In this presentation, I discuss some basic concepts in electrical engineering that are not well known to students like the concept of Live, Neutral and Earth wiring; Active, Reactive and Apparent power and Foundation of 3-phase system. I have brought in a new perspective to understand these concepts. I also discuss some misconceptions that students face and resolve it the right way.

2. Conceptions & Misconceptions
1. The term ‘Electricity’?
2. The term ‘Current Flow’.
3. The term ‘Power Flow’.
4. Concept of live, neutral and earth terminals.
5. Are capacitor and inductor linear, and do obey Ohm’s
law?
6. Active, Reactive and Apparent Power.
7. Inductor always absorb reactive power while
capacitor always deliver reactive power.
8. Distinction between ‘Source’ & ‘Load’.
9. Foundation of 3-Phase Power.
10. Significance of per unit system.

3. Misconception #1
Related to the term ‘Electricity’
What comes to our Mind?
1. Electricity is a form of Energy.
2. Electricity is the flow of electrons.
3. Electricity is made of electrons.
4. Electricity is weightless.
5. Electricity flows nearly at the speed of light.
6. Batteries and generators create electricity.
7. Electric companies sell electricity.
8. Flow of Electricity results in flow of Electrical Energy
9. Cost of Electricity / Electricity Bill.

4. Electricity is simply
‘Quantity of Electric Charge’
Its SI unit is Coulomb
Reference:
The National Institute of Standards and Technology
http://physics.nist.gov/cuu/Units/units.html
Table 3
Reality

5. Most people use the term ‘Current Flow’
to describe the direction of current in a circuit.
While current itself means flow of charges, so the term
‘Current Flow’ would mean ‘Flow of Flow of Charges’
For example, we say ‘Current Flows from A to B’
It should really be said as ‘Current has a direction from A to B’
Misconception #2
The term ‘Current Flow’
A B
I

6. Misconception #3
The term ‘Power Flow’.
Most people use the term ‘Power Flow’ which is a misnomer.
Power itself means transfer of energy per unit time, so the term
‘Power Flow’ would mean ‘Flow of Transfer of Energy’
IT IS THE ENERGY WHICH IS TRANSFERRED, NOT POWER!
SO THE TERM POWER FLOW IS COMPLETELY WRONG
For example, we say ‘Power Flows from A to B’
It should really be said as ‘Energy Transfer has a direction from A to B’
A B
E

7. How & Why the concept of
Live, Neutral and Earth
came into existence?
Misconception #4 (if any)

8. Floating Voltage (No Live & Neutral)
Distribution
Transformer

9. continued

10. Disadvantages of Floating Voltage

11. continued

12. How Live and Neutral eliminate the problem

13. Not safe to touch!

14. Case of equipment connected to Neutral
What problem would occur?

15. Continued (Non Ideal Grounding)

16. Introduction of Separate Earth Wire
Safe to Touch!

17. Definitions
• Live Wire : It is ungrounded wire of the
secondary side of distribution transformer.
• Neutral Wire: It is the grounded wire of the
secondary side of the distribution transformer
and it carries the load current
• Earth Wire: It is a thick separate wire which is
grounded and it carries leakage current, and
expected to have zero voltage.

18. Misconception #5
Are Capacitors and Inductors linear?
𝑖 𝐶 = 𝐶
𝑑𝑣 𝐶
𝑑𝑡
𝑓(𝑡) = 𝐶
𝑑(𝑣 𝐶)
𝑑𝑡
𝑣1(t)
𝑣2(t)
𝑖1(t)
𝑖2(t)
𝑣1 t + 𝑣2(t) 𝑖1 t + 𝑖2(t)
𝑓(𝑡) = 𝐶
𝑑(𝑣 𝐶)
𝑑𝑡
𝑓(𝑡) = 𝐶
𝑑(𝑣 𝐶)
𝑑𝑡
Current through capacitors
is given by the relation
1. Capacitor as a System

