Lecture Notes 3
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Transcript of Lecture Notes 3
3.0 Electric Current
EPF 0024: Physics II 1
OutlineOutline
3.1 Production of direct electric current3 2 Oh ’ L3.2 Ohm’s Law3.3 Resistivity3 4 El t i E d P3.4 Electric Energy and Power
EPF 0024: Physics II 2
Today's lecture Include:Today s lecture Include:
Production of direct electric current
Ohm’s Law
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ObjectivesObjectives
To explain the basic principles of a simplecell and define the electric current.
State and explain Ohm’s law.
Solve Problems.
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3.1 Principle of a Simple Cellp p
A simple cell (Fig. 3.1) + − Zinc Carbon
+ Terminal − Terminal
A simple cell (Fig. 3.1)consists of two rods(electrodes) of carbon andzinc immersed in solution
electrode (−)electrode (+)
zinc immersed in solution,e.g. diluted sulfuric acid(electrolyte).
The terminals of the cell isthe portion of the electrode
t id th l t l t A
Acid
Electrical symbol for a celloutside the electrolyte. ABattery are several cellsconnected together (in
i )
Electrical symbol for a cell
+ −
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series). Fig. 3.1 A Simple Electric Cell
3 1 1 Operation of a simple cell3.1.1 Operation of a simple cell
Electrolyte dissolves zinc and each zinc atomyleaves 2 electrons behind and enters theelectrolyte as positive ion making zinc electrode–vely charged and electrolyte +vely charged.y g y y gThe +ve electrolyte pulls off electrons from thecarbon electrode making it +vely charged and ap d now exists between the two terminalsp.d. now exists between the two terminals.
The p.d. (voltage) that exists between terminalsp ( g )is called the electromotive force (emf). Allowingcharges to flow externally results in more zincbeing dissolved to maintain constant voltage at
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being dissolved to maintain constant voltage atterminals. Eventually, the zinc will be used up.
3.1.2 The Electric Current
A simple electric circuit ispshown in (Fig. 3.2 (a)).Closing the switch S results ina net motion of electrons fromthe negative terminal to thepositive terminal.
Motion of charges isrepresented as flow ofconventional electric currentconventional electric currentfrom +ve terminal to negativeas shown in Fig. 3.2 (b). Fig. 3.2 (a) flashlight (simple
electric circuit) & (b) direction
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electric circuit) & (b) direction of current & electron flow.
The electric current I is defined as the netamount of charge that passes through a givencross-section of a conductor per unit time:
(3.1) tQI = Constant current
tQIΔΔ
= Variable current
The SI unit of I is coulomb per second (C/s)and is known as the ampere (symbol: A)
tΔ
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and is known as the ampere (symbol: A).
ExampleExample
A steady current of 2.5 A flows in a wire for4 0 minutes (a) How much charge pass4.0 minutes. (a) How much charge passthrough any point in the circuit. (b) How manyelectrons would this beelectrons would this be.
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SolutionSolution
(a) ( )( ) C 600s 604C/s 5.2 =×== Itq
(b)( )C600
= neq( )
( ) electrons. 108.3C101.6
C600 2119 ×=
×== −e
qn
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3 5 Ohm’s Law3.5 Ohm s Law
States: The current flowingth h d t i di tl
I as a function of V is astraight line through theorigin (Fig 3 3)through a conductor is directly
proportional to the potentialdifference applied to its ends. RV
I 1slope =ΔΔ
=
I (A)
origin (Fig. 3.3).
IR VRVI == or (3.2)
I (A)
ΔI
Where the proportionalityconstant R is called resistance( it h (Ω)) V (V)
ΔV
(units = ohm (Ω)).
Fig. 3.3: I versus V for
V (V)
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Circuit symbol for Rg
conductors
Resistors are an indispensable part of allelectronic components (Fig. 3.4).
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Fig. 3.4 Electronic components
Table 3.1 shows the convention to determinethe value of a resistor using color codes
Table 3.1: Color code for resistorsColor 1st digit 2nd digit Multiplier Tolerance (%)
the value of a resistor using color codes.
Color 1 digit 2 digit Multiplier Tolerance (%)Black 0 0 1Brown 1 1 10Red 2 2 102Red 2 2 102
Orange 3 3 103
Yellow 4 4 104
G 5 5 105Green 5 5 105
Blue 6 6 106
Violet 7 7 107
8Grey 8 8 108
White 9 9 109
Gold 0.1 5
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Silver 0.01 10No color 20
Fig. 3.5 is an example indicating the decodingof the actual value of a resistor
Using the color code in Table 3.1 the resistor
of the actual value of a resistor.
1st digit (red)
gvalue is determined to be 25 kΩ ± 10%.
