2. Chapter 6 – Power, Energy and Efficiency
Lesson Objectives
Upon completion of this topic, you should be able to:
Calculate power and energy in electrical circuits..
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3. Chapter 6 – Power, Energy and Efficiency
Specific Objectives
1. State the units for :
Quantity of electricity
Energy
Power
2. Establish the following equations :
Q=Ixt
E=Pxt
3. Establish the following equations of power
P = I2 x R = V2/R = V x I
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4. Chapter 6 – Power, Energy and Efficiency
Specific Objectives
4. Solve simple problems involving a maximum of three
resistors relating to :
Quantity of electricity
Electrical energy
Electrical power
4. Calculate energy consumption of an electrical load in
kWh.
5. Calculate the cost of energy consumed by an electrical
load for a given power and time.
6. Convert the unit from kWh to Joules and vice versa.
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5. Chapter 6 – Power, Energy and Efficiency
Units
Recall
units
Unit of Voltage : Volts (v)
Unit of Current : Ampere
(A)
Unit of Resistance : Ohms
(Ω)
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6. Chapter 6 – Power, Energy and Efficiency
Units
Unit of Charge: Coulomb (Q)
Unit of Energy : Joules (J)
Unit of Power : Watts or J/s (W)
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7. Chapter 6 – Power, Energy and Efficiency
Alternative Unit for Energy
Energy (Joules, J) can also be expressed in kilowatt-
hours (kWh).
1 kWh = 1000 Watt-hours
= 1000 x 3600 watt-seconds Watt-seconds=Joules
= 3,600,000 Joules
= 3.6 MJ
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8. Chapter 6 – Power, Energy and Efficiency
What is Energy
Energy is ability or capacity for doing work.
Work cannot be done without energy being used.
Mechanical, electrical and heat energy are all
measured in joules (J)
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9. Chapter 6 – Power, Energy and Efficiency
Energy
Energy may exist in several forms and may be
changed from one form to another.
A lead-acid cell changes chemical energy to electrical
energy on discharge and vice-versa on charge.
A generator changes mechanical energy to electrical
energy;
An electric radiator converts electrical energy to heat
energy, and etc
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10. Chapter 6 – Power, Energy and Efficiency
What is Energy
When current passes through a resistance, the
collisions of the electrons gives off heat, a form of
energy.
heat produced by
current through
resistance
is energy loss
VS
+ -
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11. Chapter 6 – Power, Energy and Efficiency
What is Energy
Heat energy produced in the resistor is proportional
to :
a) square of the current - I 2
b) resistance of the resistor - R
c) duration of time - t in seconds
Energy = I2Rt joules
= Power x Time
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12. Chapter 6 – Power, Energy and Efficiency
Cost of Energy
The cost of electrical energy is calculated by
multiplying the number of units of energy consumed in
kWh and the cost per unit.
1 unit = 1 kWh
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13. Chapter 6 – Power, Energy and Efficiency
Example 6-1 (Cost of Energy)
A colour television rated at 1600 W operates for 5 hours
a day for 30 days.
(a) What is the total energy consumed?
(b) What is the monthly consumption cost if the cost per
unit is 18 cents?
Solution :
(a) Energy = 1600 x 5 x 30 = 240 kWh
(b) cost = 240 x $ 0.18 = $ 43.2
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14. Chapter 6 – Power, Energy and Efficiency
What is Electrical Power?
Power is the rate of doing work.
Power is measured by how fast energy is
being used.
Power = work done in joules/time taken in
seconds
One watt of power is obtained when a
current of one ampere passes through a
potential difference of one volt.
