Electric Machines I Transformers
Dr. Firas Obeidat
1 Table of contents
1 • Introduction 2 • Working Principle of Transformer 3 • Flux in a Transformer 4 • Ideal Transformer 5 • E.M.F. Equation of Transformer 6 • Turns Ratio of Transformer 7 • Rules for Referring Impedance 8 • Equivalent Circuit of a Transformer
2 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Table of contents • Transformer Phasor Diagram with 9 Different Loads 10 • Exact Equivalent Circuit 11 • Approximate Equivalent Circuit 12 • Voltage Regulation of a Transformer 13 • Efficiency of a Transformer 14 • Transformer Tests 15 • Autotransformer 16 • Instrument Transformers
3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Introduction
A transformer is a device that converts one AC voltage to another AC voltage at the same frequency. It consists of one or more coil(s) of wire wrapped around a common ferromagnetic core. These coils are usually not connected electrically together. However, they are connected through the common magnetic flux confined to the core.
Transformer characteristics
• It has no moving parts, • No electrical connection between the primary and secondary windings, • Windings are magnetically coupled, • Rugged and durable in construction, • Efficiency is very high i.e., more than 95 %, • Frequency is unchanged.
4 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Working Principle of Transformer
There are two types of windings in a single-phase transformer. These are called primary and secondary windings or coils. The primary winding is connected to the alternating voltage source and the secondary winding is connected to the load. A sinusoidal current flows in the primary winding when it is connected to an alternating voltage source. This current establish a flux u which moves from the primary winding to the secondary winding through low reluctance magnetic core. About 95 % of this flux moves from the primary to the secondary through the low reluctance path of the magnetic core and this flux is linked by the both windings and a small percent of this flux links to the primary winding.
According to the Faradays laws of electromagnetic induction, a voltage will be induced across the secondary winding as well as in the primary winding. Due to this voltage, a current will flow through the load if it is connected with the secondary winding. Hence, the primary voltage is transferred to the secondary winding without a change in frequency.
5 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Flux in a Transformer
The current in the primary winding establishes a flux. The flux that moves from primary to secondary and links both the windings is called the mutual flux and its maximum value is represented by ϕm.
Flux which links only the primary winding and completes the magnetic path through the surrounding air is known primary leakage flux ϕ1l.
The secondary leakage flux ϕ2l is that flux which links only the secondary winding and completes the magnetic path through the surrounding air.
6 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Ideal Transformer
An ideal transformer is one which does not supply any energy to the load i.e., the secondary winding is open circuited.
The main points of an ideal transformer are • No winding resistance, • No leakage flux and leakage inductance, • Self-inductance and mutual inductance are zero, • No losses due to resistance, inductance, hysteresis or eddy current, • Coefficient of coupling is unity.
A small magnetizing current Im will flow in the primary winding when it is connected to the alternating voltage source, V1. This magnetizing current lags behind the supply voltage, V1 by 90° and produces the flux ϕ, which induces the primary and secondary emfs. These emfs lag behind the flux, ϕ by 90°. The magnitude of primary induced emf E1 and supply voltage V1 is the same, but are 180° out of phase.
7 Dr. Firas Obeidat Faculty of Engineering Philadelphia University E.M.F. Equation of Transformer The primary winding draws a current when it is connected to an alternating voltage source. This primary sinusoidal current produces a sinusoidal flux u that can be expressed as
Instantaneous emf induced in the primary winding is,
Instantaneous emf induced in the secondary winding is,
The maximum value of e1 is,
The rms value of the primary emf is,
8 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Turns Ratio of Transformer
The turns ratio is used to identify the step-up and step-down transformers.
According to Faraday’s laws, the induced emf in the primary e1 and the secondary e2 windings are,
where a is the turns ratio of a transformer.
