1. ELL100: INTRODUCTION TO ELECTRICAL ENGG.
Course Instructors:
J.-B. Seo, S. Srirangarajan, S.-D. Roy, and S. Janardhanan
Department of Electrical Engineering, IITD

2. Generation, Transmission, & Distribution
1. Power Plant
generates 3-phase AC power
2. Transmission Substation
steps the voltage up to 69, ,000 V
3. High Voltage Transmission Lines
moves power over long distances on the transmission grid
6. Distribution Power Lines
Moves power over distribution grid to points-of-use
steps the voltage down to ,000V
5. Distribution Substation
4. Subtransmission
Steps the voltage down to 22,000 – 69,000 V

3. Generation, Transmission, & Distribution
1. Power Plant
generates 3-phase AC power
2. Transmission Substation
steps the voltage up to 69, ,000 V
3. High Voltage Transmission Lines
moves power over long distances on the transmission grid
6. Distribution Power Lines
Moves power over distribution grid to points-of-use
steps the voltage down to ,000V
5. Distribution Substation
4. Subtransmission
Steps the voltage down to 22,000 – 69,000 V

4. Generation, Transmission, & Distribution
1. Power Plant
generates 3-phase AC power
2. Transmission Substation
steps the voltage up to 69, ,000 V
3. High Voltage Transmission Lines
moves power over long distances on the transmission grid
6. Distribution Power Lines
Moves power over distribution grid to points-of-use
steps the voltage down to ,000V
5. Distribution Substation
4. Subtransmission
Steps the voltage down to 22,000 – 69,000 V

5. Generation, Transmission, & Distribution
1. Power Plant
generates 3-phase AC power
2. Transmission Substation
steps the voltage up to 69, ,000 V
3. High Voltage Transmission Lines
moves power over long distances on the transmission grid
6. Distribution Power Lines
Moves power over distribution grid to points-of-use
steps the voltage down to ,000V
5. Distribution Substation
4. Subtransmission
Steps the voltage down to 22,000 – 69,000 V

6. Generation, Transmission, & Distribution
1. Power Plant
generates 3-phase AC power
2. Transmission Substation
steps the voltage up to 69, ,000 V
3. High Voltage Transmission Lines
moves power over long distances on the transmission grid
6. Distribution Power Lines
Moves power over distribution grid to points-of-use
steps the voltage down to ,000V
5. Distribution Substation
4. Subtransmission
Steps the voltage down to 22,000 – 69,000 V

7. Generation, Transmission, & Distribution
1. Power Plant
generates 3-phase AC power
2. Transmission Substation
steps the voltage up to 69, ,000 V
3. High Voltage Transmission Lines
moves power over long distances on the transmission grid
6. Distribution Power Lines
Moves power over distribution grid to points-of-use
steps the voltage down to ,000V
5. Distribution Substation
4. Subtransmission
Steps the voltage down to 22,000 – 69,000 V

8. Distribution Power Lines
Small Substation
Distribution Power Lines
Industrial Uses
1200 A
2400 – 15,000 Volts
6. Distribution Power Lines
Commercial Uses
480Y/277 Volts
Moves power over distribution grid to points-of-use
Residential Uses
120/240 Volts

9. Distribution Power Lines
Small Substation
Distribution Power Lines
Industrial Uses
1200 A
2400 – 15,000 Volts
6. Distribution Power Lines
Commercial Uses
480Y/277 Volts
Moves power over distribution grid to points-of-use
Residential Uses
120/240 Volts

10. Distribution Power Lines
Small Substation
Distribution Power Lines
Industrial Uses
1200 A
2400 – 15,000 Volts
6. Distribution Power Lines
Commercial Uses
480Y/277 Volts
Moves power over distribution grid to points-of-use
Residential Uses
120/240 Volts

11. Power Dissipated in an Electricity Distribution System
AC Electricity
Why high voltage?
Power Dissipated in an Electricity Distribution System
150 miles
120 Watt
Light bulb
Power Plant
on Colorado River
12 Volt
Connection Box
Estimate resistance of power lines:
- say Ohms per meter, times 200 km = W/m  2105 m = 20 Ohms
The current required by a single bulb using P = VI so I = P/V = 120 Watts/12 Volts = 10 Amps (!)
Power in transmission line:
P = I2R = 102  20 = 2,000 Watts!!
“Efficiency” is e = 120 Watts/4120 Watts = 0.3%!!!
Lecture 7

