Lecture 9. DC Current, Resistance, Ohm’s Law CH1 average: 58% ο. Let there be current! (statics is boring…) 1 Charging a Capacitor equipot. @ V=V + + βV=V βπ = 0 - equipot. @ V=0 βV=V - π− π− π‘=0 π=0 βπ = 0 wires are not equipotential! π=0 equipot. @ V=V + βV=V - π− π = +πΆπ ∗ < π βπ = π ∗ < π π− π‘ = π‘1 π = −πΆπ ∗ + βV=V - π = π = +πΆπΆ βπ = π π = −π = −πΆπΆ equipot. @ V=0 π‘ = "π‘πππππ " 1. Electrons are driven by the electric field in the wires (potential difference). 2. How much time does it take (how long is π‘πππππ )? – in two lectures. 3. To keep current running, we need to maintain the potential difference along a wire. 2 Electrons in Metals: “Classical” Microscopic Picture E=0 Electrons in random motion, colliding with static (= defects) and dynamic (= vibrations) imperfections of crystal lattice. Average distance between scattering events = the mean free path, m.f.p., π (depends on temperature, density of defects, etc.). Typically, π~10 nm @ 300K (still, about 100 lattice periods). Electron speed Wrong estimate: 1 3 ππ£ 2 = π π£π₯ 2 2 2 3 = ππ΅ π 2 π£π₯ = 1.38 β 10−23 π½/πΎ × 300πΎ ≈ 7 β 104 π/π −31 9.1 β 10 ππ Quantum mechanics : the mobile electrons, being fermions, cannot have the same energy if their wavefunctions overlap in space. As a result, their average kinetic energies are much (~102 times) greater than ππ΅ π! The current carriers move with π ≈ πππ π/π that is T-independent! v<<c - we still can apply non-relativistic mechanics. Time intervals between elastic (energy-conserving) collisions: πππ π 10−8 π ≈ = 6 = 10−14 π π£ 10 π/π 3 π¬ ≠ π: Electron Drift πΈ≠0 Under a gentle “breeze” of electric field, the electron “mosquito cloud” drifts slowly inside the wire. π£βπ Why the drift velocity, not constant acceleration??? πΈ=0 Because there are processes of energy dissipation (compare to air friction). The terminal velocity is reached when the rate of energy gained from the electric field becomes equal to the rate of energy loss. … τel τin drift velocity ≡ terminal velocity 4 Current πΈ≠0 ππ πΌ= ππ Current: the charge carried by the current carriers through a wire cross section in unit time. πΈ=0 Units: Amperes disregard random motion ππ π£π β 1π β π΄ πΌ= 1π πΌ πΌ = πππ£π π΄ 1πΆ 1π = 1π΄ n – concentration of charge carriers e – their charge π΄ β π£π β 1π – the volume that passes through cross section in 1s πΌ π = = πππ£π π΄ - current density 5 Estimates and Comments Estimate: a copper wire (n∼1029 m-3, cross section 1 mm2 = 10-6m2) carries current 1 A. Find π£π . πΌ 1π΄ −4 π/π π£π = ≈ 29 −3 = 10 πππ΄ 10 π β 10−19 πΆ β 10−6 π2 The drift velocity: π£π βͺ π£ π£ ≈ 106 π/π Regardless of the nature of charge carriers (both positive and negative carriers exist in different types of conductors), the current is defined as a directional motion of positive charges, it always flows from higher potential to lower potential. DC current resembles a flow of incompressible fluid (continuity equation, no accumulation of charges). 6 Ohm’s Law For metals πΌ ∝ βπ over a broad range of βπ (experimental observation) – Ohm’s Law 1 πΌ = π π or π =π πΌ Units: Ohms, Ω the coefficient of proportionality ≡ the resistance = Georg Ohm Of course, the resistance depends on neither V nor I (in the linear regime, where there is no overheating, etc.) 11 Voltmeters and Ammeters A good voltmeter should have π―π―π―π― π‘π‘π‘π‘ πΉππ and should be connected in parallel with the circuit element being measured. A good ammeter should have π―π―π―π― π₯π₯π₯ πΉππ and should be connected in series with the circuit element being measured. 12 Ohm’s Law and Resistivity Ohm’s Law in terms of the resistivity: π πΈπΏ πΈ πΌ= = = π΄ πΏ π π π π΄ πΌ πΈ π≡ = π΄ π - current density A/m2 Iclicker Question Consider two wires. Wire A is 10 cm long, and wire B is 5 cm long. Both wires are otherwise identical, and both have the same electric field acting in them. How do the currents in these wires compare? A. πΌπ΄ = 4πΌπ΅ B. πΌπ΄ = 2πΌπ΅ C. πΌπ΄ = πΌπ΅ D. πΌπ΄ = 0.5πΌπ΅ E. πΌπ΄ = 0.25πΌπ΅ 13 Typical Resistivities metals semi-metal semiconductor dielectric Resistance of a copper wire: cross section 1 mm2, length 1 m: πΏ 1π −8 π = π ≈ 10 Ω β π β −6 2 = 0.01Ω π΄ 10 π 17 Temperature Dependence of a Metallic Resistance Non-superconducting metals (e.g., Cu) ρ ρres – residual (T=0) resistance, due to static defects less pure ρt ≈ ρ 0 (1 + α T) - high-T approximation very pure, no lattice defects ρres ρ0 0 100 200 300 T, K Platinum resistance thermometer Superconducting materials 18 Conclusion DC current: flow of charge carriers, requires E ≠ 0 in the conductor. Microscopic picture: electron “mosquito cloud” slowly drifting in the field. Linear regime: Ohm’s Law. Resistance: the coefficient of proportionality between V and I, depends on materials parameters. Next time: Lecture 10. Resistor circuits, EMF and Batteries. §§ 24.4 - 24.5 19 Appendix 1. Mobility and Resistance Ohm’s Law implies that the drift velocity is proportional to the electric field: the wire length π πΈπΈ πΌ≡ = π π ↔ πΌ = πππππ Simplistic microscopic model for the mobility: π£π ≈ ππ = ππ π = ππ π after each scattering, the electron loses its kin. energy ππ π= π 1 πΏ π πΏ π = = πππ π΄ ππ 2 π π΄ π£π = ππ µ - the electron mobility … τel a – carrier acceleration τin m – carrier mass τ ≡ τin – time between inelastic collisions 2 10−19 × 10−14 π π≈ ≈ 10−3 −30 10 π βπ Resistance Resistivity (resistance of a cube 1 m3) π = π πΏ πΏ = π ππ 2 π π΄ π΄ π π= 2 ππ π πΌ = ππππ΄π΄ Units: Ohm Units: Ohm β m 20