N2>N1, the transformer is called a step-up transformer. N1>N2, the transformer is called a step-down transformer. 9 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Turns Ratio of Transformer
The losses are zero in an ideal transformer. In this case, the input power of the transformer is equal to its output power and this yields,
The above equation can be rearranged as
The input and output power of an ideal transformer is,
For an ideal condition, the angle ϕ1 is equal to the angle ϕ2
푉1 푃 = 푉 퐼 푐표푠휙 = 푎퐼 푐표푠휙 = 푃 표푢푡 2 2 2 푎 1 1 𝑖푛
10 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Turns Ratio of Transformer Example: The number of turns in the secondary coil of a 22 kVA, 2200V/220V single-phase transformer is 50. Find (i) number of primary turns, (ii) primary full load current, (iii) secondary full load current. Neglect all kinds of losses in the transformer.
11 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Turns Ratio of Transformer Example: A 25 kVA single-phase transformer has the primary and secondary number of turns of 200 and 400, respectively. The transformer is connected to a 220 V, 50 Hz source. Calculate (i) turns ratio, (ii) mutual flux in the core.
12 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Rules for Referring Impedance
For developing equivalent circuit of a transformer, it is necessary to refer the parameters from the primary to the secondary or the secondary to the primary. These parameters are resistance, reactance, impedance, current and voltage. The impedances in the primary and secondary windings are,
The impedance ratio is equal to the square of the turns ratio. The important points for transferring parameters are 2 (i) R1 in the primary becomes (R1/a ) when referred to the secondary, 2 (ii) R2 in the secondary becomes (a R2) when referred to the primary, 2 (iii) X1 in the primary becomes (X1/a ) when referred to the secondary, 2 (iv) X2 in the secondary becomes (a X2) when referred to the primary. 13 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Rules for Referring Impedance Example: The number of primary and secondary turns of a single-phase transformer are 300 and 30, respectively. The secondary coil is connected with a load impedance of 4Ω. Calculate (i) turns ratio, (ii) load impedance referred to the primary, (iii) primary current if the primary coil voltage is 220 V.
14 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Equivalent Circuit of a Transformer
The losses that occur in real transformers have to be accounted for in any accurate model of transformer behavior. The major items to be considered in the construction of such a model are
Copper (I2R) losses. Copper losses are the resistive heating losses in the primary and secondary windings of the transformer. They are proportional to the square of the current in the windings.
Eddy current losses. Eddy current losses are resistive heating losses in the core of the transformer. They are proportional to the square of the voltage applied to the transformer.
Hysteresis Losses. Hysteresis losses are associated with the rearrangement of the magnetic domains in the core during each half-cycle. They are a complex, nonlinear function of the voltage applied to the transformer. Leakage flux. The fluxes which escape the core and pass through only one of the transformer windings are leakage fluxes. These escaped fluxes produce a leakage inductance in the primary and secondary coils, and the effects of this inductance must be accounted for. 15 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Equivalent Circuit of a Transformer
Windings of a transformer are not connected electrically. The windings are magnetically coupled with each other. In the equivalent circuit, the related parameters need to be transferred either from the primary to the secondary or vice versa.
Copper losses are resistive losses in the primary and secondary windings of the
transformer core. They are modeled by placing a resistor R1 in the primary circuit of the transformer and a resistor R2 in the secondary circuit.
The leakage flux will be modeled by primary inductor X1 and secondary inductor X2.
The magnetization current im is a current proportional (in the unsaturated region) to the voltage applied to the core and lagging the applied voltage by o 90 , so it can be modeled by a reactance XM connected across the primary voltage source.
The core-loss current ih+e is a current proportional to the voltage applied to the core that is in phase with the applied voltage, so it can be modeled by a
resistance Rc connected across the primary voltage source. 16 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Equivalent Circuit of a Transformer
The figure below is an accurate model of a transformer, it is not a very useful one. To analyze practical circuits containing transformers, it is normally necessary to convert the entire circuit to an equivalent circuit at a single voltage level.
