Transformer Phasing: The Dot Notation and Dot Convention

Posted by: Electrical Technology 12/02/2013 in Transformer 2 Comments

Transformer Phasing: The Dot Notation and Dot Convention
The Dot Notation
Generally, when we study about Transformers, we assume that the primary and secondary
voltage and currents are in phase. But, such is not always the case. In Transformer, The
phase relation between primary and secondary currents and voltages depends on how
each winding is wrapped around the core.

Refer to fig (1) and (2), you may see that the primary sides of both transformers are
identical i.e. primary windings of both transformers wrapped in the same direction around
the core.
But in fig (2) you may notice that the secondary winding is wound around the core in the
opposite direction from the secondary winding in fig (1).
Consequently, the voltage induced in the Secondary winding in fig (2) is 180° out of phase
as compared with the induced voltage in secondary in fig (1) and the direction of secondary
current (IS) is opposite from the primary current (IP)

So we see that
1. The primary and secondary voltage and current are in phase in fig (1)
2. The primary and secondary voltage and current are 180° out of phase in fig (2)
Dot Convention
To eliminate any confusion in the phase relation between primary and secondary voltage
and current, a dot convention has been adopted for transformer schematic diagrams. Dots
are placed on the top of primary and secondary terminals as shown in fig (3) and (4)
In fig (3), we see that dots are placed at the top in both primary and secondary terminals. It
shows that the primary and secondary current and voltages are in phase. Moreover, the
primary and secondary voltages (VP and VS) have similar sine wave, also the primary and
secondary (IP and IS) currents are same in direction.
The story is opposite in fig (4). We can see that one dot is positioned at the top in primary
terminal and the other one (dot) is placed at bottom of secondary terminal. It shows that
the primary and secondary current and voltages are 180° out of phase. In addition, the
primary and secondary voltages (VP and VS) sine waves are opposite to each other. Alsothe
primary and secondary currents (IPand IS) are opposite in direction.
How to Calculate/Find the Rating of Transformer in kVA (Single
Phase and Three Phase)?
Posted by: Electrical Technology 07/13/2013in AC Fundamentals, Alternating Current, Basic
Concepts, Basic/Important Electrical Formulas, How To,Questions/Answers (Electrical), Single Phase AC
Circuits, Three Phase AC Circuits, Transformer35 Comments
How to Calculate/Find the Rating of Transformer in kVA (Single Phase and Three Phase)?
We know that, Transformer always rated in kVA. Below are two simple formulas to find the
rating of Single phase and Three phase Transformers.

Rating of Single Phase Transformer:

P = V x I.

Rating of a single phase transformer in kVA

kVA= (V x I) / 1000

Rating of a Three Phase Transformer:

P = √3. V x I

Rating of a Three phase transformer in kVA

kVA = (√3. V x I) /1000

But Wait, A question is raised here… Look at the General nameplate rating of a
100kVAtransformer.

Did you notice something????Anyway, I don’t care what is your answer  but lets me try to
explain.

Here is the rating of Transformer is 100kVA.

But Primary Voltages or High Voltages (H.V) is 11000 V = 11kV.

And Primary Current on High Voltage side is 5.25 Amperes.

Also Secondary voltages or Low Voltages (L.V) is 415 Volts

And Secondary Current (Current on Low voltages side) is 139.1 Amperes.

In simple words,

Transformer rating in kVA = 100 kVA

Primary Voltages = 11000 = 11kV

Primary Current = 5.25 A

Secondary Voltages = 415V

Secondary Current = 139.1 Amperes.

Now calculate for the rating of transformer according to

P=V x I (Primary voltage x primary current)

P = 11000V x 5.25A = 57,750 VA = 57.75kVA

Or P = V x I (Secondary voltages x Secondary Current)

P= 415V x 139.1A = 57,726 VA = 57.72kVA

Once again we noticed that the rating of Transformer (on Nameplate) is 100kVA but according
to calculation…it comes about 57kVA…

The difference comes due to ignorance of that we used single phase formula instead of three
phase formula.

Now try with this formula

P = √3 x V x I

P=√3 Vx I (Primary voltage x primary current)

P =√3 x 11000V x 5.25A = 1.732 x 11000V x 5.25A = 100,025 VA = 100kVA

Or P = √3 x V x I (Secondary voltages x Secondary Current)

P= √3 x 415V x 139.1A = 1.732 x 415V x 139.1A= 99,985 VA = 99.98kVA

Consider the (next) following example.

