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By SY Cheung
EL / IVE (HW)
││││LECTURE 1││││
Fundamental of Electricity
EEE3404 – Electrical Engineering Principles 1
2© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
1. Atom Model Nucleus
Electron
Fundamental of Electricity
3© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
2. Atom
• Protons
• Neutrons
• Electrons
Fundamental of Electricity
“Each atom has the same number of
protons and electrons.”
4© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
3. Proton and Electron
Protons carries positive charge
– it is relatively large mass
– does not play active part in electrical current flow
Electrons carries negative charge
– light mass
– play an important role in electrical current flow
Fundamental of Electricity
5© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
4. Unit of Charge
• Unit of Charge is called Coulomb (C)
• An electron and a proton have exactly
same amount of charge
• One coulomb of charge is equal to
approximately 628 x 1016 electron charge
Fundamental of Electricity
6© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
5. Free Electrons
Free Electrons
Applying Heat or Light
Fundamental of Electricity
7© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
6. Electrical Materials
All material may be classified into three
major classes:
Fundamental of Electricity
• Conductors (copper, aluminum, siliver,
platinum, bronz, gold)
• Semiconductors (germanium, silicon)
• Insulators (glass, rubber, plastic, air,
varnish, paper, wood, mica, ceramic,
certain oils)
8© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
• Conductors have many free electrons
which will be drifting in a random
manner within the material
6. Electrical Materials
Fundamental of Electricity
• Insulators have very few free electrons
• Semiconductors falls somewhere between
these two extremes
9© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
7. Electric current
• Electric current is the movement, or flow
of electrons through a conductive material
• It is measured as the rate at which the
charge is moved around a circuit, its unit
is ampere (A)
I=Q/t or Q=It
Fundamental of Electricity
10© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
8. Electromotive Force
• In order to cause the 'free' electrons to
drift in a given direction an electromotive
force must be applied
• The emf is the 'driving' force in an
electrical circuit
• The symbol for emf is E and the unit of
measurement is the volt (V)
Fundamental of Electricity
11© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
8. Electromotive Force
Typical sources of emf are cells, batteries
and generators
The amount of current that will flow through
a circuit is related to the size of the emf
applied to it
Fundamental of Electricity
12© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
9. Potential Difference (p.d.)
• Whenever current flows through a circuit
element in a circuit such as resistor, there
will be a potential difference(p.d.)
developed across it
• The unit of p.d. is volts(V) and is
measured as the difference in voltage
levels between two points in a circuit
Fundamental of Electricity
13© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
9. Potential Difference (p.d.)
• Emf (being the driving force) causes
current to flow
• Potential difference is the result of
current flowing through a circuit element
• Thus emf is a cause and p.d. is an effect
Fundamental of Electricity
14© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
9. Potential Difference (p.d.)
LOAD
P.D.=1.4 V
E.m.f.=1.5V, Rint.
Equivalent circuit of a battery
Fundamental of Electricity
15© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
10. Resistance
• Resistance is the 'opposition' to the current
flow measured in ohms (Ω)
• Conductors have a low value of resistance
• Insulators have a very high resistance
• Load in DC/AC circuits
Fundamental of Electricity
16© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
Circuit convention
+
-
+
-
I
E VR
11. Ohm’s Law
Fundamental of Electricity
17© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
11. Ohm’s Law
It states that current in a resistive circuit is directly
proportional to its applied voltage and inversely
proportional to its resistance provided that all other
factors (e.g. temperature) remain constant.
IRVor
I
V
Ror
R
V
I.e.i ===
V
I
R
Slope = 1/R
V
I
V α I
Fundamental of Electricity
18© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
12. Power
Power (P) is defined as the rate of doing work (W).
havewe
t
Q
Iand
Q
W
VSince
Wwatts
t
W
Pei
==
= )(..