19. 𝑣 𝐶 =
1
𝐶
𝑡 𝑜
𝑡
𝑖 𝐶 𝑑𝑡 + 𝑉𝐶(𝑡 𝑜)Similarly, voltage across capacitor
is given by the relation
𝑓 𝑡 =
1
𝐶
𝑡 𝑜
𝑡
𝑖 𝐶 𝑑𝑡 + 𝑉𝐶(𝑡 𝑜) 𝑣1(t) + 𝑉𝐶 (𝑡 𝑜)𝑖1(t)
𝑖2(t)
𝑣1 t + 𝑣2 t
+
𝑉𝐶(𝑡 𝑜)
𝑖1 t + 𝑖2(t)
𝑣2(t) + 𝑉𝐶 (𝑡 𝑜)
𝑓 𝑡 =
1
𝐶
𝑡 𝑜
𝑡
𝑖 𝐶 𝑑𝑡 + 𝑉𝐶(𝑡 𝑜)
𝑓 𝑡 =
1
𝐶
𝑡 𝑜
𝑡
𝑖 𝐶 𝑑𝑡 + 𝑉𝐶(𝑡 𝑜)
∴Capacitor as a system will be linear only when its initial condition, 𝑉𝐶(𝑡 𝑜) = 0

20. When Charge-Voltage (Q-V) characteristic is considered
it is a straight line passing through origin,
so it is a Linear element!
Charge
Voltage
𝑄 = 𝐶𝑉
2. Capacitor as an Element

21. When Voltage Current (V-I) characteristic is considered
it is an ellipse,
so Capacitor is a Non Linear element!
Current
Voltage
𝑖 𝐶 = 𝐶
𝑑𝑣 𝐶
𝑑𝑡

22. Question:
If V-I characteristics of
capacitor and inductor are Non-Linear
then how Ohm’s Law is defined for them?

23. Answer:
“Capacitor” and “Inductor” do not obey Ohm’s Law!
Generalized Statement of Ohm’s Law is
“At any time t, voltage across any element
is in direct proportion to current through it”
v(t) a i(t)
BUT, “Capacitive Reactance” and “Inductive Reactance” obey Ohm’s Law!
𝑽 t =
1
jω𝐶
𝑰 t , where
𝟏
𝐣𝛚𝐂
is Capacitive Reactance, 𝑋 𝐶
𝑽 t = jω𝐿 𝑰 t ,where 𝒋𝝎𝑳 is Inductive Reactance, 𝑋 𝐿
Where 𝑽 t and 𝑰 t are complex voltage and complex current respectively.

24. Misconception #6
Active, Reactive and Apparent Power.
How instantaneous power is measured?
+
-
i(t)
v(t)
a) First we assign sign for voltage & current by making use of passive sign convention.
b) If v(t) × i(t) is positive, then p(t) is the power consumed/absorbed by the element.
c) If v(t) × i(t) is negative, then p(t) is the power deliveredby the element.
𝑃𝑜𝑤𝑒𝑟, 𝑝(𝑡) =
𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑑𝑡
𝑇𝑖𝑚𝑒, 𝑑𝑡
=
𝑷𝒐𝒕𝒆𝒏𝒕𝒊𝒂𝒍 𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝑨 𝒂𝒏𝒅 𝑩 𝒂𝒕 𝒕𝒊𝒎𝒆 𝒕
× 𝑪𝒉𝒂𝒓𝒈𝒆 𝒕𝒓𝒂𝒏𝒔𝒇𝒆𝒓𝒓𝒆𝒅 𝒇𝒓𝒐𝒎 𝑨 𝒕𝒐 𝑩 𝒊𝒏 𝒕𝒊𝒎𝒆 𝒅𝒕
𝑻𝒊𝒎𝒆, 𝒅𝒕
= 𝑣(𝑡) ×
dq
𝑑𝑡
𝑻𝒉𝒖𝒔 𝒊𝒏𝒔𝒕𝒂𝒏𝒕𝒂𝒏𝒆𝒐𝒖𝒔 𝒑𝒐𝒘𝒆𝒓 𝒄𝒐𝒏𝒔𝒖𝒎𝒆𝒅 𝒂𝒄𝒄. 𝒕𝒐 𝒄𝒐𝒏𝒗𝒆𝒏𝒕𝒊𝒐𝒏, 𝒑(𝒕) = 𝒗(𝒕) × 𝒊(𝒕) watts
A B

25. Example 1: Purely Resistive Load
i(t)
v(t)
+
-
Source
Purely
Resistive
Load
R
Passive Sign Convention
A
V
𝑣 𝑡 = 𝑉𝑚 𝑆𝑖𝑛 𝜔𝑡
i 𝑡 =
𝑉𝑚
𝑅
𝑆𝑖𝑛 𝜔𝑡
i 𝑡 = 𝐼 𝑚 𝑆𝑖𝑛 𝜔𝑡
Phase Difference, 𝜑 = 0°
For periodic signal, instead of defining instantaneous power,
We define average power
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑃𝑜𝑤𝑒𝑟, 𝑃𝑎𝑣𝑔 =
𝑊𝑜𝑟𝑘 𝐷𝑜𝑛𝑒 𝑖𝑛 𝑇𝑖𝑚𝑒 𝑃𝑒𝑟𝑖𝑜𝑑 𝑇
𝑇𝑖𝑚𝑒 𝑃𝑒𝑟𝑖𝑜𝑑, 𝑇