Voltage (Line to line) = 208 V.

Current (Line Current) = 139 A

Now rating of the three phase transformer

P = √3 x V x I

P = √3 x 208 x 139A = 1.732 x 208 x 139

P = 50077 VA = 50kVA

Transformer Efficiency, All day Efficiency & Condition for

Maximum Efficiency
Posted by: Electrical Technology 12/01/2013 in Transformer Leave a comment
Transformer Efficiency, All day Efficiency & Condition for maximum Efficiency
Click image to enlarge

Efficiency of Transformer
Transformer efficiency may be defined as the ratio between Output and Input.
Efficiency = Output/Input
On specified Power factor and load, the Transformer efficiency can be found by dividing its
output on Input (Similar to other Electrical Machines i.e. motors, generators etc). But the
values of both input and Output should be same in unites (i.e. in Watts, kilowatts,
megawatts etc)
But note that a transformer has very high efficiency because the losses occur in
transformer is very low. Since the Input and Output almost equal, therefore measurement
of input and output is not possible practically. The best way to find the transformer
 efficiency is that, first determine the losses in transformer and then calculate the
transformer efficiency with the help of these losses.
Efficiency = η= Output / Input
Efficiency = η= Output / (Output + Losses) …….… (As Input = Output +Losses)
Efficiency = η= Output / (Output +Cupper Losses + Iron Losses)
You may also find the Efficiency by the following formula
Efficiency = η= Output / Input
Efficiency = η = (Input – Losses) / Input …….… (As Output = Input – Losses)
Taking LCM
Efficiency = η = 1 – (Losses /Input)
As we know that the rating of Transform is expressed in kVA not in kW. But the efficiency
doesn’t depend on VA i.e. it would be expressed in Power Watts (kW) not in kVA. Although,
the Losses are directly proportional to VA (Volt-Amperes), thus, efficiency depends on
Power factor on every kind of VA load. And the efficiency would be maximum on unity (1)
Power factor.
Good to Know
 We can also find Transformer Efficiency by determining:
 Core losses by open circuit test or No Load test, and
 Copper losses by short circuit test.
Condition for Maximum Efficiency
We know that,
Copper Loss = WC = I12. R1or I22R2
Iron Loss=Wi = Hysteresis Loss + Eddy Current Loss
WI = WH + WE
Suppose to Primary Side…
Primary Input = P1 = V1I1 Cosθ1
Efficiency = η = Output / Input
Efficiency = η = (Input – Losses) / input ….. (As Output = Input – Losses)
Efficiency = η = (Input – Copper losses – Iron Losses)/Input
Efficiency = η = (P1 - WC - WI) / P1
Efficiency = η = (V1 I1 Cosθ1 - I12. R1 - WI)/ V1 I1 Cosθ1
Taking LCM
Efficiency = η = 1- (I12. R1 /V1I1 Cosθ1) –(WI/ V1 I1Cosθ1)
Or
Efficiency = η = 1- (I1. R1 /V1Cosθ1) – (WI/ V1 I1 Cosθ1)
Differentiate both sides with respect to I1
Dη/ dI1 = 0 – ( R1 /V1 Cosθ1) + (WI/V1 I12 Cosθ1)
Dη/ dI1= – ( R1 /V1 Cosθ1) + (WI/V1 I12 Cosθ1)
For Maximum Efficiency, the value of (Dη/ dI1) should be Minimum i.e.
Dη/ dI1 = 0
The above Equation can be written as
R1 / (V1 Cosθ1) = (WI/V1 I12 Cosθ1)
Or
WI = I12. R1 or I22R2
Iron Loss = Copper Loss
The value of Output current (I2) on which Maximum efficiency can be gained
 I2 = √ (WI/ R2)
The value of Output current (I2) is the actor who equals the value of Copper Loss and Iron
Loss (i.e. Copper Loss = Iron Loss)
Doing so, the maximum efficiency can be gained. Therefore, with proper designing,
maximum efficiency can be attained at any desired load i.e. Copper loss and Iron Loss can
be equaled.
Good to Know
 Efficiency is usually less than 1 and it is often expressed as a percent (%).
 Ideal Transformer is 100% efficient i.e. the efficiency of ideal transformer is 1.
 Practical Transformers efficiencies are generally quite high in compression to other
electrical machines and electronics devices (i.e. Motors, Generators etc) on the ordure of 90
t0 98%.
All Day Efficiency
As we know that the commercial or typical efficiency of a transformer is the ratio of Output
and Input in watts
Efficiency = Output (in Watts)/Input (in Watts)
But there are number of transformers whose performance can’t be monitored according
the above general efficiency.
Those distribution transformers which supply electrical energy to lighting and other
general circuits, their primary energize for 24 hours, but the secondary windings does not
energize all the time. In other words, Secondary windings only energize at the night time