1st digit (red)
2nd digit (green)
Tolerance (silver)
Multiplier (orange)
( )
EPF 0024: Physics II 14Fig. 3.5: Decoding of actual resistor value
Example 1p
A small flashlight bulb draws 300 mA from its1 5 V b tt ( ) Wh t i th i t f th1.5 V battery. (a) What is the resistance of thebulb? (b) If the voltage is dropped to 1.2 V,how would the current change?how would the current change?
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SolutionSolution
(a) Applying Ohm’s law we find:(a) Applying Ohm s law we find:
( ) Ω05V 5.1===
VR
(b) If the voltage drops to 1 2 V assuming the
( ) Ω0.5A 0.3
===I
R
(b) If the voltage drops to 1.2 V, assuming theresistance stayed constant, then
( )( )( ) mA. 60 of drop aor A 24.0
Ω 5.0V2.1
===RVI
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3 6 Resistivity3.6 Resistivity
For conductors:
(3 3)LRLR (3.3)
Wh i i ti it L l th d A
AρR
AR =∝ or
Where ρ is resistivity, L length and A cross-sectional area. From equation (3.3) wededuce the SI unit of resistivity to be Ω.m.deduce the SI unit of resistivity to be Ω.m.
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ExampleExample
Given that the resistivity of a copper wire is1 7 × 10−8 Ω m Find (a) the diameter of a 20 m1.7 × 10 8 Ω.m. Find (a) the diameter of a 20-mcircular wire if the resistance of the wire is0.10 Ω. (b) What is the voltage drop across( ) g p
the wire if the current flowing through the wireis 12 A.
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Solution
(a) 4==
ρρ LLR( )
4
2
=
==
ρπ
RLd
dAR
( )( )( )( ) mm 2.1m101.2
Ω0 10m 20Ω.m107.14 3
8
=×=×
= −−
πR
( )Ω0.10π
(b) Using Ohm’s
( )( )Ω 10.0A 12 == IRV
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V 1.2 =
Today's lecture Include:Today s lecture Include:
Temperature Effect on Resistance.
SuperconductivitySuperconductivity.
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Objectivesj
E plain the effect of increasing temperat re onExplain the effect of increasing temperature onresistance.
Explain superconducting effect.
Consider some applications related to theseconceptsconcepts.
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3 7 Temperature Effect on Resistance3.7 Temperature Effect on Resistance
Resistivity of metals increases linearly withResistivity of metals increases linearly withincreasing temperature (for moderatetemperatures of up to 300oC) according to:
)1( ToT Δ+= αρρ (3.4)
where ρo is the resistivity at 0oC and α istemperature coefficient of resistivity. The rapidvibration of atoms at higher temperaturesvibration of atoms at higher temperaturescauses a conductor to have higher resistance.
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3.7.1 Resistance Thermometer
Variation of R with temperature is used forpprecise temperature measurement. If at 0°C Rfor Pt is 164.2 Ω. When placed in a solution, Rincreases to 187.4 Ω. What is the temperature ofthe solution if α for copper is 3.927 × 10−3 (oC)−1?
Since R0 = ρ0L/A , we can write equation 3.4 as( )
( ) C35.9Ω 2.164Ω 4.187C0 ,1
o1
0
0
=−
=−
=
=−=ΔΔ+=RRT
TTTTRR oα
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( ) ( )C35.9
Ω 2.164C103.927 1o30 ×
−−RT
α
3 7 2 Superconductivity3.7.2 Superconductivity
At low temperatures the resistance of certainpmetals and their alloys drop to zero. The effectis termed superconductivity and materialsexhibiting the phenomenon are calledsuperconductors.
It was first observed by Onnes in 1911, whenmercury was cooled down to below 4 2 K Inmercury was cooled down to below 4.2 K. Ingeneral materials become superconductingwithin a few degrees of absolute zero
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within a few degrees of absolute zero.
Resistivity of superconductors is practicallyy p p yzero. Current in a ring-shaped superconducting coil has been observed to flow for
f ffyears in the absence of a potential difference.
Earlier, the highest temperature at whichsuperconductivity is achieved was 23 K andso requires liquid hydrogen cooling Currentlyso requires liquid hydrogen cooling. Currentlysome alloys have been developed that canbe superconducting at 90 K requiring cooling
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be superconducting at 90 K requiring coolingin boiling liquid nitrogen.
Advantages of superconductivity :
(i) Smaller electric motors and generators.Electric cars will be practical.
(ii) Less power lost on transmission lines anduse of thinner wires feasible (cost saving).
(iii) Faster computer and more efficient highspeed train levitation.
EPF 0024: Physics II 26