26. Time, t
Amplitude
 Voltage Waveform v(t) at 50Hz
 Current Waveform i(t) at 50 HZ
 Power Waveform p(t) at 100HZ
 Average Power, 𝑷 𝒂𝒗𝒈
𝑽 𝒎
𝑰 𝒎
𝑷 𝒎
𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎
𝑷 𝒂𝒗𝒈 =
𝑽 𝒎×𝑰 𝒎
𝟐
= 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 watts

27. Example 2: Purely Inductive Load
𝑣 𝑡 = 𝑉𝑚 𝑆𝑖𝑛 𝜔𝑡
i 𝑡 =
𝑉𝑚
𝜔𝐿
𝑆𝑖𝑛 𝜔𝑡 − 90°
i 𝑡 = 𝐼 𝑚 𝑆𝑖𝑛 𝜔𝑡 − 90°
Phase Difference, 𝜑 = −90°
 Voltage Waveform v(t) at 50Hz
 Current Waveform i(t) at 50 HZ
 Power Waveform p(t) at 100HZ
 Average Power, 𝑷 𝒂𝒗𝒈
𝑷 𝒂𝒗𝒈 = 𝟎

28. Conclusion:
 Average power consumed by a purely resistive
load is 𝑷 𝒂𝒗𝒈 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 𝒘𝒂𝒕𝒕𝒔
 Average power consumed by a purely inductive or
capacitive load is 𝑷 𝒂𝒗𝒈 = 𝟎 𝒘𝒂𝒕𝒕𝒔
So why bother the presence of Inductor or Capacitor in load circuit?
Reasons:
1. For same values of RMS voltage and RMS current magnitude, their active power
output is less than that of a purely resistive load
2. For same active power load, they draw more RMS current than purely resistive
load
3. This excess current increases copper loss in the transmission line
and increases current and thermal rating of all components in the power system.
For these reasons there is need to define some quantity
which can account the effect of presence of reactive elements in the load.

29. Power Factor:
It is the measure of the degree -- to which -- a given load--
matches to that of – a pure resistance.

30. 𝑃𝑜𝑤𝑒𝑟 𝐹𝑎𝑐𝑡𝑜𝑟 =
𝑃𝐿𝑂𝐴𝐷
𝑃𝑅
𝑃𝐿𝑂𝐴𝐷 = 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑃𝑜𝑤𝑒𝑟 𝐶𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑏𝑦 𝑎 𝑔𝑖𝑣𝑒𝑛 𝑙𝑜𝑎𝑑
𝑃𝑅 = 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑃𝑜𝑤𝑒𝑟 𝐶𝑜𝑛𝑠𝑢𝑚𝑒𝑑 𝑏𝑦 𝑎 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑒 𝑙𝑜𝑎𝑑
𝑓𝑜𝑟 𝑠𝑎𝑚𝑒 𝑅𝑀𝑆 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑎𝑛𝑑 𝑐𝑢𝑟𝑟𝑒𝑛𝑡.
Generalized formula of Power Factor
𝑃. 𝐹. =
𝑃𝐿𝑂𝐴𝐷
𝑃𝑅
=
517
1750
≅ 0.3

31. Circuit with low power factor draws more current from the supply
causing more copper loss in the transmission line
and increases current and thermal rating of all components in the power system

32. i(t)
v(t)
+
-
RL
Load
V
A
Consider an R-L load
Concept of Reactive and Apparent Power
A Graphical Approach

33. 𝝋 = −𝟕𝟎°
𝝋 = 𝟎°
𝝋 = −𝟗𝟎°
𝑰 𝒎𝒂𝒙
𝑰 𝒎𝒂𝒙 𝐜𝐨𝐬 𝟕𝟎°
𝑰 𝒎𝒂𝒙 𝐬𝐢𝐧 𝟕𝟎°
 In Phase Component
of Current
 Phase Quadrature
Component of
Current
 Load Voltage
 Load Current