Additional relationships are obtained by substituting
V=IR and I=V/R, we have:
P = VI
R
V
PandRIP
2
2
==
Fundamental of Electricity
19© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
13. Resistors in Series
E
I
V1 V2 V3
R1 R2 R3
Fundamental of Electricity
20© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
By Ohm's law
• V1 = IR1 volts;
• V2 = IR2 volts; and
• V3 = IR3 volts
• E = V1 + V2 + V3
• E = I (R1 + R2 + R3)
13. Resistors in Series
Fundamental of Electricity
21© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
• E = IReq and
• Req = R1 + R2 + R3 ohm
• where Req is the total circuit
resistance
when resistors are connected in series the
total resistance is found simply by adding
together the resistor values
13. Resistors in Series
Fundamental of Electricity
22© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
Fundamental of Electricity
23© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
14. Potential Divider
The voltage drops shared by the resistors
are given as follows:
• V1=E x R1/ (R1+ R2+ R3)
• V2=E x R2 / (R1+ R2+ R3)
• V3=E x R3 / (R1+ R2+ R3)
Fundamental of Electricity
24© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
15. Resistors in Parallel
R1
R2
R3
I1
I2
I3
I
E
Fundamental of Electricity
25© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
By Ohm’s Law
I1=E/ R1
I2=E/ R2
I3=E/ R3
15. Resistors in Parallel
Fundamental of Electricity
26© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
The total current of the circuit I is the sum
of I1, I2 and I3 , thus
I = I1 + I2 +I3
15. Resistors in Parallel
Fundamental of Electricity
27© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
The total resistance or the equivalent
resistance(Req) of the circuit is defined to
be
Req= E/I
15. Resistors in Parallel
Fundamental of Electricity
28© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
By substituting the above expression for
the currents, we have
E/Req=I=E(1/R1+1/R2+ 1/R3)
Thus we found
1/Req=(1/R1+1/R2+ 1/R3)
15. Resistors in Parallel
Fundamental of Electricity
29© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
Quiz
Calculate the total resistance of the
following resistors connected in parallel:
a)30 Ω, 60 Ω
b)180 Ω, 90 Ω, 60 Ω, 60 Ω
Fundamental of Electricity
30© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
16. Current Divider
The current in each branch of the circuit is
given as
• I1= I x Req/ R1
• I2= I x Req/ R2
• I3= I x Req/ R3
Fundamental of Electricity
31© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
17. Power and Energy in a resistive
circuit
Power is equal to the current multiplied by
the voltage and the unit of power is watt (W)
P = I x E (W)
Energy is equal to the power multiplied by
the time for the circuit being energized.
The unit is Joule (J).
Energy = P x t (J)
Fundamental of Electricity
32© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
18. Power in a resistive circuit
By Ohm's law E=IR, the above equation
can be modify to be
P = I2R
Power is equal to the current squared,
multiplied by the resistance.
Fundamental of Electricity
33© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
Use Ohm's law again, where I =E/R, we
have
P=E2/R
Power is equal to the voltage squared,
divided by the resistance.
18. Power in a resistive circuit
Fundamental of Electricity
34© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
Example 1
RBC
E=10V
I
VAB VBC VCD
2Ω 5Ω 3Ω
A B C D
RAB
RCD
Fundamental of Electricity
35© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
•VAB + VBC + VCD is exactly equal to the
emf=10V
•The total resistance of the circuit is
2+5+3=10Ω
•By Ohm's law V=IR, the current I
should be equal to 1A
Example 1 - Answer
Fundamental of Electricity
36© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
• VAB = IRAB =1 x 2 = 2V.
• VBC = I RBC =1 x 5 = 5V.
• VCD = IRCD =1 x 3 = 3V.
• Power dissipation in RAB = I2RAB = 12 x 2 = 2W.
• Power dissipation in RBC = I2RBC = 12 x 5 = 5W.
• Power dissipation in RCD = I2RCD = 12 x 3 = 3W.
• Total power dissipated = 2+5+3 =10W
Example 1 - Answer
Fundamental of Electricity
37© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
Example 2
R1
R2
R3
I1
I2
I3
I
E=6V
3Ω
6Ω
2Ω
Fundamental of Electricity
38© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
–The potential difference across each of the
three resistors is equal to the battery emf 6V
–Apply Ohm's Law
• E=I1 R1 ; I1=E/R1 = 6/2 = 3A
• E=I2 R2 ; I2=E/R2 = 6/3 = 2A
• E=I3 R3 ; I3=E/R3 = 6/6 = 1A
Example 2 - Answer
Fundamental of Electricity
39© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2
• The total current I is equal to the sum of
currents
• I1+I2+I3 = 3+2+1 =6A.
• Power dissipation in R1 = I1
2R1 = 32 x 2 = 18W.
• Power dissipation in R2 = I2
2R2 = 22 x 3 = 12W.
• Power dissipation in R3 = I3
2R3 = 12 x 6= 6W.