Primary Primary No-load Secondary Secondary voltage resistance current resistance voltage
Primary Primary Magnetization Secondary reactance Secondary current reactance reactance current Ideal transformer
Hysteresis & eddy current
Core resistance Magnetization current
17 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Transformer Phasor Diagram with Different Loads
퐸2 = 푉2 + 퐼2푍2
푉1 = −퐸1 + 퐼1푍1
ϕm
I1 Io I1R1 -I2 -E1 E1 E2 X1 I1 I1Z1 R Load
I2Z2 V1 I2 I2X2 V2
I2R2 18 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Transformer Phasor Diagram with Different Loads
ϕm
I1
Io
I1R -I2 I1X1 1 I -E E1 E2 1Z1 1 RL Load I2 V2 Z2 V1
I2X I2R 2 I2 2
ϕm
I1 Io
I1R1 -I2 I1X1 -E1 E1 E2 RC I1Z1
I2Z2 Load V1 I2
V2 I2X2
I2R2 19 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Exact Equivalent Circuit
The figure is the exact equivalent circuit referred to the primary where all the parameters are transferred from the secondary to the primary and these parameters are
20 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Exact Equivalent Circuit
The figure is the exact equivalent circuit referred to the secondary where all the parameters are transferred from the primary to the secondary and these parameters are
′ 퐼ℎ+푒 = 푎퐼ℎ+푒
21 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Approximate Equivalent Circuit
The no-load current is very small as compared to the rated primary current.
Therefore, there is a negligible voltage drop due to R1 and X1. In this condition, it can be assumed that the voltage drop across the no-load circuit is the same as the applied voltage without any significant error. The approximate equivalent circuit can be drawn by shifting the no-load circuit across the
supply voltage, V1.
The figure is the approximate equivalent circuit referred to the primary where all the parameters are transferred from the secondary to the primary and these parameters are
′ 2 푅푒푞 = 푅1 + 푅2 = 푅1 + 푎 푅2
′ 2 푋푒푞 = 푋1 + 푋2 = 푋1 + 푎 푋2
푍푒푞 = 푅푒푞 + 푗푋푒푞
푉1 푉1 푅푐 = 푋푚 = 퐼ℎ+푒 퐼푚
22 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Approximate Equivalent Circuit
The figure is the approximate equivalent circuit referred to the secondary where all the parameters are transferred from the primary to the secondary and these parameters are
푅1 푅 = 푅 + 푅 ′ = 푅 + 푒푞 2 1 2 푎2
푋1 푋 = 푋 + 푋 ′ = 푋 + 푒푞 2 1 2 푎2
푍푒푞 = 푅푒푞 + 푗푋푒푞
′ ′ 푉1 푅푐 = ′ 퐼ℎ+푒
′ ′ 푉1 푋푚 = ′ 퐼푚
23 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Approximate Equivalent Circuit Example: A 2.5 kVA, 200V/40V single-phase transformer has the primary resistance and reactance of 3 and 12 Ω, respectively. On the secondary side, these values are 0.3 and 0.1 Ω, respectively. Find the equivalent impedance referred to the primary and the secondary. 푉1 200 푎 = = = 5 푉2 40 The total resistance, reactance and impedance referred to the primary can be determined as, 2 푅푒푞 = 푅1 + 푎 푅2 = 3 + 25 × 0.3 = 10.5Ω 2 푋푒푞 = 푋1 + 푎 푋2 = 12 + 25 × 0.1 = 14.5Ω
푍푒푞 = 푅푒푞 + 푗푋푒푞 = 10.5 + 푗14.5 The total resistance, reactance and impedance referred to the secondary are calculated as,
푅1 3 푅 = 푅 + = 0.3 + = 0.42Ω 푒푞 2 푎2 25
푋1 12 푋 = 푋 + = 0.1 + = 0.58Ω 푍푒푞 = 푅푒푞 + 푗푋푒푞 = 0.42 + 푗0.58 푒푞 2 푎2 25 24 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Voltage Regulation of a Transformer
The voltage regulation (VR) of a transformer is defined as the difference
between the no-load terminal voltage (V2NL) to full load terminal voltage (V2FL) and is expressed as a percentage of full load terminal voltage. It is therefore can be expressed as,
푉2푁퐿 − 푉2퐹퐿 푉표푙푡푎푔푒 푅푒푔푢푙푎푡푖표푛 = × 100% 푉2퐹퐿
Since at no load, V2 = V1/a, the voltage regulation can also be expressed as
푉1 − 푉2퐹퐿 푉표푙푡푎푔푒 푅푒푔푢푙푎푡푖표푛 = 푎 × 100% 푉2퐹퐿
Usually it is a good practice to have as small a voltage regulation as possible. For an ideal transformer, Voltage Regulation = 0 percent. It is not always a good idea to have a low-voltage regulation, sometimes high-impedance and high-voltage regulation transformers are deliberately used to reduce the fault currents in a circuit.
