In an ideal two-winding transformer excited by the primary winding, all of the magnetizing flux is within the core and both the primary and secondary windings are linked by the same flux. The situation in a real transformer is somewhat different. The main difference is that all of the magnetic flux is not contained in the core. This is because the load currents in the primary and secondary windings are considerably greater than magnetizing current, so we cannot ignore the magnetic fields induced by these currents in the spaces surrounding the winding conductors. These magnetic fields circulate around the windings in the spaces between the conductors.
The local magnetic flux circulating around the primary winding are approximately equal to the local magnetic flux circulating around the secondary winding because the primary ampere turns are approximately equal to the secondary ampere turns. Also, the magnetic flux is in phase with the currents that produce it. The composite of all flux in the spaces around and between the conductors is called leakage flux, because this can be visualized as flux ‘‘leaking’’ out of the core through empty space. Because the permeability of iron is finite, there will always be some flux that leaks from the core, even without any load current. However, the leakage flux increases considerably as the currents in the windings increase. Since the leakage flux travels through empty space instead of through iron, there is no hysteresis or saturation and the permeability in this region is a constant.
Let in primary winding, leakage flux øL1 completes it magnetic circuit by passing through air, rather than core. This leakage flux, flux øL1, links with only primary amp-turn and induces an e.m.f. eL1 in primary and none in secondary.
Similarly, in secondary winding, leakage flux øL2 produces a self induced e.m.f. eL2 in secondary but none in primary.


As the magnitude of the leakage flux increases, the magnitude of flux in the core decreases. The drop in secondary voltage is therefore directly proportional to the rate of change of the load current. The relationship between load current and secondary voltage is equivalent to placing physical impedance in series with the output of an ideal transformer. This equivalent impedance is called the leakage reactance of the transformer because it is an external manifestation of the effect of leakage flux on secondary voltage.
Leakage reactance limits short-circuit currents, and is therefore useful, but it also reduces the secondary voltage under load (regulation) and consumes reactive power. Varying the winding dimensions and the spacing between the windings can control the amount of leakage reactance.
So, as per figure, we can say, primary impedance, Z1= eL1/I1, and secondary impedance, Z2= eL2/I2. Where I1 & I2 represents primary and secondary current.
From the equation, Z12= R12 + X12 and Z22= R22 + X22, we can calculate leakage reactances, X1 and X2.


For an ideal transformer, the magnetizing current is assumed to be negligible. For a real transformer, some magnetizing current must flow when voltage is applied to the winding in order to establish a flux in the core. The voltage induced in the winding by the flux restrains the magnetizing current. The magnetizing current is not really sinusoidal, but contains many odd harmonics in addition to the fundamental frequency. If we neglect the harmonics and concentrate on the fundamental frequency, the magnetizing current in the winding lags the applied voltage by 90°. In a two-winding transformer, this is equivalent to placing a reactance Xm, called the magnetizing reactance, in parallel with the transformer terminals.
The peak value of the magnetizing current is determined from the B-H curve of the core, which is very nonlinear. Therefore, the magnetizing reactance is not a constant but is voltage dependent. It is generally desirable to maximize Xm in order to minimize the magnetizing current. Inductance is inversely proportional to the reluctance of the core along the flux path and the reluctance of an air gap is several thousand times the reluctance of the same distance through the steel.
Therefore, even tiny air gaps in the flux path can drastically increase the core’s reluctance and decrease Xm. A proper core design must therefore eliminate all air gaps in the flux path. All ripples or waves must be eliminated by compressing the core laminations together tightly.


If the spaces between the windings are kept small, there will be fewer flux lines contained in these spaces and the leakage reactance will be small. Placing the windings as close together as possible increases the ‘‘coupling’’ between them, which is the same thing as reducing the leakage reactance.

In case of low or light loads, the primary and secondary amp-turns are small, hence leakage fluxes are negligible. But if we increase the load, primary and secondary windings carry load current which is huge in magnitude. Hence, large m.m.f. is set up which, while acting on leakage paths, increase the leakage flux.


It refers to voltage drop inside the transformer due to resistance. Cu-loss represents as I2R loss. So, cu loss increases with increase of load current.
The resistance of the winding is proportional to the conductor length divided by the conductor cross-sectional area. When an alternating magnetic field is applied to any conductor, eddy currents are induced around the paths surrounding the lines of magnetic flux that penetrate the conductor. These currents generate local I2R losses even if the conductor itself is not carrying any net electrical current.
Large amounts of leakage flux can occur when a transformer is heavily loaded. The magnetic fields associated with leakage flux not only penetrate the winding conductors themselves, but can involve other metallic parts. The eddy currents that are induced by these fields are proportional to the leakage flux, which in turn is proportional to the load currents. Therefore, the square of eddy currents and the eddy-current losses are both proportional to the square of the load current. These eddy losses are increases the effective resistance of the conductors.
When an AC current flows in a conductor, the flux density increases near the outer edges of the conductor. The concentration of current toward the edge of a conductor is called the skin effect, reducing the area of the conductor that actually carries current and increasing the effective resistance of the conductor.
RAC = winding resistance + eddy current resistance + skin resistance.
The conductor losses are equivalent to placing a lumped resistance in series with the terminals of an ideal transformer. Conductor losses are commonly referred to as load losses, because they result only from load currents. Load losses are sometimes referred to as copper losses.


