Operation of Transformer On load

Figure (1) shows a transformer with a load connected across the secondary winding. The load current I₂ flowing through the secondary turns sets up its own m.m.f N₂I₂ which produces the flux ϕ₂.

According to Lenz’s law this flux is in such a direction that it opposes the flux ϕ, produced by the m.m.f N₁I₀ which is set up by the no-load current I₀ flowing throw the primary turns. Consequently the flux is momentarily reduced due to opposing flux ϕ. This in turn causes reduction in induced e.m.f (E₁) in primary according to Faraday’s law E₁ reduces, the difference between applied voltage (V₁) and E₁ increases.

Consequently, the primary will draw more current. Consider \({{{I}’}_{1}}\) to be this additional primary current. It is also known as counter balancing current as it balances between applied voltage and primary e.m.f or it is known as load component of primary current and it is antiphase with secondary current I₂. Now this current \({{{I}’}_{1}}\) sets up its own m.m.f N₁\({{{I}’}_{1}}\) which produces the flux and it is equal in magnitude in such a direction that it opposes the flux ϕ₂. Hence ϕ’₁ and ϕ₂ cancel each other and only flux ϕ flows in the core. Therefore the total flux produced during loaded condition is approximately equal to the flux at no-load.

\[{{\phi }_{2}}=-{{{\phi }’}_{1}}\]

As secondary ampere turns of I₂ are neutralized by primary ampere turns of \({{{I}’}_{1}}\).

\[{{N}_{2}}{{I}_{2}}={{N}_{1}}{{{I}’}_{1}}\]

\[{{{I}’}_{1}}=\frac{{{N}_{2}}}{{{N}_{1}}}{{I}_{2}}\]

The net primary current is the vector sum of primary counter balancing current \({{{I}’}_{1}}\) and the no-load current I₀.

\[{{I}_{1}}={{{I}’}_{1}}+{{I}_{0}}\]

Since the no-load current I₀ is very small compared to the counter balancing current \({{{I}’}_{1}}\), therefore the net primary current is approximately equal to the current \({{{I}’}_{1}}\).

\[{{I}_{1}}={{{I}’}_{1}}\]

\[=\frac{{{N}_{2}}}{{{N}_{1}}}{{I}_{2}}=K{{I}_{2}}\]

‘K’ represents transformation ratio.

\[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{{{{{I}’}}_{1}}}{{{I}_{2}}}=\frac{{{N}_{2}}}{{{N}_{1}}}=K\]

\[{{I}_{1}}=K{{I}_{2}}={{{I}’}_{1}}\]

Therefore, the primary and secondary currents are inversely proportional to their turns ratio. The total primary current is in anti-phase with I₂ and K times the current I₂.

Phasor Diagram of Transformer with Resistive Load

The phasor diagram for resistive load is drawn as shown in the following figure (2). For purely resistive load, the secondary load current I₂ is in phase with the secondary’ terminal voltage V₂. The counter balancing current \({{{I}’}_{1}}\) is in opposition and equal in magnitude with the secondary load current I₂. The primary current I₁ is the vector sum of \({{{I}’}_{1}}\) and no-load current I₀ respectively. I₀ lags behind V₁ by no-load power factor angle ϕ₀ and I₁ lags behind the voltage V₁ by primary power factor angle ϕ₁.

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When the transformer is on no-load, the secondary winding is opened as in figure (a) and current I₂ is zero. In this condition, the primary winding draws a no-load current ‘I₀‘ which has two components i.e.,

1. Magnetizing component (I_µ), and
2. Working component (I_w).

1. Magnetizing Component (I_µ) :

It lags behind the applied voltage on primary winding ‘V₁‘ by 90º. It is also called as reactive or wattless component of no-load current and is responsible to develop an e.m.f to maintain the flux ‘ϕ’ in the core. It is expressed as \({{I}_{\mu }}=\text{ }{{I}_{0}}\sin {{\phi }_{0}}\).

2. Working Component (I_w) :

It is in phase with the primary applied voltage ‘V₁‘. The component is also called as active component or iron loss component, and is used for describing the core losses such as hysteresis loss and eddy current loss. It is expressed as \({{I}_{w }}=\text{ }{{I}_{0}}\cos {{\phi }_{0}}\).