when they supply electrical energy to lighting circuits. I.e. secondary windings supply
eclectic power for very small load or no load for maximum time in 24 hours. It means that
core loss occurs for 24 hours regularly but copper loss occurs only when transformer is on
loaded.
Therefore it realizes the necessity to design a transformer in which the core loss should be
low. As copper loss depends on load, therefore, they should be neglected. In this type of
transformers, we can track their performance only by all day efficiency. All day efficiency
may be also called “Operational efficiency”. On the base of usable energy, we estimate the
all day efficiency for a specific time (During the 24 hours =one day). And we can find it by
the following formula
All Day Efficiency = Output (in kWh)/Input (in kWh)
To understand about the all day efficiency, we must know about the load cycle i.e. how
much load is connected, and for how much time (in 24 hours).
Transformers MCQs With Explanatory Answers
Transformers (MCQs With Explanatory Answers)
1. A Transformer is designed to be operated on both 50 & 60 Hz frequency.For the
Same rating, which one will give more out put; when,
1. Operates on 50 Hz
2. Operates on 60 Hz
Show Explanatory Answer
Answer: 1. operates on 50 Hz
Suppose,
When Transformer operates on 50 Hz Frequency
Transformer = 100kVA, R=700Ω, L=1.2 H, f= 50 Hz.
XL = 2πfL = 2 x 3.1415 x 50 x 1.2 = 377 Ω
impedance Z = √ (R2+XL2) = √ (7002 + 3772) = 795 Ω
Power factor Cos θ = R/Z = 700/795 =0.88
Transformer Output (Real Power)
kVA x Cos θ = 100kVA x 0.88
88000 W = 88kW
Now,
When Transformer operates on 60 Hz Frequency
Transformer =100kVA, R=700Ω, L=1.2 H, f= 60 Hz.
XL = 2πfL = 2 x 3.1415 x 60 x1.2 = 452.4 Ω
impedance Z = √ (R2+XL2) = √ (7002 + 452.4 2) = 833.5 Ω
Power factor = Cos θ = R/Z = 700/833.5 =0.839
Transformer Output (Real Power)
kVA x Cos θ 100kVA x 0.839
=83900W = 83.9kW Output
Now see the difference (real power i.e., in Watts)
88kW- 83.9kW = 4100 W = 4.1kW
If we do the same (As above) for the power transformer i.e, for 500kVA
Transformer, the result may be huge, as below.
(Suppose everything is same, without frequency)
Power Transformer Output (When operates on 50 Hz)
500kVA x 0.88 = 44000 = 440kW
Power Transformer Output (When operates on 60 Hz)
500kVA x 0.839 = 419500 = 419.5kW
Difference in Real power i.e. in Watts
440kW – 419.5kW = 20500 = 20kVA
2. In a Transformer , The primary flux is always _________ the secondary ( flux).
1. Greater then
2. Smaller then
3. Equal
4. Equal in both step up and Step down Transformer
Show Explanatory Answer
Answer: 4. Equal in both step up and Step down Transformer
Flux in Primary and Secondary Winding is always equal.
Explanation:
Given Data;
Primary Number of Turns N1 = 524,
Secondary Number of Turns N2 = 70
Primary Input Voltage V1= 3300 Volts.
Secondary Current I2= 250 A.
Find/Calculate?
Secondary Voltage V2 =?
Primary Current I1=?
Φm 1 = Φm2
We Know that,
N2/N1 = V2/V1 ====> V2 = (N2 x V1)/N1
Putting the Values
V2 = (70 x 3300)/525 = 440 Volts Ans.
Now if Neglecting Losses,
V1I1= V2I2 ====> I1/I2 = V2/ V1 …..Or…..I1 = (V2 x I2) / V1
Putting the Values,
I1 = 440 x 250/3300 = 33.3 Amp Ans.
Now turn around the Transformer equation.
E1 = 4.44 f N1 φm1
φm1 = E1 / 4.44 f N1
Putting the Values
Φm 1 = 3300 / (4.44 x 50 x 525) = 0.0283 Weber’s
Φm 1 = 28.3mWeber’s = Flux in Primary Windings
Same is on the other side,
E2 = 4.44 f N2 φm2
Φm2 = E2 / 4.44 f N2
Putting the values,
Φm2 = 440 / (4.44 x 50 x 70) = 0.0283 Weber’s
Φm2 = 28.3mWeber’s = Flux in secondary Windings
So You can see the flux (Φm) produced in Both Primary and Secondary Winding
is same.
3. What would happen if we operate a 60 Hz Transformer on 50 Hz Source of
Supply.(and how can we do that?
1. Current will decrease (so increase the current)
2. Current will increase ( so decrease the current)
3. Current will be same in both cases.
4. No Effect ( We can do that without changing anything)
5. We can’t perform such an operation.
Show Explanatory Answer
Answer: 2. Current will increase ( so decrease the current)
Explanation: Suppose this is a 60 Hz transformer
4. A Step-Up Transformer which has 110/220 turns.What would happen if we
replace it with 10/20 turns? (because Turns ratio would be same in both cases)
1. induced E.M.F wold be same
2. Induced E.M.F would be decreased

