LECT11_212216 Coefisien Alfa

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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.