PHY2054: Chapter 1712Electrons in the WireÎ If the electrons move so slowly through the wire, why does the light go on right away when we flip a switch?1. Household wires have almost no resistance2. The electric field inside the wire travels much faster3. Light switches do not involve currents4. None of the aboveThink of what happens when you turn on a hose full of water. Water atend of hose comes out immediately because of push by pressure wave.PHY2054: Chapter 1713Electrons in the Wire, Part 2 Î Okay, so the electric field in a wire travels quickly. But, didn’t we just learn that E = 0 inside a conductor?1. True, it can’t be the electric field after all!!2. The electric field travels along the outside of the conductor3. E = 0 inside the conductor applies only to static charges4. None of the abovePHY2054: Chapter 1714Resistance and Ohm’s LawÎOhm’s law is an empirical observation: for most materials the current is proportional to the applied voltageÎWe write the constant of proportionality as R and call it the “resistance”, measured in ohms (Ω)ÎExample, 120 V applied to a material gives I = 15 A. R = 120/15 = 8ΩÎMost materials are “ohmic”, i.e. obey Ohm’s law over a very wide range of applied voltages Common “nonohmic” materials are semiconductors such as silicon & germanium for which current rises exponentially with ΔVIV∝ΔVIRΔ=PHY2054: Chapter 1715Resistivity and ResistanceÎResistance depends on Fundamental properties of the material (i.e., resistivity) Geometry, e.g. size and shapeÎFor materials with an “extruded” shape (constant cross sectional area), we can write resistance as ρ = resistivity (property of material) A = cross sectional area L = length of materialÎExample: Find R of nichrome wire, r = 1mm, L = 4mLRAρ=()()82150 10 4/ 3.14 0.001 6.0R−=× × × =ΩPHY2054: Chapter 1716Resistivities of Common Materials0.00455.6 x 10-8Tungsten–0.0480.46Germanium–0.075640Silicon1013Hard rubber1010–1014Glass7.5 x 1017Fuzed quartz0.0004150 x 10-8Nichrome0.003922 x 10-8Lead0.005010 x 10-8Iron0.00342.82 x 10-8Aluminum0.00342.44 x 10-8Gold0.00391.7 x 10-8Copper0.00381.59 x 10-8SilverTemp. Coeff. / °Cρ (Ω⋅m) @ 20° CMaterialPHY2054: Chapter 1717Dependence of R on GeometryÎExample: Material has R = 2Ω for r = 2 mm, L = 3 m Reshape same volume of material to r = 1 mm, L = 12 m (Area shrinks by x 4 so L increases by x 4 to maintain volume) Calculate new resistance (ρ cancels)ÎSo R’ = 2 x 16 = 32 Ω()2222/4216RL LLrRLrrrρρππ′′ ′⎛⎞⎛⎞⎛⎞====⎜⎟⎜⎟⎜⎟′′⎝⎠⎝⎠⎝⎠RR'PHY2054: Chapter 1718Dependence of ρ & R on TemperatureÎGenerally, ρ increases with T because of extra thermal motion of atoms in material Fractional increase per °C is called temperature coefficient α α is normally measured at T0= 20 °CÎExample involving copper wire Resistance of Cu wire is 20 Ω at T = 20 °C Find R at T = 180 °C α = 0.0039/°C, ΔT = 180 – 20 = 160()()0011TRRTρραα=+Δ=+Δ()32020 1 0.0039 160 32.5R =+ ×=ΩAssumes α is constant over this T range!PHY2054: Chapter 1719Electrical Energy and PowerÎEnergy delivered to circuit by EMF source of voltage VÎPower deliveredÎPower dissipated by resistor (heat) Voltage drop is IRÎIn all circuits, power delivered = power dissipated EMF source is needed to replenish power lost by resistors Dissipated power shows up as heatUQVΔ=ΔUQPVIVttΔΔ== =ΔΔ2RQPIR I RtΔ=×=ΔPHY2054: Chapter 1720Power ExampleÎV = 120, R = 15 Ω I = 120 / 15 = 8 Delivered: P = IV = 8 × 120 = 960 W Dissipated: PR= I2R = 82× 15 = 960 WVRPHY2054: Chapter 1721Another Power ExampleÎV = 120, R1= 20 Ω, R2= 40 Ω I = 120 / 60 = 2 Delivered: P = IV = 2 × 120 = 240 W Dissipated R1: P1= I2R1= 22× 20 = 80 W Dissipated R2: P2= I2R2= 22× 40 = 160 W Total dissipated: 240
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