Show Explanatory Answer

Ans: 2. Induced E.M.F would be decreased
Explanation:
 Click image to enlarge
5. The rating of transformer may be expressed in ____________.
1. kW
2. kVAR
3. kVA
4. Horse power.
Show Explanatory Answer
Answer: 3. kVA
Explanation:
There are two type of losses in a transformer;
1. Copper Losses
2. Iron Losses or Core Losses or Insulation Losses
Copper losses ( I²R )depends on Current which passing through transformer winding
while Iron Losses or Core Losses or Insulation Losses depends on Voltage.
That’s why the rating of Transformer is in kVA,Not in kW.
6. What will happen if the primary of a transformer is connected to D.C supply?
1. Transformer will operate with low efficiency
2. Transformer will operate with high efficiency
3. No effect
4. Transformer may start to smoke and burn
Show Explanatory Answer
Answer: 4. Transformer may start to smoke and burn.

Explanation:
7. What would happen if a power transformer designed for operation on 50 Hz
(frequency) were connected to a 500 Hz (frequency) source of the same voltage?
1. Current will be too much high
2. Transformer may start to smoke and burn
3. Eddy Current and Hysteresis loss will be excessive
4. No effect

Show Explanatory Answer

Answer: 3. Eddy Current and Hysteresis loss will be excessive
8. What would happen if a power transformer designed for operation on 50 Hz
(frequency) were connected to a 5 Hz (frequency) source of the same voltage?
1. Current will be too much low
2. Transformer may start to smoke
3. Eddy Current and Hysteresis loss will be excessive
4. No effect
Show Explanatory Answer
Answer: 2. Transformer may start to smock

Explanation:
9. A Step Up transformer _____________.
1. Step Up the level of Voltage
2. Step down the level of current
3. Step up level the power
4. Step up the level of Frequency
5. 1 and 2 only
Show Explanatory Answer
Answer: 5. 1 and 2 only.
Explanation:
A Step up transformer only step up the level of voltage and step down the level of
current.
because the input power is same.
So according to P=VI→ I = P/V…. We can see that, when Voltage increases, current
decreases.
So in Step up transformer, input power is same, therefore, when voltage increases, then
current decreases.
10. Under what condition is D.C supply applied safely to the primary of a
transformer?
1. We can connect directly to DC. No condition required
2. We can’t connect to DC Supply
3. A High resistance should be connect in series with primary, but circuit will be useless.
4. The above statement is wrong
Show Explanatory Answer
Answer: 3. A High resistance should be connect in series with primary, but
circuit will be useless.
Explanation:
11. An Auto-transformer (which has only one winding) may be used as a ______?
1. Step-Up Transformer
2. Step-Down Transformer
3. Both Step-Up and Step-Down transformer
4. None of the above
Show Explanatory Answer
Answer: 3. Both Step-Up and Step-Down transformer

