lect11_212216 coefisien alfa
TRANSCRIPT
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Temperature coefficient of resistivity
)]TT(1[ oo +=
)]TT(1[RR oo +=
T0 = reference temperature
= temperature coefficient of resistivity, units of(C)-1
For Ag, Cu, Au, Al, W, Fe, Pt, Pb: values of are ~3-510-3 (C)-1
T
slope=
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)]TT(1[RR oo +=
Example: A platinum resistance thermometer uses thechange in R to measure temperature. Suppose R0 = 50at T0=20 C.for Pt is3.9210-3 (C)-1 in this temperature range.
What is R when T = 50.0 C?
R = 50[1 + 3.92 10-3 (C)-1 (30.0 C)] = 55.9
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Temperature coefficient of resistivity
Example: A platinum resistance thermometer has aresistance R0 = 50.0at T0=20 C.for Pt is 3.9210
-3
(C)-1. The thermometer is immersed in a vesselcontaining melting tin, at which point R increases to91.6. What is the melting point of tin?
)]TT(1[RR oo +=
91.6 = 50[1 + 3.9210-3 (C)-1 (T20C)]
1.83 = [1 + 3.9210-3 (C)-1 (T20C)]
0.83 = 3.9210-3 (C)-1 (T20C)
212C = T20C
T = 232 C
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Light bulbs
Inert gas (typically Ar) is used to make sure that filamentis housed in an O-free environment to preventcombustion reaction between W and O.
Englishman Sir Joseph Swan (1878) &American Thomas Edison (1879).
Filament: The atoms are heated to4000 F to emit visible light. Tungsten is
durable under such extremetemperature conditions. (In weaker, lessdurable metals, atomic vibrations breakapart rigid structural bonds, so materialbecomes molten/liquid)
http://www.myhomeus.com
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Typical tungsten filament: ~1 m long, but 0.05mm inradius.
Calculate typical R.
A = (5x10-5m)2 = 7.9x10-9 m2
= 5.6x10-8m (Table 21.1)
R = L/A = (5.6x10-8m) (1m)/ 7.9x10-9 m2 = 7.1
Note: the resistivity value used above is valid only at atemperature of 20C, so this derived value of R holds only forT=20C.
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Calculate at T=2000C, assuming a linear-T relation:
For tungsten, = 4.5x10-3/C
= 0[1+(T-T0)] = 5.5x10-7m
R = L/A = 70 .
(note I suspect the -T relation in reality may not be strictly
linear over such a wide range of temperature; my guesswould be that the above value of may only be valid for
temperatures of tens to hundreds of C. In a few slides, wederive R from the power consumption and get R=144 ,
which is probably more realistic)
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21.3: Superconductors
For some materials, as temperaturedrops, resistance suddenly plummets to0 below some Tc.
Once a current is set up, it can persistwithout any applied voltage becauseR0!
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Superconductors
Applications:
Energy storage at power plantsSuper conducting distribution power lines couldeliminate resistive lossesSuperconducting magnets with much strongermagnetic fields than normal electromagnets
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More recently: As the field has advanced,
materials with higher values of Tc get discovered21.3
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21.4: Electrical Conduction: A Microscopic View
Re-cap: Microscopic motions of charge carriers are random,v of ~ 106 m/s; collisions with molecules
When applied E-field is 0, net velocity is zero, and I=0
e
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e
E
When an E-field is applied,
electrons drift opposite to fieldlines. Average motion is vdrift,typically tenths to a few mm/s
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e
E
Stronger applied E-field
means larger vdrift,
vdrift prop to E
I prop to vdrift
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The excess energy acquired by the electrons in thefield is lost to the atoms of the conductor during thecollision
The energy given up to the atoms increases their
vibration and therefore the temperature of theconductor increases
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21.5 Electrical Energy and Power
Power dissipated in a R is due to collisions of chargecarriers with the lattice. Electrical potential energy isconverted to thermal energy in the resistor--a light bulb filament thus glowsglowsor toaster filaments give off heat (and turn orange)
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Power = work / time = qV/t
P = I * V
P= I2 R
P = V2 / R
UNITS:P = I V = Amp * Volt = C/s * J/C = J/s = WATT
Power dissipated in a resistor
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Example: A typical household incandescent
lightbulb is connected to a 120V outlet. The poweroutput is 100 Watts. What's the current through thebulb? Whats R of the filament?
V = 120 V (rel. to ground)
P=IV I = P/V = 100W/120V = 0.83 A
P = V2 / R ----> R = V2 / P = (120V)2 / 100 W =144
Note -- a few slides earlier, wed estimated the typical resistance of a tungsten
light bulb filament at 2000C -- that estimate of ~70 assumed for simplicity a
constant coefficient of resistivity from 20C to 2000C, which might not be the
case in reality. If the actual value of increases as T increases, then thedependence ofon T will also be non-linear
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Electric Range
A heating element in an electric range is rated at2000 W. Find the current required if the voltage is240 V. Find the resistance of the heating element.