Heat generated by these losses increase temperatures. Therefore, it is highly desirable to reduce the load losses as much as possible by reducing the AC resistance of the conductor. Reducing winding resistance as well as Reddy current and Rskin can do this. Reducing winding resistance can be done by shortening the conductor length and/or by increasing the conductor cross-sectional area. Shortening the conductor length can only be achieved to a point, and increasing the conductor cross-sectional area have the unfortunate effect of increasing both the eddy-current losses and the skin effect losses. These losses can be reduced by special conductor designs. Subdividing the conductors into strands that are insulated from each other to break up the eddy current paths can reduce eddy-current and skin effect losses.
By subdividing one large-area conductor into a number of small-area conductors, the skin effect is substantially reduced as well. If a number of strands were simply bundled together, and the strands are properly transposed, each strand is exposed to the same amount of leakage flux with equal voltages induced. Transposition allows the current to divide equally among the strands to optimize the use of the entire cross section area of the conductor. After seven complete transpositions, all the strands return to their original positions.


Alternating magnetic flux produces both hysteresis losses and eddy-current losses in the steel. Hysteresis losses depend on several factors including the frequency, the peak flux density, the type of core steel used, and the orientation of the flux with respect to the ‘‘grain’’ of the steel. All of the above factors, except the frequency, are under the control of the transformer design. Core losses are sometimes referred to as iron losses and are commonly referred to as no load losses, because core losses do not require load currents.
So, the core losses = hysteresis loss + eddy current loss.
Hysteresis loss, Wh = PBmax1.6f.
And eddy current loss We = QBmax2f2t2v2. Where t = thickness of lamination, and v = volume.
So, if we know volume and thickness of core lamination from core design, we can easily calculate no load losses.
Way of reducing core losses is to use various types of low-loss core steels that are now available


In a practical transformer, primary and secondary windings are different and there is no electrical connection between it. Both primary and secondary winding has own resistance and reactance. In load condition, as voltage and current are different in primary and secondary, equivalent impedance are calculated in either primary or secondary side by transferring any one side voltage, current and impedance to other side.

Equivalent resistance:

Suppose a transformer having primary and secondary resistances are R1 and R2 respectively. This resistance can be transferred to any one of the two windings.
Let, equivalent secondary resistance as referred to primary is R2’.
Now, copper loss in secondary is I22R2. This loss is supplied by primary which takes current I1.
So, I12R2’= I22R2.
Or, $$ R_{ 2 }^{ / }=(I_{ 2 }/I_{ 1 })^{ 2 }R_{ 2 } $$.
As we know that, I2/I1 = V1/V2 = 1/K = 1/ transformation ratio.
So, R2’ = R2/K2.
Similarly, equivalent primary resistance as referred to secondary is R1’= R1X K2.
So, overall equivalent resistance of transformer as referred to primary is R01=R1 + R2’= R1+(R2/K2).
Similarly, equivalent resistance of transformer as referred to secondary is R02=R2+R1’ = R2+R1K2.

Equivalent reactance:

In case of reactance, same rule is applied for calculation.
As, X2’= X2/K2 and X1’= X1K2.
So, equivalent reactance of transformer as referred to primary is X01= X1+ X2’= X1 +X2/K2.
And equivalent reactance as referred to secondary, X02= X2+X1’ = X2+ K2.
Now we can easily calculate the equivalent impedance of transformer as referred to primary or secondary.
(Z01)2= (R01)2 + (X01)2.
Or, (Z02)2=(R02)2+ (X02)2.

Calculation of equivalent circuit is necessary to calculate total internal impedance of transformer viewing from primary or secondary side. Another important factor is percentage impedance. For parallel operation of transformer, percentage impedance should be properly matched. So, installing a new transformer with existing system, percentage impedance carries a great value. From equivalent impedance, we can calculate the percentage impedance.


Percentage resistance is the ratio of voltage drop due to resistance, at full load current to the normal voltage either referred all resistance to primary or secondary side.
As % R =( I1R01)/V1, or % R =( I1R02)/V2.
Percentage reactance is the ratio of voltage drops due to leakage reactance at full load current to the normal voltage as referred all reactances either to primary or secondary side.
So, % X = (I1X01)/V1, or % X = ( I1X02)/V2.
Similarly, percentage impedance is the ratio of impedance volt (I1Z01) and rated voltage as referred to primary or secondary.
% Z = (I1Z01)/V1, or % Z = ( I1Z02)/V2.