Explanation:
12. E.M.F Equation of the Transformer is _________.
1. E1 = 4.44 f N1 Øm , E2=4.44 f N2 Øm
2. E1= 4.44 f N1 Bm A , E2 = 4.44 f N2 BmA
3. E1= 4.44 N1 Øm/T , E2=4.44 N2 Øm/T
4. All of the above
5. None of the above
Show Explanatory Answer
Answer: 4. All of the above
Explanation:
Take the basic Equation of the transformer (Option 1) E1 = 4.44 f N 1 Øm , E2=4.44
f N2 Øm ,
and then, first put the value of Øm = Bm A. So the equation becomes as in Option 2.
Now put the value of Frequency ( f = 1/T ) in Equation on Option (1). So the equation
becomes as in Option 3.
13. The friction losses in Real Transformers are _________?
1. 0%
2. 5%
3. 25%
4. 50%
Show Explanatory Answer
Answer: 1. 0%
Explanation: Transformer is a Static Devise. So, no rotation, No Friction losses.
14. In Three Phase Transformer, The load Current is 139.1A, and Secondary
Voltage is 415V. The Rating of the Transformer would be ___________.
1. 50kVA
2. 57.72kVA
3. 100kVA
4. 173kVA
Show Explanatory Answer
Answer: 3. 100kVA
Explanation:
Rating of a Three Phase Transformer:
P = √3. V x I
Rating of a Three phase transformer in kVA
kVA = (√3. V x I) /1000
Now
P = √3 x V x I (Secondary voltages x Secondary Current)
P= √3 x 415V x 139.1A = 1.732 x 415V x 139.1A= 99,985 VA = 99.98kVA=100kVA
For more Detail
How to Calculate/Find the Rating of Transformer (Single Phase and Three Phase)?
15 In Single Phase Transformer, The Primary Current and Primary Voltage is 4.55
and 11kV respectively. The Rating of the transformer would be________?
1. 50kVA
2. 86kVA
3. 100kVA
4. 150kVA
Show Explanatory Answer
Answer: 1. 50kVA
Explanation:
Rating of a Single Phase Transformer:
P= VxI
Rating of a Single phase transformer in kVA
kVA = (V x I) /1000
Now
P = V x I (Primary voltages x Primary Current)
P = 11000V x 4.55A = 50,050VA = 50 kVA
For more Detail .. Read the rating of transformer post in MCQs No 14 explanatory
section titled as
“How to Calculate/Find the Rating of Transformer (Single Phase and Three Phase)”?
16. An Isolation Transformer Has Primary to Secondary turns ratio of
__________.
1. 1 : 2
2. 2 : 1
3. 1 : 1
4. Can be any ratio
Show Explanatory Answer
Answer: 3. 1:1
Explanation: Isolation Transformer is used for isolation purpose only. Isolation
transformer transfer electrical power from the source circuit to another circuit with
connecting electrically (but magnetically) for preventing electric shock and also used in
sensitive devices (like medical equipment etc). Thus, isolation between two electrical
circuit can be done by Isolation transformer with turns ratio of 1:1.
17. In an Auto Transformer, The Primary and Secondary are__________Coupled.
1. Only Magnetically
2. Only Electrically
3. Magnetically as well as Electrically
4. None of the above
Show Explanatory Answer
Answer: 3. Magnetically as well as Electrically
Explanation: As we know that in a Transformer, Primary and Secondary winding are
magnetically coupled. But in case of Auto transformer, there is only one winding (which
is used both as a Primary and Secondary). Thus, in an In an Auto Transformer, The
Primary and Secondary are Magnetically as well as Electrically Coupled.
for More detail: Check MCQs No 11 with diagram.
18. A Transformer______________.
1. Changes ac to DC
2. Changes dc to AC
3. Steps up or down DC Voltages & Current
4. Steps up or down AC Voltages & Current
Show Explanatory Answer
Answer: 4. Step up or Step down AC Voltage & Current
Explanation: A Transformer does not work on DC and operates only and only on AC,
therefore it Step up of Step down the level of AC Voltage or Current.
For More detail: Check MCQs No 9
19. Transformer is a device which:________________.
1. Transfer Electrical power from one electrical circuit to another Electrical circuit
2. It’s working without changing the frequency
3. Work through on electric induction.
4. When, both circuits take effect of mutual induction
5. Can step up or step down the level of voltage.
6. Its Working without changing the Power.
7. All of the above
Show Explanatory Answer
Answer: 7. All of the above
Explanation: Transformer