P = IV I = P/V = 2000W/240V = 8.3 A
R = V2 / P = (240V)2/2000W = 28.8
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Cost of electrical power
1 kilowatt-hour = 1000 W * 1 hour = 1000 J/s (3600s) = 3.6x106 J.
1kWh costs about $0.13, typically
How much does it cost to keep a single 100W light bulb on for 24
hours?(100W)*24hrs = 2400 W-hr = 2.4kWh2.4kWh*$0.13 = $0.31
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Power Transmission
Transmitting electrical power is done much more efficientlyat higher voltages due to the desire to minimize (I2R) losses.
Consider power transmission to a small community which is 100mi from the power plant and which consumes power at a rate of
10 MW.
In other words, the generating station needs to supply whateverpower it takes such that Preq =10 MW arrives at the end user(compensating for I2R losses): Pgenerated = Ploss + Preq
Consider three cases:A: V=2000 V; I=5000 A (Preq = IV = 10
7 W)B: V=20000 V; I=500 A (Preq = IV = 10
7 W)C: V=200000 V; I=50 A (Preq = IV = 10
7 W)
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Power Transmission
Resistance/length = 0.0001 / foot.Length of transmission line = 100 mile = 528000 feet.Total R = 52.8 .
A: Ploss = I2R = (5000A)2(52.8) = 1.33x103 MW
Pgenerated = Ploss + Preq = 1.33x103 MW + 10 MW = 1.34x103 MW
Efficiency of transmission = Preq / Pgenerated = 0.75%
B: Ploss = I2R = (500A)2(52.8) = 13.3 MW
Pgenerated = Ploss + Preq = 13.3 MW + 10 MW = 23.3 MWEfficiency of transmission = Preq / Pgenerated = 43%
C: Ploss = I2R = (50A)2(52.8) = 0.133 MWPgenerated = Ploss + Preq = 0.133 MW + 10 MW = 10.133 MWEfficiency of transmission = Preq / Pgenerated = 98.7% (most reasonable)
Lower current during transmission yields a reduction in Ploss!
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You can do the same exercise for local distribution lines(assume Preq = 0.1 MW), which are usually a few miles long
(so the value of R is ~ a few) and need to distribute powerfrom substations to local neighborhoods at a voltage of atleast a few thousand volts (keeping currents under ~30A,roughly) to have a transmission efficiency above ~90%.
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Household circuits
Circuits are in parallel. All deviceshave same potential. If one devicefails, others will continue to work atrequired potential.
V is 120 V above ground potential
Heavy-duty appliances (electricranges, clothes dryers) require 240V. Power co. supplies a line which
is 120V BELOW ground potential soTOTAL potential drop is 240 V
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Circuit breakers or fuses are
connected in series.
Fuses: melt when I gets toohigh, opening the circuit
Circuit breakers: opens circuitwithout melting. So they canbe reset.
Many circuit breakers useelectromagnets, to bediscussed in future chapters
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Example: Consider a microwave oven, a toaster, and aspace heater, all operating at 120 V:
Toaster: 1000 WMicrowave: 800 WHeater: 1300 W
How much current does each draw? I = P/VToaster: I = 1000W/120V = 8.33AMicro: I = 800W/120V = 6.67AHeater: I = 1300W/120V = 10.8 A
Total current (if all operated simultaneously)= 25.8 A(So the breaker should be able to handle this level ofcurrent, otherwise it'll trip)
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Electrical Safety
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Rskin(dry) ~ 105
So forV = 10,000V:
I = V/R = 10,000V/105 = 0.1 = dangerous.
But Rskin(wet) is much, much lower, ~103:
So in this case, when V = 120V, I is also ~ 0.1 A =
dangerous
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21.6-21.9: Direct-Current Circuits
EMF
Resistors in Series & in Parallel
Kirchoffs Junction & Loop Rules forcomplex circuits
RC Circuits
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21.6: Sources of EMF
In a closed circuit, the sourceof EMF is what drives andsustains the current.
EMF = work done per charge:Joule / Coulomb = Volt
Assume internal resistance rof battery is negligible.
Here, = IR
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R
A
D
B
C
From A to B: Potential
increases by V = +From B to A: Potential
decreases by V = .
From C to D: Potentialdecreases by V = IR
=
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If circuit is grounded: V at points A & D will bezero.
R
A
D
B
C
From A to B: Potential
increases by V = +From B to A: Potential
decreases by V = .
From C to D: Potentialdecreases by V = IR
=
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Why is this useful?
The middle voltage can be 'tailored' to
any voltage we desire (between 0 and
) by adjusting R1 and R2!
R1
R2
B
A
C
D
E
FA B C D E F
V
0
IR1
IR2
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A batterys internal
resistance rIn reality, there will be some V
lost within the battery due tointernal resistance: Ir
Terminal Voltage actuallysupplied: V = Ir.