Unit: J / s or watts
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15. Chapter 6 – Power, Energy and Efficiency
Low power
consumption
High power
consumption
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16. Chapter 6 – Power, Energy and Efficiency
Equations of Power
As power is work done per sec therefore
P = IR2t
t
P = I2R
P= V 2
R = V2
Ohm’s law
R R
I = V/ R
R = V/ I
= I
2 V = VI
I
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17. Chapter 6 – Power, Energy and Efficiency
Total Power
The total power for both series and parallel circuits,
is equal to the sum of the powers in each resistor :
PT = P1 + P2 + P3 + … … + Pn
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18. Chapter 6 – Power, Energy and Efficiency
Comparison
SERIES PARALLEL
IT = I1 = I2 = I3 = … = In IT = I 1 + I 2 + I 3 + … + I n
VS = V1 + V2 + V3 + … + Vn VS = V1 = V2 = V3 = … = Vn
RT = R1 + R2 + R3 + … + Rn 1 1 1 1 1
= + + …+
RT R1 R2 R3 Rn
PT = P1 + P2 + P3 + … … + Pn
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19. Chapter 6 – Power, Energy and Efficiency
Efficiency
The efficiency of an electric circuit is the ratio of output to
the input expressed in percentage. It is symbolized by η.
Efficiency of electric circuit,
Output power is always smaller than input due to losses
Efficiency is always < 100%
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20. Chapter 6 – Power, Energy and Efficiency
Problem Solving
Power
Calculations
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21. Chapter 6 – Power, Energy and Efficiency
Example 6-2 (Efficiency)
The output of a generator is 1500W and the input is
equivalent to 1900W.
Calculate its percentage efficiency.
η= (1500/1900)x100 = 78.9%
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22. Chapter 6 – Power, Energy and Efficiency
Example 6-3
Calculate the total power consumed by the circuit
below (i.e. the power dissipated in R).
Vs +
R
R=20Ω
60V
-
Power dissipated in R
V2 602V2
P = = = 180W
R 20Ω
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23. Chapter 6 – Power, Energy and Efficiency
Example 6-3
… cont”d
Vs R
60V 20Ω
V 60V
I = = = 3A
R 20Ω
Power dissipated in R
P = I2 x R = 32A x 20Ω = 180W
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24. Chapter 6 – Power, Energy and Efficiency
Example 6-4
Calculate the power dissipated in R1 , R2 and
the total power consumed by the circuit below.
R1 = 10Ω R2 = 20Ω
+ -
VS = 60V
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25. Chapter 6 – Power, Energy and Efficiency
Example 6-4
…….cont”d
R1 = 10Ω R2 = 20Ω
+ -
VS = 60V
RT = R1 + R2 = 10Ω + 20Ω = 30 Ω
IT = VS = 60V = 2A
RT 30Ω
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26. Chapter 6 – Power, Energy and Efficiency
Example 6-
4
….cont”d R1 = 10Ω R2 = 20Ω
+ -
VS = 60V
Power dissipated in R1
P1 = IT2 x R1 = 22A2 x 10Ω = 40W
Power dissipated in R2
P2 = IT2 x R2 = 22A2 x 20Ω = 80W
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27. Chapter 6 – Power, Energy and Efficiency
Example 6-4
….. Cont”d
R1 = 10Ω R2 = 20Ω
+ -
VS = 60V
Total power dissipated
PT = IT x VS = 2A x 60V = 120W
or PT = P1 + P2 = 40W + 80W = 120W
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28. Chapter 6 – Power, Energy and Efficiency
Example 6-5
Calculate the power dissipated in R1 , R2 and the
total power consumed by the circuit below.
+
VS R1 R2
60V - 10Ω 20Ω
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29. Chapter 6 – Power, Energy and Efficiency
Example 6-5
…. Cont”d
VS + R1 R2
60V - 10Ω 20Ω
Power dissipated in R1
V1 2 VS2 602V2 = 360W
P1 = = = 10Ω
R1 R1
Power dissipated in R2
V22 VS 2 602V2
P2 = = = = 180W
R2 R2 20Ω
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30. Chapter 6 – Power, Energy and Efficiency
Example 6-5
c
o + R1 R2
n VS
10Ω 20Ω
t 60V -
i
n R1 x R 2 10Ω x 20Ω
u RT = = = 6.67 Ω
R1 + R 2 10Ω + 20Ω
e
Total power dissipated
VS 2 602V2
… PT = =
RT 6.67Ω = 540W
.O PT = P1 + P2 = 360W + 180W = 540W
.R
.
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31. Chapter 6 – Power, Energy and Efficiency
Next Lesson
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