From the phasor diagram of figure (b),

\[\sin {{\phi }_{0}}=\frac{{{I}_{\mu }}}{{{I}_{0}}}\]

Thus, \({{I}_{\mu }}={{I}_{0}}\sin {{\phi }_{0}}\) is the reactive component of no-load current I₀ and

\[\cos {{\phi }_{0}}=\frac{{{I}_{w }}}{{{I}_{0}}}\]

Thus, \({{I}_{w }}={{I}_{0}}\cos {{\phi }_{0}}\) is the active component of no load current I₀.

Hence,

\[{{I}_{0}}=\sqrt{I_{w}^{2}+I_{\mu }^{2}}\]

cosϕ₀ is the no-load power factor and ϕ₀ is the hysteresis angle of advance.

The post Transformer on No Load – Circuit Diagram & Phasor Diagram appeared first on Electrical and Electronics Blog.

Consider the two winding single-phase transformer shown in figure (1).

\({{{I}}_{1}}\) = Current in the primary

\({{{E}}_{1}}\) = Induced e.m.f in the primary

\({{{V}}_{1}}\) = Voltage applied to the primary

\({{{I}}_{2}}\) = Current in the secondary

\({{{E}}_{2}}\) = Induced e.m.f in the secondary

\({{{V}}_{2}}\) = Terminal voltage of secondary

Here, the primary current \({{{I}}_{1}}\) has two components, one is no-load primary current, \({{{I}}_{0}}\) and the other one is load component of primary current \({{{I}’}_{2}}\). The function of current \({{{I}’}_{2}}\) is to counter balance the secondary current \({{{I}}_{2}}\). The no-load primary current \({{{I}}_{0}}\) leads to the production of losses in the core while magnetizing the core of the transformer. The no- load primary current \({{{I}}_{0}}\) can be resolved into two components i.e., active (or) working component I_w and reactive (or) magnetizing component ‘I_µ‘. The working component ‘I_w’ of no-load current \({{{I}}_{0}}\) leads to the core loss, hence it can be represented by a resistance R₀. The magnetizing current ‘I_µ‘ produces flux which induces e.m.f E₁.

The reactance due to flux is represented by X₀. To account for the core loss and the magnetizing current, an equivalent circuit can be represented by a shunt branch in the primary side as shown in the figure (2).

\[\text{Core loss = }I_{w}^{2}{{R}_{0}}=\frac{E_{1}^{2}}{{{R}_{0}}}\]

To make transformer calculations simpler, transfer voltage, current and impedance either to the primary or secondary

Equivalent Circuit of Transformer as Referred to Primary Side

Secondary parameters transferred to primary side are given as follows,

\[{{{R}’}_{2}}=\frac{{{R}_{2}}}{{{K}^{2}}}\]

\[{{{X}’}_{2}}=\frac{{{X}_{2}}}{{{K}^{2}}}\]

\[{{{Z}’}_{2}}=\frac{{{Z}_{2}}}{{{K}^{2}}}\]

\[{{{I}’}_{2}}=K{{I}_{2}}\]

\[{{{E}’}_{2}}=\frac{{{E}_{2}}}{K}\]

\[{{{V}’}_{2}}=\frac{{{V}_{2}}}{K}\]

Where,

\[K=\frac{{{N}_{2}}}{{{N}_{1}}}\]

We know that,

\[{{R}_{01}}={{R}_{1}}+{{{R}’}_{2}}={{R}_{1}}+\frac{{{R}_{2}}}{{{K}^{2}}}\]

\[{{X}_{01}}={{X}_{1}}+{{{X}’}_{2}}={{X}_{1}}+\frac{{{X}_{2}}}{{{K}^{2}}}\]

\[{{Z}_{01}}=\sqrt{R_{01}^{2}+X_{01}^{2}}=\frac{{{Z}_{02}}}{{{K}^{2}}}\]

The equivalent circuits referred to primary side are as shown in figures (3) and (4).

Equivalent Circuit of Transformer Referred to Secondary Side

Primary parameters transferred to secondary side are given as follows,

\[{{{R}’}_{1}}={{K}^{2}}{{R}_{1}}\]

\[{{{X}’}_{2}}={{K}^{2}}{{X}_{1}}\]

\[{{{Z}’}_{1}}={{K}^{2}}{{Z}_{1}}\]

\[{{{E}’}_{1}}=K{{E}_{1}}\]

\[{{{V}’}_{1}}=K{{V}_{1}}\]

\[{{{I}’}_{1}}=\frac{{{I}_{1}}}{K}\]

\[{{{I}’}_{0}}=\frac{{{I}_{0}}}{K}\]

\[{{{R}’}_{0}}=\frac{{{R}_{0}}}{{{K}^{2}}}\]

\[{{{X}’}_{0}}=\frac{{{X}_{0}}}{{{K}^{2}}}\]

We know that,

\[{{R}_{02}}={{R}_{2}}+{{{R}’}_{2}}={{R}_{2}}+{{K}^{2}}{{R}_{1}}\]

\[{{X}_{02}}={{X}_{2}}+{{{X}’}_{1}}={{X}_{2}}+{{K}^{2}}{{X}_{1}}\]

\[{{Z}_{02}}=\sqrt{R_{02}^{2}+X_{02}^{2}}={{K}^{2}}{{Z}_{01}}\]

The equivalent circuit diagrams referred to secondary side are shown in figure (5) and figure (6).