12. Power Dissipated in an Electricity Distribution System
AC Electricity
Why high voltage?
Power Dissipated in an Electricity Distribution System
150 miles
120 Watt
Light bulb
Power Plant
on Colorado River
12 Volt
Connection Box
Estimate resistance of power lines:
- say Ohms per meter, times 200 km = W/m  2105 m = 20 Ohms
The current required by a single bulb using P = VI so I = P/V = 120 Watts/12 Volts = 10 Amps (!)
Power in transmission line:
P = I2R = 102  20 = 2,000 Watts!!
“Efficiency” is e = 120 Watts/4120 Watts = 0.3%!!!
Lecture 7

13. AC Electricity
Why high voltage?
The thing that kills us most is the high current through the (fixed resistance) transmission lines
Need less current to reduce power: I2R
But our appliance needs a certain amount of power
P = VI so less current demands higher voltage
Solution is high voltage transmission
Repeating the previous calculation with 12,000 Volts delivered to the house draws only
I = 120 Watts/12 kV = 0.01 Amps for one bulb, giving
P = I2R = (0.01)220 = 2010-4 Watts, so
P = Watts of power dissipated in transmission line
Efficiency in this case is e = 120 Watts/ = %
Lecture 7

14. AC Electricity
Why high voltage?
The thing that kills us most is the high current through the (fixed resistance) transmission lines
Need less current to reduce power: I2R
But our appliance needs a certain amount of power
P = VI so less current demands higher voltage
Solution is high voltage transmission
Repeating the previous calculation with 12,000 Volts delivered to the house draws only
I = 120 Watts/12 kV = 0.01 Amps for one bulb, giving
P = I2R = (0.01)220 = 2010-4 Watts, so
P = Watts of power dissipated in transmission line
Efficiency in this case is e = 120 Watts/ = %
Lecture 7

15. three-phase “live” wires
AC Electricity
Why high voltage?
three-phase “live” wires
to house
500, , , ,000 7–13,000
long-distance neighborhood
But having high voltage in each household is a recipe for disaster
sparks every time you plug something in/risk of fire/not cat-friendly
Need a way to step-up/step-down voltage at will
can’t do this with DC, so go to AC
Lecture 7

16. Instantaneous and average power
 Instantaneous power absorbed by the element:
where

17. Instantaneous and average power
 Average power absorbed by the element:
and

18. Instantaneous and average power
 Average power absorbed by the element:
and

19. Instantaneous and average power
Inductive load: current lags 90 degree behind voltage

20. Average power for a general impedance load

21. Example Find the power absorbed by each element and
The average power at the voltage source

22. Example Find the power absorbed by each element and
The average power at the voltage source

23. Example Find the power absorbed by each element and
The average power at the voltage source

24. Example Find the power absorbed by each element
 The voltage across the resistor
The average power at the resistor
 The voltage across the current source
The average power at the current source

25. Example Find the power absorbed by each element
 The voltage across the resistor
The average power at the resistor
 The voltage across the current source
The average power at the current source

26. Example Find the power absorbed by each element
 The voltage across the capacitor
The average power at the inductor
 The voltage across the inductor
The average power at the inductor

27. Example Find the power absorbed by each element
 The voltage across the capacitor
The average power at the inductor
 The voltage across the inductor
The average power at the inductor

28. Maximum power transfer
 What impedance will absorb the maximum amount of power?
circuit

29. Maximum power transfer 

30. Maximum power transfer
When the load impedance is a complex conjugate of the Zth, the maximum power is transferred to the load

31. Maximum power transfer
 What impedance will absorb the maximum amount of power?
circuit

32. Example – 1

33. Example – 1

34. Example – 1

35. Example – 1

36. Example – 1

37. Example – 1

38. Example – 2
Find the load impedance that absorbs the maximum average power. Calculate that maximum average power.

39. Example – 2
Find the load impedance that absorbs the maximum average power. Calculate that maximum average power.

40. Example – 2
Find the load impedance that absorbs the maximum average power. Calculate that maximum average power.

41. Effective or RMS value
 Effective value of a periodic current is the dc current that delivers the same average power to a resistor as the periodic current
— Effective value of a periodic signal is its root mean square (rms) value.
— For a sinusoidal , its rms value is

42. Effective or RMS value The average power in terms of rms value:
The average power absorbed by a resistor R:
Determine the rms value of the current waveform shown below
If the current is passed through a 2-Ohm resistor, find the average power absorbed by the resistor

