Lecture 1 fundamental of electricity
Transcript of Lecture 1 fundamental of electricity
By SY CheungEL / IVE (HW)
LECTURE 1
Fundamental of Electricity
EEE3404 – Electrical Engineering Principles 1
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1. Atom Model Nucleus
Electron
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2. Atom
• Protons
• Neutrons
• Electrons
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“Each atom has the same number of protons and electrons.”
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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
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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
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5. Free Electrons
Free Electrons
Applying Heat or Light
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6. Electrical Materials
All material may be classified into three major classes:
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• Conductors (copper, aluminum, siliver,
platinum, bronz, gold)
• Semiconductors (germanium, silicon)
• Insulators (glass, rubber, plastic, air,
varnish, paper, wood, mica, ceramic,
certain oils)
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• Conductors have many free electrons which will be drifting in a random manner within the material
6. Electrical Materials
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• Insulators have very few free electrons
• Semiconductors falls somewhere between
these two extremes
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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
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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)
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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 emfapplied to it
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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
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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
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9. Potential Difference (p.d.)
LOAD
P.D.=1.4 V
E.m.f.=1.5V, Rint.
Equivalent circuit of a battery
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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
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Circuit convention
+
-
+
-
I
E VR
11. Ohm’s Law
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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.
IRVorI
VRor
R
VI.e.i ===
V
IR
Slope = 1/R
V
I
V α I
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12. Power
Power (P) is defined as the rate of doing work (W).
havewet
QIand
Q
WVSince
Wwattst
WPei
==
= )(..
Additional relationships are obtained by substituting
V=IR and I=V/R, we have:
P = VI
R
VPandRIP
22
==
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13. Resistors in Series
E
I
V1
V2
V3
R1
R2
R3
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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
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• 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
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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)
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15. Resistors in Parallel
R1
R2
R3
I1
I2
I3
I
E
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By Ohm’s Law
I1=E/ R1
I2=E/ R2
I3=E/ R3
15. Resistors in Parallel
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The total current of the circuit I is the sum of I1, I2 and I3 , thus
I = I1 + I2 +I3
15. Resistors in Parallel
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The total resistance or the equivalent resistance(Req) of the circuit is defined to be
Req= E/I
15. Resistors in Parallel
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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
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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
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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
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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)
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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.
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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
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Example 1
RBC
E=10V
I
VAB
VBC
VCD
2Ω 5Ω 3Ω
A B C D
RAB
RCD
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•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
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• 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
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Example 2
R1
R2
R3
I1
I2
I3
I
E=6V
3Ω
6Ω
2Ω
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–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
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• The total current I is equal to the sum of currents
• I1+I2+I3 = 3+2+1 =6A.
• Power dissipation in R1 = I12R1 = 32 x 2 = 18W.
• Power dissipation in R2 = I22R2 = 22 x 3 = 12W.
• Power dissipation in R3 = I32R3 = 12 x 6= 6W.
• Total power dissipated =18+12+6 =36W
Example 2 - Answer
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