The post Equivalent Circuit of Transformer – Circuit Diagram & Derivation appeared first on Electrical and Electronics Blog.

Open Circuit (O.C.) Test on Transformer

Set-up:

Fig. 1 : Set-up for Open Circuit Test on Transformer.

This test is performed so as to calculate the no load losses (core losses) of a transformer and the values of I₀, R₀ and X₀ of the equivalent circuit. The set up for O.C. test of a transformer is shown in Fig. 1. The primary or secondary winding of the transformer is connected to the rated ac voltage by means of using a variac. The other winding is left open. Generally the high voltage winding is open circuited, and the voltage is applied to the low voltage winding. Assume that this is a step up transformer. A voltmeter is connected across the primary winding to measure the primary voltage. An ammeter is used for measuring the no load primary current (I₀) and the wattmeter is connected to measure the input power. The secondary is open circuited because it is an open circuit (O.C.) test. Sometimes a voltmeter is connected across the secondary to measure V₂ = E₂. Note that the ac supply voltage is applied generally to the low voltage side and the higher voltage side is used as secondary.

Procedure:

• Connect the circuit as shown in Fig. 1.
• Keep the variac at its minimum voltage position.
• Switch on the ac power supply and adjust the variac to get the rated primary voltage as measured by voltmeter V across the primary.
• Now measure the primary current (I₀) and power (W₀) using the ammeter and wattmeter respectively.
• The ammeter reads the no load primary current I₀ whereas the wattmeter measures the no load input power W₀.
• The observation table for the O.C. test is as follows.

• The two components of no load current I₀ are,

\[{{I}_{m}}={{I}_{0}}\sin {{\phi }_{0}}\]

\[{{I}_{c}}={{I}_{0}}\sin {{\phi }_{0}}\]

• The no load power factor is given by cosϕ₀ and the input power at no load is given by,

\[{{W}_{0}}={{V}_{1}}{{I}_{0}}\cos {{\phi }_{0}}\]

• The phasor diagram on no load showing the two components of I₀ is shown in Fig. 2.

Fig. 2 : Phasor Diagram for Open Circuit Test on Transformer.

• The no load current I₀ is very small as compared to the full load primary current. The no load current I₀ is about 3 to 5 % of the full load value.
• As I₂ is zero, the secondary copper loss is zero. The primary copper loss will be negligible because I₀ is small.
• Therefore the total copper loss is very small and can be assumed to be equal to zero. Hence the wattmeter reading W₀ represents the iron losses.

\[{{W}_{0}}={{P}_{i}}=\text{ Iron losses}\]

Calculation of parameters:

The two parameters which can be calculated from the open circuit test are R₀ and X₀. They are calculated as follows.

Step 1: Calculate no load power factor cosϕ_0,

The wattmeter reads the real input power.

\[{{W}_{0}}={{V}_{1}}{{I}_{0}}\cos {{\phi }_{0}}\]

And

\[\cos {{\phi }_{0}}=\frac{{{W}_{0}}}{{{V}_{1}}{{I}_{0}}}\]

Calculate ϕ₀ from this.

Step 2: Calculate I_m and I_c,

\[{{I}_{m}}={{I}_{0}}\sin {{\phi }_{0}}\]

\[{{I}_{c}}={{I}_{0}}\cos {{\phi }_{0}}\]

So calculate I_m and I_c from the above equations.

Step 3: Calculate R₀ and X_0,

\[{{R}_{0}}=\frac{{{V}_{1}}}{{{I}_{c}}}\Omega \]

\[{{X}_{0}}=\frac{{{V}_{1}}}{{{I}_{m}}}\Omega \]

The value of cosϕ₀ is very small. Therefore it is necessary to use the low power factor type wattmeter to avoid any possibility of error in measurements.

Short Circuit (S.C.) Test on Transformer

Set-up:

Fig. 3 : Set-up for short Circuit Test on Transformer.