• Total power dissipated =18+12+6 =36W
Example 2 - Answer
Fundamental of Electricity

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Lecture 1 fundamental of electricity

  • 1. By SY Cheung EL / IVE (HW) ││││LECTURE 1││││ Fundamental of Electricity EEE3404 – Electrical Engineering Principles 1
  • 2. 2© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 1. Atom Model Nucleus Electron Fundamental of Electricity
  • 3. 3© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 2. Atom • Protons • Neutrons • Electrons Fundamental of Electricity “Each atom has the same number of protons and electrons.”
  • 4. 4© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 3. Proton and Electron Protons carries positive charge – it is relatively large mass – does not play active part in electrical current flow Electrons carries negative charge – light mass – play an important role in electrical current flow Fundamental of Electricity
  • 5. 5© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 4. Unit of Charge • Unit of Charge is called Coulomb (C) • An electron and a proton have exactly same amount of charge • One coulomb of charge is equal to approximately 628 x 1016 electron charge Fundamental of Electricity
  • 6. 6© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 5. Free Electrons Free Electrons Applying Heat or Light Fundamental of Electricity
  • 7. 7© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 6. Electrical Materials All material may be classified into three major classes: Fundamental of Electricity • Conductors (copper, aluminum, siliver, platinum, bronz, gold) • Semiconductors (germanium, silicon) • Insulators (glass, rubber, plastic, air, varnish, paper, wood, mica, ceramic, certain oils)
  • 8. 8© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 • Conductors have many free electrons which will be drifting in a random manner within the material 6. Electrical Materials Fundamental of Electricity • Insulators have very few free electrons • Semiconductors falls somewhere between these two extremes
  • 9. 9© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 7. Electric current • Electric current is the movement, or flow of electrons through a conductive material • It is measured as the rate at which the charge is moved around a circuit, its unit is ampere (A) I=Q/t or Q=It Fundamental of Electricity
  • 10. 10© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 8. Electromotive Force • In order to cause the 'free' electrons to drift in a given direction an electromotive force must be applied • The emf is the 'driving' force in an electrical circuit • The symbol for emf is E and the unit of measurement is the volt (V) Fundamental of Electricity
  • 11. 11© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 8. Electromotive Force Typical sources of emf are cells, batteries and generators The amount of current that will flow through a circuit is related to the size of the emf applied to it Fundamental of Electricity
  • 12. 12© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 9. Potential Difference (p.d.) • Whenever current flows through a circuit element in a circuit such as resistor, there will be a potential difference(p.d.) developed across it • The unit of p.d. is volts(V) and is measured as the difference in voltage levels between two points in a circuit Fundamental of Electricity
  • 13. 13© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 9. Potential Difference (p.d.) • Emf (being the driving force) causes current to flow • Potential difference is the result of current flowing through a circuit element • Thus emf is a cause and p.d. is an effect Fundamental of Electricity
  • 14. 14© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 9. Potential Difference (p.d.) LOAD P.D.=1.4 V E.m.f.=1.5V, Rint. Equivalent circuit of a battery Fundamental of Electricity
  • 15. 15© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 10. Resistance • Resistance is the 'opposition' to the current flow measured in ohms (Ω) • Conductors have a low value of resistance • Insulators have a very high resistance • Load in DC/AC circuits Fundamental of Electricity
  • 16. 16© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 Circuit convention + - + - I E VR 11. Ohm’s Law Fundamental of Electricity
  • 17. 17© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 11. Ohm’s Law It states that current in a resistive circuit is directly proportional to its applied voltage and inversely proportional to its resistance provided that all other factors (e.g. temperature) remain constant. IRVor I V Ror R V I.e.i === V I R Slope = 1/R V I V α I Fundamental of Electricity
  • 18. 18© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 12. Power Power (P) is defined as the rate of doing work (W). havewe t Q Iand Q W VSince Wwatts t W Pei == = )(.. Additional relationships are obtained by substituting V=IR and I=V/R, we have: P = VI R V PandRIP 2 2 == Fundamental of Electricity
