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 (I2 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, V1 = E1 and E2 = V2.
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 V1 (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 V1. 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 I0 .
I = I0 cos f0 and I = I0 sin f0 c m
Core loss = P0 = V1 I0 (cos f0)
where f0 is the phase angle between V1 and I0, and, (cos f0) is the no load power factor.The phase relationship between applied voltage V1, no-load current I0, and its componentsIc, 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 V1 and Xm being an inductive reactance, the current Im in this branch lags V1 by 90o as shown in the phasordiagram of Figure 4.3(a). (The representation in Figure 4.3, assumes that V1 = E1 (equal to and in opposition to V1).This implies that the primary winding resistance and leakage reactance are neglected.Similarly, in the secondary winding of transformer mutually induced emf is antiphase withV1 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 Rc 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, I0= 0.
where, V1 is supply voltage and V2 is voltage across load terminals.When load is applied, let the impedance of load be ZL, as shown in Figure 4.4.Sinusoidal current i2 flows through the secondary.Therefore, secondary winding creates an mmf F2 = N2 i2 which opposes the flux f.But mutual flux f is invariant with respect to load (otherwise v1 = e1 balance is disturbed).As a result, the primary winding starts drawing a current i1 from the source so as to create mmf F1 = N1 i1 which at all times cancels out the load-caused mmf N2 i2 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. V1 I1 = V2 I2
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
V2 = I2 ZL
Zl’=(N1/N2)^2*Zl is said to be the load impedance referred to the primary side.
From V2 = I2 ZL we can also easily obtain V2¢ = I2¢ 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 V1 = 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 I1) 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 Xl1, Xl2.Xl1 and Xl2 are called leakage reactances of the primary and secondary respectively. These are also denoted as X1, X2.The windings of the transformer have resistance R1, R2. These resistances cause a voltage drop I1 R1 and I2 R2, 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 V1 less the drop across the resistance R1 . Moreover, the copper loss in the primary winding is ( I1 R1) and in the secondary winding (I2 R2) .
(b) A Real Transformer has some leakage flux, as shown in the Figure 4.4. These fluxes, as discussed earlier, lead to self-reactancesX L1and X L2for the two windings respectively.
(c) The magnetizing current cannot be zero. Its value is determined by the mutual flux fm. The mutual flux also determines core-loss taking place in the iron parts of the transformer. The value of Io 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
Z2= R2+ j XL2 can be referred to the primary side as Z2’=(N1/N2)^2*Z2,where Z2’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′ 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 V1 » E1 (V1 is approximately equal to E1 ) 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 Circuits of Transformer Referred to Primary Side
Denoting R1 + R2′ = 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)
⇒ɸ0 = 83.76
Magnetizing current, Im = I0 sin f0
= 2 sin (83.76)
= 1.988 A
Core-loss current Ic = I0 cos ɸ0= 2 × 0.108
= 0.216 A
Gc=50/230⇒0.945 10 (mho)
Bm=Im/V1=1.988 3/230⇒ 8.64 X10^-3 (mho)
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