The set up for carrying out the shown circuit (SC) test on a transformer is shown in Fig. 3. Generally the high voltage side is connected to the ac supply and the low voltage high current side is shorted. Variac is used to adjust the input voltage precisely to the rated voltage. We assume that the transformer used here is a step down transformer. Hence the secondary is shorted and primary is connected to the variac. The voltmeter is connected to measure the primary voltage The ammeter measure the short circuit rated primary current I_sc and the wattmeter measures the short circuit input power. The secondary is short circuited with the help of thick copper

Procedure:

• Connect the circuit as shown in Fig. 3.
• Shown circuit the secondary which is a low voltage high current, low resistance winding.
• Keep the variac at its minimum voltage position and switch on the ac supply voltage.
• Increase the primary voltage very gradually: and adjust it to get the primary current equal to the rated value I_sc. Do not increase the primary voltage further.
• Note down the wattmeter, voltmeter and ammeter readings. The observation table is as shown below in Table.

Parameter calculations:

The primary and secondary currents are the rated currents. Therefore the total copper loss is the full load copper loss. If we adjust the primary current to half the full load current then we get the copper loss at half load. The iron losses are a function of applied voltage. As the applied voltage in S.C. test is small, the iron losses will be negligibly small. Hence the wattmeter reading W_sc corresponds almost entirely to the full load copper loss.

\[{{W}_{SC}}=\text{ Full load copper loss}\]

\[={{P}_{cu(FL)}}\]

We can calculate the parameters R_IT, X_IT and Z_IT of the equivalent circuit from the short circuit (S.C.) test.

We know that

\[{{W}_{SC}}={{\text{V}}_{SC}}{{I}_{SC}}\cos {{\phi }_{SC}}\]

Hence the short circuit power factor is given by,

\[\cos {{\phi }_{SC}}=\frac{{{W}_{SC}}}{{{V}_{SC}}{{I}_{SC}}}\]

But the wattmeter reading W_sc indicates the full load copper loss.

\[{{W}_{SC}}=\text{Copper loss}\] \[=I_{SC}^{2}\times {{R}_{1T}}\]

\[{{R}_{1T}}=\frac{{{W}_{SC}}}{I_{SC}^{2}}\]

Similarly,

\[{{Z}_{1T}}=\frac{{{V}_{SC}}}{I_{SC}^{{}}}=\sqrt{R_{1T}^{2}+X_{1T}^{2}}\]

\[{{X}_{1T}}=\sqrt{Z_{1T}^{2}+R_{1T}^{2}}\]

In this way the parameters R_IT, X_IT and Z_IT can be calculated from the S.C. test. If the transformation ratio K, it is possible to obtain the parameters referred to the secondary side.

Efficiency Calculation from O.C. and S.C. Test:

The expression for full load efficiency IS given by,

\[{{\eta }_{FL}}=\frac{{{V}_{2}}{{I}_{2(FL)}}\cos \phi }{{{V}_{2}}{{I}_{2(FL)}}\cos \phi +{{P}_{i}}+{{P}_{cu(FL)}}}\]

The iron loss P_i can be obtained from the O.C. test because,

\[ {{W}_{0}}={{P}_{i}}=\text{ Total iron loss}\],

And the full load copper loss is obtained from the S.C. test because,

\[{{\text{W}}_{SC}}={{P}_{cu(FL)}}=\text{ Copper loss}\]

Voltage Regulation Calculation from O.C. and S.C. Test:

The percentage regulation (%R) is given by,

\[\%R=\frac{{{I}_{2}}{{R}_{2T}}\cos \phi \pm {{I}_{2}}{{X}_{2T}}\sin \phi }{{{V}_{2}}}\times 100\]

\[\%R=\frac{{{I}_{1}}{{R}_{1T}}\cos \phi \pm {{I}_{1}}{{X}_{1T}}\sin \phi }{{{V}_{1}}}\times 100\]

We can obtain the parameters such as R_IT, X_IT, R_2T, X_2T from the S.C. test of the transformer. Whereas the rated voltages V₁,V₂ and the rated currents I₁ and I₂ in the above expressions are known from the given transformer

The post Open Circuit and Short Circuit Test on Transformer – Experimental Set-up & Procedure appeared first on Electrical and Electronics Blog.

Fig. 1 : Connection of Current Transformer (CT) to measure current.