43. Effective or RMS value The average power in terms of rms value:
The average power absorbed by a resistor R:
Determine the rms value of the current waveform shown below
If the current is passed through a 2-Ohm resistor, find the average power absorbed by the resistor

44. Apparent power and power factor
(volt-amperes, VA)
 Power factor (pf)
Inductive element, lagging power factor (the current lags the voltage)
Capacitive element, leading power factor (the current leads the voltage)
Recall the impedance
power factor is the angle of the load impedance

45. Example
 Determine the power factor of the entire circuit below. Find the average power delivered by the source.
Total impedance is
 Power factor:
 The rms value of the current:
 The average power supplied by the source:

46. Example
 Determine the power factor of the entire circuit below. Find the average power delivered by the source.
Total impedance is
 Power factor:
 The rms value of the current:
 The average power supplied by the source:

47. Example
 Determine the power factor of the entire circuit below. Find the average power delivered by the source.
Total impedance is
 Power factor:
 The rms value of the current:
 The average power supplied by the source:

48. Complex power  The complex power absorbed by the ac load and with
: the average power in watt delivered to a load (only useful power)
: a measure of energy exchange between the source and the reactive par of the load
(volt-ampere reactive: VAR)

49. Complex power  The complex power absorbed by the ac load and with
: the average power in watt delivered to a load (only useful power)
: a measure of energy exchange between the source and the reactive par of the load
(volt-ampere reactive: VAR)

50. Complex power (p.211)  The complex power absorbed by the ac load
Based on the reference voltage signal,
Leading power factor (or load):
Current leads voltage (capacitive load)
Lagging power factor (or load):
Current lags voltage (Inductive load)
Apparent power
Power triangle

51. Example – 1
A load Z draws 12 kVA at a power factor of lagging from a 120-V rms sinusoidal source. Calculate: (a) the average and reactive powers delivered to the load, (b) the peak current, and (c) the load impedance
(kW)
(kVA)

52. Example – 2
A coil is to be represented by a linear model consisting of inductance L in series with resistance R. When a 60-Hz current of 2 A (rms) is supplied to the coil, the voltmeter across the coil reads 26 V (rms). A wattmeter indicates 20 W delivered to the coil. Determine L and R.
and
Reactive power

53. Example – 3 (p.212)
An industrial load consists of the following: 30 kW of heating (resistive) and 150 kVA of induction motors operating 0.6 lagging pf. Power is supplied to the plant at 4000 V. Determine the total current and plant power factor.  
From power factor we have 
(kVA)

54. Example – 3 (p.212)
An industrial load consists of the following: 30 kW of heating (resistive) and 150 kVA of induction motors operating 0.6 lagging pf. Power is supplied to the plant at 4000 V. Determine the total current and plant power factor.  
From power factor we have 
(kVA)

55. Example – 3 (p.212)
An industrial load consists of the following: 30 kW of heating (resistive) and 150 kVA of induction motors operating 0.6 lagging pf. Power is supplied to the plant at 4000 V. Determine the total current and plant power factor.  
The power factor gives

56. Example – 3 (p.212)
An industrial load consists of the following: 30 kW of heating (resistive) and 150 kVA of induction motors operating 0.6 lagging pf. Power is supplied to the plant at 4000 V. Determine the total current and plant power factor. 
(kVA)

57. AC power conservation
 Whether the loads are connected in series or in parallel (or in general), the total power supplied by the source equals the total power delivered to the load. Thus, in general, for a source connected to N loads,
 The complex, real, and reactive powers of the sources equal the respective sums of the complex, real, and reactive powers of the individual loads.

58. Example
Find the real and reactive power absorbed by the source, the line and the load i)
ii)

59. Example
Find the real and reactive power absorbed by the source, the line and the load
iii)
iv)

60. Example
Find the real and reactive power absorbed by the source, the line and the load v)

61. Power correction factor
(Original)

62. Power correction factor
 Example: When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.

63. Example – 1
When connected to a 120-V (rms), 60-Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.

64. Example – 2 (p.213)
An industrial load consists of the following: 30 kW of heating (resistive) and 150 kVA of induction motors operating 0.6 lagging pf. Power is supplied to the plant at 4000 V.
Power factor of the previous example is to be corrected to 0.9 lagging. Specify the necessary auxiliary equipment.

65. Example – 2 (p.213)
Power factor of the previous example is to be corrected to 0.9 lagging. Specify the necessary auxiliary equipment.
The necessary correction is