Inherent regulation of a transformer is nothing but the difference of secondary voltage from no- load to full load at rated primary voltage.
At no-load condition, the impedance drop is negligible, but when transformer is loaded with constant primary voltage, due to impedance drop including its internal resistance and leakage reactance, the secondary terminal voltage decreases. The regulation is published as percentage of rated secondary voltage from no-load to full load.
If E2 be the secondary terminal voltage at no-load, and V2 be the secondary terminal voltage at full load, then, % regulation = (E2-V2)/E2 X100. It is known as % regulation (down). E2-V2 = voltage drop of transformer.
Whereas, (E2-V2)/V2 X100 known as % regulation (up).
Normally, % regulation means % regulation (down).
The value of percentage regulation is as low as good that means lower voltage drop. So, in good transformer, percentage regulation should be lesser value.


The efficiency of a transformer at a particular load and power factor is defined as ratio of output power to input power.
Efficiency = Output/Input = Output/(output + losses)
= Output/(Output+ cu loss+ core loss)
Or, ɳ = (Input-losses)/Input = 1-(losses/input).
So, efficiency depends on power output, not in volt-ampere, thus efficiency, depends on load power factor and being maximum at unity power factor.

Condition for maximum efficiency:

As per previous discussion, we know that cu-loss = I12R01 or I22R02 = Wcu.
Iron loss = Hysteresis loss + eddy current loss = Wh+ We = Wi.
Now, input power = V1I1cosø1.
So, ɳ = (V1I1cosø1 – losses)/ V1I1cosø1. Where V1, I1, cosø1 = input voltage, current and power factor.
ɳ= (V1I1cosø1 – I12R01-Wi)/ V1I1cosø1.
= 1-(I1R01/V1 cosø1)-( Wi/V1I12cosø1).
As differentiate w.r.t. I1, we get
$$ dɳ/dI_{ 1 }=0-\frac { R_{ 01 } }{ V_{ 1 }COSø_{ 1 } } +\frac { W_{ i } }{ V_{ 1 }I_{ 1 }^{ 2 }COSø_{ 1 } } $$
Efficiency will be maximum, when, $$ dɳ/dI_{ 1 }=0 ,$$
Or,$$ \frac { R_{ 01 } }{ V_{ 1 }COSø_{ 1 } } =\frac { W_{ i } }{ V_{ 1 }I_{ 1 }^{ 2 }COSø_{ 1 } } $$
Wi = I12R01 or, I22R02.
So, in case of maximum efficiency of transformer, cu-loss = core-loss.

All-day efficiency:

In case of distribution transformer, the loading throughout the day is not same. Due to domestic loads, the transformer loads increase at evening peak but little load during the day. As we already know that, transformer efficiency depends on the losses of transformer and we also know iron and cu-losses are same at light as well as peak load, but cu-loss varies by the flow of load current. That means, core-loss occurs throughout the day whereas, cu-loss occur when transformer is loaded.
So, in case of distribution transformer, we need to design it as low iron or core loss, whereas cu-losses are less important.
The all day efficiency is calculated throughout a time period, usually 24 hours (a day). So, it is the ratio of output energy to the input energy for 24 hours.
ɳall day = Output in KWH/Input in KWH (for 24 hours).
It is a common sense, that all day efficiency is less than commercial efficiency of transformer. To find out all day efficiency, load flow picture throughout the day is necessary.

Keep in pocket:→

✓ The flux, which does not links with two coils, known as leakage flux.
✓ Leakage flux increases leakage reactance, and can be controlled by proper winding design(reducing air path between two coils).
✓ Cu- losses mainly depends on conductor resistivity, beside this it also depends on eddy current and skin effect.
✓ No -load losses or core losses or iron losses are nothing but the hysteresis and eddy current losses.
✓ Equivalent circuit calculation is necessary to calculate the percentage impedance of transformer.
✓ Regulation of transformer is the ratio between voltage drop from no-load to full load, and no-load voltage.
✓ Transformer efficiency is very high ranging from 0.96 to 0.99 depending on load power factor.
✓ Efficiency is maximum when core loss and copper loss are same.
✓ For distribution transformer, all day efficiency is needed for design.

Short questions related to this topic:

Q. How leakage reactance minimize ?

A. By placing two winding without leaving air gap.

Q. Low load or full load, when leakage flux is maximum ?

A. At full load.

Q. Core loss or cu loss, which one is constant?

A. Core loss.

Q. Why percentage impedance is necessary?

A. For paralleling the transformer.

Q. Regulation high or low, which one is good?

A. Low (low voltage drop).

Q. When transformer efficiency is high?

A. Core loss = cu-loss.

Q. Transformer efficiency depends on_____?

A. Load power factor.

Q. What is all day efficiency?

A. Transformer efficiency through out the day.

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