We have to take an important precaution while operating with a CT. The precaution is that the CT should never be operated with its secondary open circuited. The secondary should be either shorted or a small when it is not connected to the meter resistance should be connected across it. If the secondary is open circuited, then the secondary ampere turns will become zero. These secondary ampere turns generally oppose the primary turns. So as the secondary ampere turns become zero, this opposition is reduced to zero and the primary ampere turns (mmf) will produce a large flux in the core. The excessive flux has two effects on the operation. First effect is increase in core loss resulting in core heating beyond safe limits. The second effect is that a large emf is induced on the primary and secondary sides, which can damage the insulation of windings. The excessive secondary induced voltage is dangerous for the user as well. Hence operation of a CT with open circuited secondary should always be avoided. The secondary of CT is generally connected to ground to protect the user against the possible shocks.

Connection Diagram of Current Transformer (CT)

The construction of a current transformer is shown in Fig. 1 (a) and its equivalent circuit is shown in Fig. 1 (b). CT has a primary coil which is of one or more turns. In Fig. 1 (a), the bar acts as the primary. The primary of CT carries the large current I₁ which is to be measured, so the bar is of large cross sectional area. The secondary of a CT is made up of large number of turns. It is wound on a core. The secondary winding is a low current winding hence its cross sectional area is small. An ammeter of small range (typically 0-5 A) is connected across the secondary shown as Fig. 1 (b).

Working Principle of Current Transformer (CT)

CT is basically a step up transformer. Hence the secondary is a high voltage low current winding. The current is stepped down. The secondary current is given by,

\[{{I}_{2}}={{I}_{1}}\times \frac{{{N}_{1}}}{{{N}_{2}}}\]

Where,

N₂ >> N₁

The current I₂ is measured by the ammeter. So if we know the turns ratio N₁ / N₂ then it is possible to measure I₁ (i.e. the large current). The ammeter can be calibrated directly to measure the current I₁, if the turns ratio is known. Thus a CT can be used to measure a high current without actually connecting the ammeter directly in series with the high current.

\[{{I}_{1}}=\frac{{{N}_{1}}}{{{N}_{2}}}\times {{I}_{2}}\]

Advantages of Current Transformer (CT)

Some of the advantages of a CT are as follows :

1. It is possible to use the ammeters of normal current range to measure very high currents.
2. We can calibrate the same ammeter to measure different currents by adjusting the turns ratio of the CT
3. The user is safe due to the isolation provided by the current transformer between high current primary and low current secondary.
4. The CT can be used to operate indicating and protecting devices such as relays.
5. The same CT can be used to provide the measured current to several instruments.

Disadvantages  of Current Transformer (CT)

1. It can be used to measure only the ac current. The DC current cannot be measured.
2. CT cannot be operated with open circuited secondary

The post What is a Current Transformer (CT)? Working Principle, Connection Diagram, Construction & Advantages appeared first on Electrical and Electronics Blog.

Fig. 1 : 3 Phase Autotransformer.

Construction and Working Principle of Three Phase Autotransformer

Using a rotary contact which moves over each winding, we can change the 3-phase output voltage. Thus it is possible to vary the three phase output with a single rotary knob. (See Fig. 1(a)). Fig. 1(b) shows the electrical connections of the three windings. One end of all the windings is connected together. The other ends are connected to the 3 phase ac supply (R, Y, B). The variable point of each winding is brought out as output (r, y, b points in Fig. 1(b)). The dotted line in Fig. 1(b) shows that all the rotary (sliding) contacts are mechanically coupled to each other so that only one controlling knob can control the output voltage. As we move the sliding contact on each winding, the voltage XV will change. It is possible to use this as a step up or step down transformer.

Fig. 2 : Connection Diagram of Three Phase Autotransformer.

Sometimes instead of three windings, only two windings are used in the open delta connection as shown in Fig. 2.

Advantages of Three Phase Autotransformer

1. Reduction in the copper required.
2. The size and cost is reduced as compared to conventional transformer.
3. Losses in the windings reduce so efficiency is higher.
4. Better voltage regulation.
5. Variable output voltage.

Disadvantages of Three Phase Autotransformer

1. No electrical isolation between primary and secondary windings. It can prove to be dangerous.
2. If the secondary circuit is shorted then a large current will flow on the secondary side.

Applications of 3 Phase Autotransformer

1. To get variable output voltage.
2. For smooth starting of ac machines.
3. To vary the supply voltage (as per requirement) of a furnace.

The post What is 3 Phase Autotransformer? Working Principle, Connection Diagram & Applications appeared first on Electrical and Electronics Blog.

