactance (called magnetizing reactance). Note that the current in the resistance is in phase with *V*1 and *Xm *being an inductive reactance, the current *Im *in this branch lags *V*1 by 90o as shown in the phasordiagram of Figure 4.3(a). (The representation in Figure 4.3, assumes that *V*1 = *E*1 (equal to and in opposition to *V*1).This implies that the primary winding resistance and leakage reactance are neglected.Similarly, in the secondary winding of transformer mutually induced emf is antiphase with*V*1 and its magnitude is proportional to the rate of change of flux and the number ofsecondary turns. (You will learn about the concept of leakage reactance when you studyabout the equivalent circuit at load).The equivalent circuit parameters *R*c and *Xm *can also be expressed as conductance andsusceptance *Gc*, *Bm *such that* c*

*Rc=V1^2/P0 ,Ic=V/Rc ,P0=Ic^2*Rc*

Also, Xm=V1/Im or Im=V1/Xm

**Equivalent Circuit of an Ideal Transformer on Load**

Under certain conditions the transformer can be treated as an ideal transformer. Theidealizing assumptions are listed below :

(a) Both primary and secondary windings have zero resistance. This means,no ohmic power loss and no resistive voltage drop.

(b) No leakage flux, i.e. all the flux produced is confined to the core and links both the windings

(c) Infinite permeability of the core. This means no zero magnetizing current is needed to establish the requisite amount of flux in the core, i.e. *Im *= 0.

(d) Core-loss (hysteresis as well as eddy-current loss) is zero, i.e. *Ic *= 0.

Assumptions (a), (b) and (d) mean that copper losses, and iron losses being zero, the efficiency of the transformer is 100%. Since *Im *= *Ic *= 0, *I*0= 0.

where, *V*1 is supply voltage and *V*2 is voltage across load terminals.When load is applied, let the impedance of load be *ZL*, as shown in Figure 4.4.Sinusoidal current *i*2 flows through the secondary.Therefore, secondary winding creates an mmf *F*2 = *N*2 *i*2 which opposes the flux f.But mutual flux f is invariant with respect to load (otherwise *v*1 = *e*1 balance is disturbed).As a result, the primary winding starts drawing a current *i*1 from the source so as to create mmf *F*1 = *N*1 *i*1 which at all times cancels out the load-caused mmf *N*2 *i*2 so that f is maintained constant independent of the magnitude of the load current which flows in the secondary winding. This implies that for higher load, more power will be drawn from the supply.

Thus, N1*i1= N2 **i 2*= *N2/N1=i2/i1=v2/v1*⇒ v1*i1=v2*i2

(Instantaneous power into primary) = (Instantaneous power out of secondary)

In terms of rms values, i.e. *VA *output = *VA *input, i.e. *V*1 *I*1 = *V*2 *I*2

Since ,V1/V2=N1/N2

*so,I1/I2=V2/V1=N2/N1*

The circuit representation of Figure 4.4, can be simplified by referring the load impedance and secondary current to the primary side. From Figure 4.4, we see that

*V*2 = *I*2 *ZL*

or *V1*N2/N1=I1*N1/N2*Zl*

orV1=I1*(N1/N2)^2 *Zl=I1*Zl’

Zl’=(N1/N2)^2*Zl is said to be the load impedance referred to the primary side.

From *V*2 = *I*2 *ZL *we can also easily obtain *V*2¢ = *I*2¢ *ZL*¢ , where

secondary terminal voltage referred to primary side, and I2’=I2*(N2/N1)

is secondary current referred to primary side. In the ideal transformer, I1=I2′ and *V*1 = *V2′* .

**Equivalent Circuit of a Real Transformer**

In real conditions, in addition to the mutual flux which links both the primary and secondary windings transformer, has a component of flux, which links either the primary winding or the secondary, but not both. This component is called leakage flux. The flux which links only with primary is called primary leakage flux, and the flux which links only with secondary is called secondary leakage flux. Figure 4.4 shows schematically the mutual and the leakage flux. From our knowledge of magnetic circuits, we know that a flux lining with a winding is the cause of inductance of the winding (Inductance = Flux linkage per ampere). Since in a transformer the flux is alternating, its flux linkage gives rise to an induced voltage in thewinding. Thus, primary leakage flux (which is proportional to *I*1) causes an induced voltage, w hich acts as a voltage drop. Similarly for the secondary leakage flux. The effect of these induced EMFs are, therefore, represented as inductive leakage reactances *Xl*1, *Xl*2.*Xl*1 and *Xl*2 are called leakage reactances of the primary and secondary respectively. These are also denoted as *X*1, *X*2.The windings of the transformer have resistance *R*1, *R*2. These resistances cause a voltage drop *I*1 *R*1 and *I*2 *R*2, as also ohmic loss I1^2*R1=I2^2*R2

To sum up, in a Real Transformer,

(a) Both the primary and secondary windings possesses resistance. As a result,the value of actual impressed voltage across the transformer is the voltage *V*1 less the drop across the resistance *R*1 . Moreover, the copper loss in the primary winding is ( *I*1 *R1) *and in the secondary winding (*I*2 *R2) *.