  • 19. 19© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 13. Resistors in Series E I V1 V2 V3 R1 R2 R3 Fundamental of Electricity
  • 20. 20© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 By Ohm's law • V1 = IR1 volts; • V2 = IR2 volts; and • V3 = IR3 volts • E = V1 + V2 + V3 • E = I (R1 + R2 + R3) 13. Resistors in Series Fundamental of Electricity
  • 21. 21© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 • E = IReq and • Req = R1 + R2 + R3 ohm • where Req is the total circuit resistance when resistors are connected in series the total resistance is found simply by adding together the resistor values 13. Resistors in Series Fundamental of Electricity
  • 22. 22© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 Fundamental of Electricity
  • 23. 23© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 14. Potential Divider The voltage drops shared by the resistors are given as follows: • V1=E x R1/ (R1+ R2+ R3) • V2=E x R2 / (R1+ R2+ R3) • V3=E x R3 / (R1+ R2+ R3) Fundamental of Electricity
  • 24. 24© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 15. Resistors in Parallel R1 R2 R3 I1 I2 I3 I E Fundamental of Electricity
  • 25. 25© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 By Ohm’s Law I1=E/ R1 I2=E/ R2 I3=E/ R3 15. Resistors in Parallel Fundamental of Electricity
  • 26. 26© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 The total current of the circuit I is the sum of I1, I2 and I3 , thus I = I1 + I2 +I3 15. Resistors in Parallel Fundamental of Electricity
  • 27. 27© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 The total resistance or the equivalent resistance(Req) of the circuit is defined to be Req= E/I 15. Resistors in Parallel Fundamental of Electricity
  • 28. 28© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 By substituting the above expression for the currents, we have E/Req=I=E(1/R1+1/R2+ 1/R3) Thus we found 1/Req=(1/R1+1/R2+ 1/R3) 15. Resistors in Parallel Fundamental of Electricity
  • 29. 29© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 Quiz Calculate the total resistance of the following resistors connected in parallel: a)30 Ω, 60 Ω b)180 Ω, 90 Ω, 60 Ω, 60 Ω Fundamental of Electricity
  • 30. 30© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 16. Current Divider The current in each branch of the circuit is given as • I1= I x Req/ R1 • I2= I x Req/ R2 • I3= I x Req/ R3 Fundamental of Electricity
  • 31. 31© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 17. Power and Energy in a resistive circuit Power is equal to the current multiplied by the voltage and the unit of power is watt (W) P = I x E (W) Energy is equal to the power multiplied by the time for the circuit being energized. The unit is Joule (J). Energy = P x t (J) Fundamental of Electricity
  • 32. 32© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 18. Power in a resistive circuit By Ohm's law E=IR, the above equation can be modify to be P = I2R Power is equal to the current squared, multiplied by the resistance. Fundamental of Electricity
  • 33. 33© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 Use Ohm's law again, where I =E/R, we have P=E2/R Power is equal to the voltage squared, divided by the resistance. 18. Power in a resistive circuit Fundamental of Electricity
  • 34. 34© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 Example 1 RBC E=10V I VAB VBC VCD 2Ω 5Ω 3Ω A B C D RAB RCD Fundamental of Electricity
  • 35. 35© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 •VAB + VBC + VCD is exactly equal to the emf=10V •The total resistance of the circuit is 2+5+3=10Ω •By Ohm's law V=IR, the current I should be equal to 1A Example 1 - Answer Fundamental of Electricity
  • 36. 36© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 • VAB = IRAB =1 x 2 = 2V. • VBC = I RBC =1 x 5 = 5V. • VCD = IRCD =1 x 3 = 3V. • Power dissipation in RAB = I2RAB = 12 x 2 = 2W. • Power dissipation in RBC = I2RBC = 12 x 5 = 5W. • Power dissipation in RCD = I2RCD = 12 x 3 = 3W. • Total power dissipated = 2+5+3 =10W Example 1 - Answer Fundamental of Electricity
  • 37. 37© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 Example 2 R1 R2 R3 I1 I2 I3 I E=6V 3Ω 6Ω 2Ω Fundamental of Electricity
  • 38. 38© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 –The potential difference across each of the three resistors is equal to the battery emf 6V –Apply Ohm's Law • E=I1 R1 ; I1=E/R1 = 6/2 = 3A • E=I2 R2 ; I2=E/R2 = 6/3 = 2A • E=I3 R3 ; I3=E/R3 = 6/6 = 1A Example 2 - Answer Fundamental of Electricity
  • 39. 39© Vocational Training Council, Hong Kong EEE3404 Electrical Engineering Principles 1 Weeks 1 ~ 2 • The total current I is equal to the sum of currents • I1+I2+I3 = 3+2+1 =6A. • Power dissipation in R1 = I1 2R1 = 32 x 2 = 18W. • Power dissipation in R2 = I2 2R2 = 22 x 3 = 12W. • Power dissipation in R3 = I3 2R3 = 12 x 6= 6W. • Total power dissipated =18+12+6 =36W Example 2 - Answer Fundamental of Electricity