Types of Autotrnasformer

There are two types of auto-transformers. They are,

1. Step-up auto-transformer
2. Step-down auto-transformer.

Construction of Autotransformer

Even though the working principle of two winding transformer and an auto-transformer is same but power transfer in both the transformers is somewhat different. In a two winding transformer the power transfer is purely inductive whereas in case of auto-transformer, the power transfer from primary to secondary winding is both inductive and conductive. In an auto-transformer the power transfer is inductive because of the mutual induction and is conductive because both the windings i.e., primary and secondary are electrically in contact with each other.

Both step-up auto-transformer and step-down auto-transformer along with direction of flow of currents are shown in figures (l) and (2) respectively. In step-up transformer, the winding from section BC is the primary winding and the winding of section AB is the secondary winding. Hence primary winding BC is a part of secondary winding AB. Therefore, the voltage at secondary is more than the primary voltage and the transformer acts as step-up transformer. Also the condition N₂ > N₁ satisfies in this transformer. In figure (2) the arrangement of step-down transformer is shown where the entire winding AB is the primary winding and BC acts as the secondary winding. Therefore, secondary winding is a part of primary winding and the transformer acts as step-down transformer. For this case (N₁ > N₂) and (V₁ > V₂).

Saving of Copper in Autotransformer

For any winding, copper’s weight is proportional to the area of cross-section and length of the conductors. But it is known that the area and length of the copper winding are in turn proportional to the current and number of turns respectively. Hence it can be concluded that the weight of the copper used for winding is proportional to the product of current and number of turns.

Rating of two types of transformers is identical i.e., both the transformers have equal voltage ratio \(\left( \frac{{{V}_{1}}}{{{V}_{2}}} \right)\), same input and output, maximum flux density and area of core are assumed to be equal.

Weight of copper ∝ Current in winding × Number of turns of winding ∝ IN

Weight of copper in two winding transformer,

w_T  ∝ (Copper weight of primary + Copper weight of secondary)

\\(\)

And also,

Weight of copper in auto-transformer,

w_A ∝ (Copper weight of AC + Copper weight of BC)

\\(\)

Now, taking the ratio of two weights, we get,

\[\frac{{{w}_{T}}}{{{w}_{A}}}=\frac{{{I}_{1}}{{N}_{1}}+{{I}_{2}}{{N}_{2}}}{{{I}_{1}}({{N}_{1}}-{{N}_{2}})+({{I}_{2}}-{{I}_{1}}){{N}_{2}}}\]

\[\frac{{{w}_{T}}}{{{w}_{A}}}=\frac{{{I}_{1}}{{N}_{1}}+{{I}_{2}}{{N}_{2}}}{{{I}_{1}}{{N}_{1}}-{{I}_{1}}{{N}_{2}}+{{I}_{2}}{{N}_{2}}-{{I}_{1}}{{N}_{2}}}\]

\[\frac{{{w}_{T}}}{{{w}_{A}}}=\frac{{{I}_{1}}{{N}_{1}}+{{I}_{2}}{{N}_{2}}}{{{I}_{1}}{{N}_{1}}+{{I}_{2}}{{N}_{2}}-2{{I}_{1}}{{N}_{2}}}\]

\[\frac{{{w}_{T}}}{{{w}_{A}}}=\frac{{{I}_{1}}{{N}_{1}}\left[ \frac{{{N}_{1}}}{{{N}_{2}}}+\frac{{{I}_{2}}}{{{I}_{1}}} \right]}{{{I}_{1}}{{N}_{1}}\left[ \frac{{{N}_{1}}}{{{N}_{2}}}+\frac{{{I}_{2}}}{{{I}_{1}}}-2 \right]}\]

We know that,

\[\frac{{{N}_{1}}}{{{N}_{2}}}=\frac{{{I}_{2}}}{{{I}_{1}}}=\frac{1}{K}\]

\[\frac{{{w}_{T}}}{{{w}_{A}}}=\frac{\frac{1}{K}+\frac{1}{K}}{\left( \frac{1}{K}+\frac{1}{K}-2 \right)}\]

\[\frac{{{w}_{T}}}{{{w}_{A}}}=\frac{\frac{2}{K}}{\frac{2-2K}{K}}\]

\[\frac{{{w}_{T}}}{{{w}_{A}}}=\frac{2}{2(1-K)}\]

\[\frac{{{w}_{T}}}{{{w}_{A}}}=\frac{1}{(1-K)}\]

\[{{w}_{T}}(1-K)={{w}_{A}}\]

\[{{w}_{A}}={{w}_{T}}(1-K)\]

The above equation is the weight of copper in auto-transformer when compared to two winding transformer. Thus,

Saving in copper = Copper weight in two winding transformer – Copper weight in auto-transformer

\[={{w}_{T}}-{{w}_{A}}\]

\[={{w}_{T}}-{{w}_{T}}(1-K)\]

\[={{w}_{T}}-{{w}_{T}}+K{{w}_{T}}\]

\[\text{Saving in copper = }K{{w}_{T}}\]

The above saving is for step down auto-transformer.