34. 𝑷 𝒂𝒗𝒈 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 𝐜𝐨𝐬 𝟕𝟎°
𝑽 𝒎
𝑰 𝒎 cos 𝟕𝟎°
𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎 cos 𝟕𝟎°
Power due to ‘in phase component’ of current

35. Power due to ‘phase quadrature component’ of current
𝑽 𝒎
𝑰 𝒎 sin 𝟕𝟎°
𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎 sin 𝟕𝟎°
𝑷 𝒂𝒗𝒈 = 𝟎

36. 𝑽 𝒎
𝑰 𝒎 sin 𝟕𝟎°
𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎 sin 𝟕𝟎°
𝑻𝒉𝒖𝒔, 𝑷 𝒂𝒗𝒈|𝒗𝒊𝒓𝒕𝒖𝒂𝒍 = 𝐑𝐞𝐚𝐜𝐭𝐢𝐯𝐞 𝐏𝐨𝐰𝐞𝐫, 𝐐 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 sin 𝟕𝟎° 𝐕𝐀𝐑
𝑷 𝒂𝒗𝒈 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 sin 𝟕𝟎°
By manually shifting the voltage and current
waveform to bring them in phase
Virtual Power
Waveform

37. Resultant Power
𝑽 𝒎
𝑰 𝒎
𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎
𝑷 𝒂𝒗𝒈 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 cos 𝟕𝟎°

38. 𝑻𝒉𝒖𝒔, 𝑷 𝒂𝒗𝒈|𝒗𝒊𝒓𝒕𝒖𝒂𝒍 = 𝐀𝐩𝐩𝐚𝐫𝐞𝐧𝐭 𝐏𝐨𝐰𝐞𝐫, 𝐒 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺 𝐕𝐀
𝑽 𝒎
𝑰 𝒎
𝑷 𝒎 = 𝑽 𝒎 × 𝑰 𝒎
𝑷 𝒂𝒗𝒈 = 𝑽 𝑹𝑴𝑺 × 𝑰 𝑹𝑴𝑺
By manually shifting the voltage and current
waveform to bring them in phase
Virtual Power
Waveform

39. Conclusion
• Only Active Power is the real & genuine power
which is obtained by taking the average of the
instantaneous power.
• ‘Reactive Power’ and ‘Apparent Power’ are not
real as they are obtained by manually shifting the
waveform to 0 phase difference and calculating
the virtual power by taking average of the virtual
power waveform.
• Since Q & S power quantities are obtained by
shifting they are totally independent quantities,
they have no physical relationship with active
power P. But they posses a mathematical
relationship.

40. Power Triangle
𝑆 = 𝑉𝑅𝑀𝑆 × 𝐼 𝑅𝑀𝑆
𝑃 = 𝑉𝑅𝑀𝑆 × 𝐼 𝑅𝑀𝑆 cos 𝜑
𝑄 = 𝑉𝑅𝑀𝑆 × 𝐼 𝑅𝑀𝑆 sin 𝜑
P
S
Q
𝝋
 𝑺 = 𝑷 𝟐 + 𝑸 𝟐
 𝑷 = 𝑺 𝐜𝐨𝐬 𝝋
 𝑸 = 𝑺 𝐬𝐢𝐧 𝝋

41. Misconception #7
Inductor always absorb reactive power while
capacitor always deliver reactive power.

42. Power, p(t)
Power, p(t)
Inductor
Capacitor
time
time

43. Misconception #8
Distinction between ‘Source’ & ‘Load’.
How to tell which is source and which is load?
At any time t,
The one which deliver power is source.
The one which absorb power is load.

44. Element Parameter Nature
Ideal Resistor Resistance, R Always a Load
Inductor Inductance, L Load/Source
Capacitor Capacitance, C Load/Source
Ind. Voltage/Current Source Voltage/Current Source/Load
Dep. Voltage/Current Source Voltage/Current Source/Load
Ideal Transformer Turns Ratio Neither load nor source
Element Parameter Nature
Ideal Viscous Damper Coefficientof viscous friction, B Always a Load
Spring Spring Constant, K Load/Source
Mass/Moment of Inertia Mass, M/Moment of Inertia, J Load/Source
Force/Torque Force, F/ Torque 𝜏 Source/Load
Ideal Lever/Gear Teeth Ratio Neither load nor source
Now what exactly the term ‘Increasing the Load’ means?
‘Increasing the Load’ means varying R, L, C in Electrical Domain and
varying B, K, J in Mechanical Domain such that apparent power
consumed by the system is increased.