25 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Efficiency of a Transformer
The efficiency (η), of the transformer can be defined as the ratio of its output power to the input power. Mathematically, it can be expressed as,
푃표푢푡 푃표푢푡 휂 = × 100% = × 100% 푃𝑖푛 푃표푢푡 + 푃푙표푠푠
푃표푢푡 = 푉2퐼2푐표푠휃2
푃푙표푠푠 = 푃푐푢 + 푃𝑖푟표푛 There are three types of losses present in transformers:
1- Copper Losses
Copper losses occur due to the primary and the secondary resistances. The full load copper losses can be determined as 2 2 푃푐푢 = 퐼1 푅1 + 퐼2 푅2 2 푃푐푢 = 퐼2 푅푒푞 26 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Efficiency of a Transformer The iron loss of a transformer is often called as core loss, which is a result of an alternating flux in the core of the transformer. The iron loss consists of the eddy current loss and the hysteresis loss.
푃𝑖푟표푛 = 푃푒 + 푃ℎ 2- Eddy Current Losses A current will flow in that parts of the transformer. This current does not contribute in output of the transformer but dissipated as heat. This current is known as eddy current and the power loss due to this current is known as
eddy current loss. The eddy current loss Pe is directly proportional to the square of the frequency (f ) times the maximum magnetic flux density Bm and the eddy current loss can be expressed as, 2 2 푃푒 = 푘푒푓 퐵푚 where ke is the proportionality constant. 3- Hysteresis Losses th The hysteresis loss Ph is directly proportional to the frequency (f ) and 2.6 power of the maximum magnetic flux density (Bm) and the expression of hysteresis loss is 푛 푃ℎ = 푘ℎ푓퐵푚 where kh is the proportionality constant. n=1.5-2.5 27 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Efficiency of a Transformer
Practically, hysteresis loss depends on the voltage and the eddy current loss depends on the current. Therefore, total losses of the transformer depend on the voltage and the current not on the power factor. That is why the transformer rating is always represented in kVA instead of kW.
푉2퐼2푐표푠휃2 푉2퐼2푐표푠휃2 휂 = × 100% = 2 2 × 100% 푉2퐼2푐표푠휃2 + 푃푐푢 + 푃𝑖푟표푛 푉2퐼2푐표푠휃2 + 퐼1 푅1 + 퐼2 푅2 + 푃𝑖푟표푛
Condition for Maximum Efficiency
푉2퐼2푐표푠휃2 휂 = 2 × 100% 푉2퐼2푐표푠휃2 + 퐼2 푅푒푞 + 푃𝑖푟표푛
To find the condition for maximum efficiency, put the derivative of the
efficiency with respect to I2 equal to zero.