Similarly for step-up auto-transformer,

\[\text{Saving in copper = }\frac{1}{K}{{w}_{T}}\]

Advantages of Autotransformer

1. High volt-ampere rating can be transferred.
2. Voltage variation is smooth and continuous.
3. Amount of copper required is less.
4. Higher voltage regulation due to less resistance and leakage reactance.
5. Efficiency is more.
6. It is economical due to its small size.

Applications of Autotransformer

1. Autotransformers are used as starter for securely starting the machines like induction motor and synchronous motors.
2. These are used as boosters to provide a small boost to a distribution cable to compensate the voltage drop.
3. These are also used as furnace transformers for providing required supply to the furnaces.
4. These are used as inter connecting transformers in 132 kV/330 kV systems.
5. These are also used as variac.
6. These are used in various control systems and home appliances.

The post What is an Autotransformer? Working Principle, Construction, Types & Applications appeared first on Electrical and Electronics Blog.

1. To stabilize the voltage
2. To control the voltage.

Tap changing transformer uses the principle of regulating the voltage is based on changing the transformation ratio by altering the number of active tums in one of the winding of transformer.

Figure 1: Tap Changing Transformer.

Tappings can be provided at either primary or secondary winding of transformer as shown in figure. Number of active tums in any winding can be changed by changing it’s tappings. This can be done by the equipment known as tap changer

If secondary active tums (N₂) are reduced by the tap changer, then the resultant output voltage (E₂) will be reduced or increase in active primary turns (N₁) by using tap changer and also results in reduced voltage at the output.

Since,

\[{{E}_{2}}=\frac{{{E}_{1}}{{N}_{2}}}{{{N}_{1}}}\]

Thus,

E₂ decreases if N₁ increases or E₂ decreases if N₂ decreases

Similarly, decrease in active primary turns or increase in active secondary turns will result in increased output voltage. It is clear that voltage regulation is achieved with the help of tap changer by changing active tums in one of the windings. The necessity for voltage regulation by the tap changer is due to the following requirements.

1. To change the secondary voltage to the desired value.
2. To match the transformer output voltage with the consumer terminal voltage.
3. To keep the output voltage constant though the input voltage is varying.
4. To control the real and reactive power flow in the network.
5. To provide the neutral point.

Off-Load Tap Changing Transformer

Figure 1: Off-Load Tap Changing Transformer.

Off-load tap charger is used to change the transformer turns ratio. In this type of tap changer, the tapping is changed when the transformer is de-energized i.e., in order to change the tap settings the transformer has to be switched-off and then settings can be changed. The generalized circuit of an offload tap changer is as shown in figure. The taps are usually provided on the H. V side of the transformer. Here different settings are provided, i.e., 25%, 50%, 75%, and 100% . Depending on the required voltage, the load can be connected between the appropriate taps. Suppose the given transformer has the ratings, 500/1000 V Now if the load requirement is 500 V then the terminals should be connected between 0% and 50% and if the load requirement is 750 V, then the load terminals should be connected between 0% and 75%. Hence, a smooth control of voltage is possible depending on the requirements. Off-load tap changers are used for seasonal voltage variations which does not come frequently.

On-Load Tap Changing Transformer

Figure 1: On-Load Tap Changing Transformer.

The main draw back of off-load tap changer is that, it can’t be operated by keeping the transformer energized i.e., the transformer has to be disconnected in order to change the settings. However, this can be overcome by employing on-load tap changer, where the settings can be changed when the transformer is in operating condition, this ensures reliability.

However care must be taken to see that short circuit of any part of tapped winding is prevented. Also, alternate path must be provided for load current using alternate circuit during the process of tap changing in main circuit otherwise huge sparking will occur and may damage the transformer. The generalized circuit of on-load tap changer is as shown in figure. The tappings are provided on the H. V side. A, B, C, D and E are the studs connected to different tappings, r is a centre tapped reactor connected through switches 1 and 2 to the stud terminals by using movable terminals q and p. The function of the reactor is to prevent the short circuit of the tapped windings. The above figure shows the arrangement of tap changer for full supply voltage when both the terminals p and q are in contact with stud E. Now in order to have a change in voltage, the terminals must move to desired stud i.e., D, C, B, A. To do this, first of all the switch 2 must be opened and q must be moved to desired stud. Now, close switch 2 and open switch 1 and move terminal p also to the desired stud.