45. Apparent Electrical Power,
𝑆 𝑒 = 𝑉𝐼 =
𝑉2
𝑍 𝑒
= 𝐼2 𝑍 𝑒,
where 𝑍 𝑒=𝑅 𝑒+j[ωL +
1
𝑗2 𝜔𝐶
]
Apparent Mechanical Power,
𝑆 𝑚 𝑠 = τ 𝑠 ω 𝑠 =
𝜏(𝑠)2
𝑍 𝑚(𝑠)
= 𝑠𝜔(𝑠)2 𝑍 𝑚(𝑠)
where, Zm s = B s + s[ J(s) +
𝐾(𝑠)
𝑠2 ]
Increase in Load means:
1. Decrease in Electrical Impedance, if applied voltage is kept constant.
2. Increase in Electrical Impedance, if current is kept constant.
3. Decreasein Mechanical Impedance, if applied torque is kept constant.
4. Increase in Mechanical Impedance, if angular velocity is kept constant.

46. Foundation of 3-Phase Power

47. 1 – Φ Generator

48. 2 – Φ Generator

49. 2 – Φ Generator under load

50. 3 – Φ Generator

51. Total Instantaneous Power
Phase B Instantaneous Power
Phase Y Instantaneous Power
Phase R Instantaneous Power
Peak Power ERIR = Pm
Er = ER ∠ 0°
Ey = EY ∠ 120°
Eb = EB ∠ -120°
Ey
Eb
Ir
Iy Ib
Peak Power EY IY = Pm
Peak Power EB IB = Pm
Output Power = 1.5 Pm
3 – Φ Generator under load
Er

52. The Star Connection
Three Phase, 6 Wire System & Corresponding Phasor Diagram

53. Three Phase, 4 Wire System & its Line Currents

54. ELine-Line (RMS) = √3 x EPhase (RMS)
Where,
Eab,Ebc and Eca are line-line voltages
Ean, Ebn and Ecn are phase voltages

55. 0
EMF
A
B
C
B
C
A
EAB
EBC
ECA
0
Voltage
Drop
The Delta Connection

56. A
C
B
0
EMF
0
Voltage
Drop
ILine (RMS) = √3 x IPhase (RMS)
Where,
IA, IB and IC are line current
IAB, IBC and ICA are phase current

57. Transformer Example
The Per-Unit System

58. Sn kVA 1 10 100 1000 400000
Enp V 2400 2400 12470 69000 13800
Ens V 460 347 600 6900 424000
Inp A 0.417 4.17 8.02 14.5 29000
Ins A 2.17 28.8 167 145 943
R1 Ω 58.0 5.16 11.6 27.2 0.0003
R2 Ω 1.9 0.095 0.024 0.25 0.354
Xf1 Ω 32 4.3 39 151 0.028
Xf2 Ω 1.16 0.09 0.09 1.5 27
Xm Ω 200000 29000 150000 505000 460
Rm Ω 400000 51000 220000 432000 317
Io A 0.0134 0.0952 0.101 0.210 52.9
ACTUAL TRANSFORMER VALUES

59. Sn kVA 1 10 100 1000 400000
Enp V 2400 2400 12470 69000 13800
Ens V 460 347 600 6900 424000
Inp A 0.417 4.17 8.02 14.5 29000
Ins A 2.17 28.8 167 145 943
Znp Ω 5760 576 1555 4761 0.4761
Zns Ω 211.6 12.0 3.60 47.61 449.4
R1 (pu) – 0.0101 0.0090 0.0075 0.0057 0.00071
R2 (pu) – 0.0090 0.0079 0.0067 0.0053 0.00079
Xf1 (pu) – 0.0056 0.0075 0.0251 0.0317 0.0588
Xf2 (pu) – 0.0055 0.0075 0.0250 0.0315 0.0601
Xm (pu) – 34.7 50.3 96.5 106 966
Rm (pu) – 69.4 88.5 141.5 90.7 666
Io (pu) – 0.032 0.023 0.013 0.015 0.0018
PER-UNIT TRANSFORMER VALUES

60. Distinction between Flux & Flux Linkage
Open Surface
Flux

61. Flux Linkage
ψ = 𝐵. 𝑑𝑠
Open Surface
having
Conductor as its Contour
Open Surface

62. References
1. Electrical Machines, Drives and Power System, Theodore Wildi
2. The National Institute of Standards and Technology
3. Power System Analysis, Grainger & Stevenson
4. Basic Electrical Technology, NPTEL
5. National Instruments

63. End of Presentation.