푑휂푚푎푥 = 0 푑퐼2
28 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Efficiency of a Transformer
2 푑휂푚푎푥 푉2퐼2푐표푠휃2 + 퐼2 푅푒푞 + 푃𝑖푟표푛 푉2푐표푠휃2 − 푉2퐼2푐표푠휃2(푉2푐표푠휃2 + 2퐼2푅푒푞) = = 0 2 2 푑퐼2 (푉2퐼2푐표푠휃2 + 퐼2 푅푒푞 + 푃𝑖푟표푛)
2 푉2퐼2푐표푠휃2 + 퐼2 푅푒푞 + 푃𝑖푟표푛 푉2푐표푠휃2 − 푉2퐼2푐표푠휃2(푉2푐표푠휃2 + 2퐼2푅푒푞) = 0
2 2 푉2퐼2푐표푠휃2 + 퐼2 푅푒푞 + 푃𝑖푟표푛 − 푉2퐼2푐표푠휃2 − 2퐼2 푅푒푞 = 0
2 푃𝑖푟표푛 − 퐼2 푅푒푞 = 0
2 푃𝑖푟표푛 = 퐼2 푅푒푞 푃𝑖푟표푛 = 푃퐶푢
푃𝑖푟표푛 퐼2휂푚푎푥 = 푅푒푞
푃𝑖푟표푛 푃𝑖푟표푛 푉퐼2휂푚푎푥 = 푉퐼2푟푎푡푒푑 2 = 푉퐼2푟푎푡푒푑 퐼2푟푎푡푒푑 푅푒푞 푃푐푢 푟푎푡푒푑
29 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Efficiency of a Transformer
Example: A 30 kVA transformer has the iron loss and full load copper loss of 350 and 650 W, respectively. Determine the (i) Full load efficiency, (II) 0.8 load efficiency, (iii) Output kVA corresponding to maximum efficiency, and (iv)Maximum efficiency. Consider that the power factor of the load is 0.6 lagging. 30 × 0.6 18 (i) 휂 = × 100% = × 100% = 94.7% 30 × 0.6 + 0.65 + 0.35 19
(ii) At half load 푃표푢푡 = 0.8 × 30 × 0.6 = 14.4푘W
푃𝑖푟표푛 = 350W = 0.35kW 2 2 푃푐표푝푝푒푟_0.8퐹퐿 = 0.8 푃푐표푝푝푒푟_퐹퐿 = 0.8 × 0.65kW = 0.416kW 14.4 14.4 휂 = × 100% = × 100% = 94.9% 14.4 + 0.416 + 0.35 15.166
푃𝑖푟표푛 350 (iii) 푉퐼2휂푚푎푥 = 푉퐼2푟푎푡푒푑 2 = 30 = 22푘푉퐴 퐼2푟푎푡푒푑 푅푒푞 650
(iv) 푃휂푚푎푥 = 22 × 0.6 = 13.2푘W At maximum efficiency iron losses is equal to copper losses, so
푃𝑖푟표푛 = 푃푐표푝푝푒푟 = 350W = 0.35kW
푉2퐼2푐표푠휃2 13.2 휂푚푎푥 = × 100% = × 100% = 94.9% 푉2퐼2푐표푠휃2 + 푃푐표푝푝푒푟 + 푃𝑖푟표푛 13.2 + 0.35 + 0.35 30 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Transformer Tests Open Circuit Test
The main objectives of the open circuit test are to determine the no-load current and iron loss. The components of the no-load current are used to determine the no-load circuit resistance and reactance. In an open circuit test, the high voltage side is considered to be open circuit, and the low voltage coil is connected to the source, where all measuring instruments are connected in the low voltage side. A specific alternating voltage is applied to the low voltage winding. Then the wattmeter will measure the iron loss and small amount of copper loss. The ammeter and voltmeter will measure the no-load current and the voltage, respectively. Since, the no-load current is very small, the copper losses can be neglected.
A W + Io LV HV Ih+e Im
Rc X V V V m
-
31 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Transformer Tests Open Circuit Test
The wattmeter reading can be expressed as
푃표 = 푉퐼표푐표푠휃표 The no load power factor can be determined as
푃표 푐표푠휃표 = 푉퐼표 The no load current components can be determined as
퐼ℎ+푒 = 퐼표푐표푠휃표
퐼푚 = 퐼표푠푖푛휃표 The no load resistance and reactance can be determined as 푉 푅푐 = 퐼ℎ+푒 푉 푋푚 = 퐼푚 32 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Transformer Tests Open Circuit Test
Example: A 200/400V, 50Hz single-phase transformer has the no-load test data of 200V, 0.6A, 80W. Calculate the no-load circuit resistance and reactance.
푃표 80 푐표푠휃표 = = = 0.67 푉퐼표 200 × 0.6 휃 = 푐표푠−10.67 =47.9
푠푖푛47.9 = 0.74
퐼ℎ+푒 = 퐼표푐표푠휃표 = 0.6 × 0.67 = 0.4퐴
퐼푚 = 퐼표푠푖푛휃표 = 0.6 × 0.74 = 0.44A
푉 200 푅푐 = = = 500Ω 퐼ℎ+푒 0.4 푉 200 푋푚 = = = 454.5Ω 퐼푚 0.44 33 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Transformer Tests Short Circuit Test
The main objectives of the short circuit test are to determine the equivalent resistance, reactance, impedance and full load copper loss.