Therefore, while changing the terminals p and q to the desired position. One switch is kept open and other is closed so that there is no interruption in supply The on-load tap changer is used for voltage variations which come frequently.

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Figure 1: Open Delta Connection of Transformer.

In delta-delta connection of three 1-ϕ transformers, if one of the transformer is damaged and removed from the circuit and if the primaries are energized by 3-ϕ supply, then the 3-phase voltages obtained at the terminals of secondary will still remain balanced, ensuring continuous supply of 3-phase load with remaining two transformers. The connection thus obtained is known as open-delta (or) V-V connection. The total kVA rating of open delta connection is less than sum of the kVA rating of the remaining two transformers i.e., the total load that can be supplied by V-V bank is less than two-third of the capacity of delta-delta connection.

Open-delta connection of transformer is shown in figure 1. In this type of connection, the line current supplied to the load I_L is equal to the phase current in each transformer (I_ph). The 3-phase power rating for open delta connection is given by,

\[{{P}_{V.V}}=\sqrt{3}{{V}_{L}}{{I}_{L}}=\sqrt{3}{{V}_{L}}{{I}_{ph}}…(1)\]

Where,

V_L = Line voltage across the load.

From delta-delta connection, the 3-phase power rating is given by,

\[{{P}_{\Delta .\Delta }}=\sqrt{3}{{V}_{L}}{{I}_{L}}=\sqrt{3}{{V}_{L}}\left( \sqrt{3}{{I}_{ph}} \right)=3{{V}_{L}}{{I}_{ph}}…(2)\]

Dividing equations (1) & (2), we get,

\[\frac{{{P}_{V.V}}}{{{P}_{\Delta .\Delta }}}=\frac{\sqrt{3}{{V}_{L}}{{I}_{ph}}}{3{{V}_{L}}{{I}_{ph}}}=\frac{\sqrt{3}}{3}=0.577\simeq 58%\]

Thus the power available in open-delta connection is only 58% of the power available in original delta-delta connection.

Advantages of V-V Connections

A V-V Connection can work as automatic standby for a Δ-Δ connection under the failure of any one winding of Δ-Δ connection.

1. Power handling capacity of V-V connection is \(\frac{1}{\sqrt{3}}\) times of Δ-Δ connection.
2. Hence for small load V-V connection can be used instead of Δ-Δ connection.
3. If the future load increases, the open-delta can be closed to increase the rating.
4. Though open delta connection leads to small imbalance in voltage yet it is generally, unnoticed by motor loads and other commercial loads.

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Figure 1.

As the primary winding of transformer is purely inductive hence during steady state operation the flux (ϕ) lags the supply voltage by 90º as shown in figure (1). From figure (1), it is seen that when t = 0 the supply voltage is zero and the flux is maximum negative.

Now consider the transient condition when the transformer is switched ON at the instant when applied voltage is zero.

Figure 2.

Ideally, the flux at this moment of switching should be maximum negative. But practically, it is impossible to develop maximum flux within zero time. Hence, the flux will start from zero and attains a value equal to twice the value of steady state flux (ϕ_m) as shown in figure (2).Such effect is known as doubling effect. Due to high amount of flux, core of transformer is saturated and high magnetizing current is drawn by primary from the source. This current is called inrush current.

Nature of Transformer Inrush Current

Figure 3.

The nature of inrush current is that it decays slowly. The magnitude of inrush current is very high i.e., of the order of 8 to 30 times the full load current of transformer. Inrush current has peaky non-sinusoidal waveform as shown in figure (3).

Problems Associated with Transformer Inrush Current

Figure 4.

The inrush current causes many problems like interruptions in fuse or circuit breaker, injection of noise and harmonics into the mains. It also leads to mechanical stress on transformer thereby affecting the magnetic property of the core of transformer. Also, high inrush current requires over-sizing of fuses or circuit breaker-s.

To prevent inrush current phenomenon the primary of transformer should be switched ON at the instant when applied voltage is maximum positive. When applied voltage is maximum positive, the flux which lags applied voltage by 90º will start from zero and attains maximum value equal to steady state flux as shown in figure (4). Thus, inrush current is absent in this situation.

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