(b) A Real Transformer has some leakage flux, as shown in the Figure 4.4. These fluxes, as discussed earlier, lead to self-reactances*X L1*and *X L2*for the two windings respectively.

(c) The magnetizing current cannot be zero. Its value is determined by the mutual flux f*m*. The mutual flux also determines core-loss taking place in the iron parts of the transformer. The value of *I*o does not depend on load and hence the iron-loss or core-loss is constant which is not zero.

**Representation of Transformer Showing Leakage Reactances**

In the form of equivalent circuit, this can be represented as in Figure 4.6.

The use of this equivalent circuit is difficult and calculations involved are laborious. For most practical purposes (like calculations of voltage regulation and efficiency) we need only a simplified form of equivalent circuit. We will now proceed to first obtain a **simplified **equivalent circuit and then to obtain an **approximate **equivalent circuit.

**Equivalent Circuits of Transformer Referred to Primary Side**

We will now refer the impedance *R 2*+ *j XL2 *to the primary side i.e. to the left hand side of the ideal transformer. We have seen earlier that a load impedance *Zl *can be referred to primary side as *Zl’* , where

Zl’=(N1/N2)^2*Zl

Similarly

*Z2*= *R2*+ *j XL2 *can be referred to the primary side as Z2’=(N1/N2)^2*Z2,where *Z*2’is said to be the secondary winding impedance referred to the primary side.

Eq. (4.13) can be re-written as *R2′*+ *j XL2’=(N1/N2)*( R2+ j XL2)*

Equating real and imaginary parts

R2’=(N1/N2)^2**R2*

Xl2’=(N1/N2)^2*Xl*2*

*R*2′ is the secondary winding resistance referred to primary, and *Xl2′*is the secondary winding leakage reactance referred to primary side.Figure 4.6 can now be modified (i.e. referring the secondary resistance and reactance to the primary side) to get the equivalent circuit shown in Figure 4.7.

**Approximate Equivalent Circuits of Transformer**

Transformers which are used at a constant power frequency (say 50 Hz), can have very simplified approximate equivalent circuits, without having a substantial effect on the performance evaluation (efficiency and voltage regulation). It should be borne in mind that higher the VA or KVA rating of the transformers, better are the approximation-based evaluation results.’

It is assumed that *V*1 » *E*1 (*V*1 is approximately equal to *E*1 ) even under conditions ofload. This assumption is justified because the values of winding resistance and leakagereactances are very small. Therefore, the exciting current drawn by the parallel combination of conductance *Gc *and susceptance *Bm *would not be affected significantly byshifting it to the input terminals. With this change, the equivalent circuit becomes as shown in Figure 4.8(a).

**(a) Equivalent Circuit s of Transformer Referred to Primary Side**

Denoting *R*1 + *R*2′ = *R’*eq

and *XL1’+XL2′*= XL’eq

The equivalent circuit becomes as shown in Figure 4.8(b) *R’*eq, *X’*eq are called the equivalent resistance and equivalent reactance referred to primary side.

**(b) Approximate Equivalent Circuits of Transformer**

If only voltage regulation is to be calculated even the excitation branch can be neglected and the equivalent circuit becomes as shown in Figure 4.9.

#### QUESTION AND ANSWER

**Q. At no-load a transformer has a no-load loss of 50 W, draws a current of2A (RMS) and has an applied voltage of 230V (RMS). Determine the(i) no-load power factor, (ii) core loss current, and (iii) magnetizing current. Also,calculate the no-load circuit parameter ( Rc, Xm) of the transformer.**

Ans: Pc = 40 W,I0 = 2 A,E1 = 230 V

Pc = V1 I0 cos (ɸ0)

⇒cos (ɸ0)=Pc/I0*E1

=50/230*2=0.108(lagging)

⇒ɸ0 = 83.76

Magnetizing current, *Im *= *I*0 sin f0

= 2 sin (83.76)

= 1.988 A

Core-loss current *Ic *= *I*0 cos ɸ0= 2 × 0.108

= 0.216 A

Pc=Vc^2/Rc=Gc*v1⇒Rc=V1^2/Pc=230^2/50=1.058 kΏ

*Gc=50/230*⇒0.945 10 (mho)

Also,

Im=Bm.V1=V1/Xm

Xm=V1/Im= 230/1.988=115.6*Ώ*

Bm=Im/V1=1.988 3/230⇒ 8.64 X10^-3 (mho)

*Please share your views regarding this in the comment box below. It would be our pleasure.*

**You may also like to see:-**

- Transmission Lines – Part 2
- Power Transformer Protection, Generator Protection Scheme, Pilot Protection
- Syllabus for Electrical Engineering IES 2013
- IES Electrical Paper

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equivalent circuit of a transformer on load as well as on no load and their respective calculations

the core flux in transformer maily dependes on……………..i.e v/f but why? and how ?

why magnetising component and working componento oftrasformer are in parallel

Why is induced emf in the secondary coil of transforner antiphase with supply ac voltage of primary?

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