The supply voltage and the measuring instruments (e. g,. wattmeter, ammeter) are connected to the high voltage side and the low voltage winding is connected with ammeter. The voltage is adjusted until the current in the low voltage winding is equal to the rated low voltage side current. Under this condition, the wattmeter will measure the full load copper loss
The full load copper loss and it can be written as A W 푃푠푐 = 푉푠푐퐼푠푐푐표푠휃푠푐 HV LV The short circuit power V A factor can be determined as V
푃푠푐 푐표푠휃푠푐 = 푉푠푐퐼푠푐
34 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Transformer Tests Short Circuit Test
The equivalent impedance can be calculated as,
푉푠푐 푍푒푞 = 퐼푠푐 The equivalent resistance and reactance can be calculated as
푅푒푞 = 푍푒푞푐표푠휃푠푐
푋푒푞 = 푍푒푞푠푖푛휃푠푐
Example: A 25kVA, 2200/220V, 50Hz single-phase transformer’s low voltage side is short-circuited and the test data recorded from the high voltage side are
P=150W, I1=5A and V1=40V. Determine the (i) equivalent resistance, reactance and impedance referred to primary, (ii) equivalent resistance, reactance and impedance referred to secondary, and (iii) voltage regulation at unity power factor.
푉푠푐 40 150 (i) 푍푒푞 = = = 8Ω 푐표푠휃푠푐 = = 0.75 퐼푠푐 5 40 × 5 35 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Transformer Tests Short Circuit Test
휃푠푐 = 41.4 푠푖푛41.4 = 0.66
푅푒푞 = 푍푒푞푐표푠휃푠푐 = 8 × 0.75 = 6Ω
푋푒푞 = 푍푒푞푠푖푛휃푠푐 = 8 × 0.66 = 5.28Ω (ii) 푉1 2200 푎 = = = 10 푉2 220 The parameters referred to the secondary
푍푒푞_푝 8 푅푒푞_푝 6 푍 = = = 0.08Ω 푅 = = = 0.06Ω 푒푞_푠 푎2 100 푒푞_푠 푎2 100 푋푒푞_푝 5.28 푋 = = = 0.0528Ω 푒푞_푠 푎2 100 (iii) 25000 퐼2 = = 113.6A 퐸 = 푉 + 퐼 푍 = 220 + 113.6 × 0.08 = 229V 220 2 2 2 푒푞_푠
퐸2 − 푉2 229 − 220 푉푅 = = = 4% 푉2 220 36 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Autotransformer
Autotransformer usually used in the educational laboratory as well as in the testing laboratory.
An autotransformer has one continuous winding that is common to both the primary and the secondary. Therefore, in an autotransformer, the primary and secondary windings are connected electrically.
Autotransformer Advantages over a two-winding transformer lower losses lower lower lower initial compared to leakage excitation investment conventional reactance current transformer
37 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Autotransformer
+ + Nc: indicates the common winding. Vp Nc N V Ns: indicates the series winding. s s - - Conventional transformer + Is Ns IL The total voltage in the primary side is V1 + Nc VL Ic From the above equation - - Step down autotransformer
+ Is Ns The load current can be written as IL VL + Nc
Ic V1 - - Step up autotransformer 38 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Autotransformer Example: A single-phase 120kVA, 2200/220V, 50Hz transformer which is connected as an autotransformer. The voltages of the upper and lower parts of the coil are 220 and 2200 V, respectively. Calculate the kVA rating of the autotransformer.
39 Dr. Firas Obeidat Faculty of Engineering Philadelphia University Instrument Transformers
The magnitude of the voltage and the current are normally high in the power system networks. To reduce the magnitude of the voltage and current, instrument transformers are used.
The current transformer is connected in series with the line to step down the high magnitude of the current to a rated value for the ammeter and the current coil of the wattmeter.
The potential transformer is used to step down to the voltage to a suitable value of the voltage at the secondary for supplying the voltmeter and the voltage coil of the wattmeter.
40 Dr. Firas Obeidat Faculty of Engineering Philadelphia University 41