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Transcript
Lecture 21. R-L and L-C Circuits. Outline: Mutual Inductance βChargingβ and βdischargingβ an inductor in R-L circuits. Oscillations in L-C circuits 1 Self-Induction and Inductance πΉπΉπΉπΉπΉπΉπΉ π π π π : β° = β Variable battery πΌ π© The back e.m.f. : So: The flux depends on the current, so a wire loop with a changing current induces an e.m.f. in itself that opposes the changes If we change the current by varying the voltage, the flux changes and an βadditionalβ e.m.f. is induced (sometimes called the back e.m.f.). Lenz: This e.m.f. will drive a current in the direction that opposes the change in the magnetic flux. πΞ¦ ππ β° =β = βπΏ ππ ππ Inductance: (or self-inductance) πΞ¦π΅ ππ Ξ¦ πΏβ‘ πΌ πΏ β the coefficient of proportionality between Ξ¦ and πΌ, purely geometrical quantity. Units: Tβ m2/A = Henry (H) The self-inductance L is always positive. This definition applies when there is a single current path (see below for an inductance of a system of distributed currents π½ π ). Inductance as a circuit element: 2 Inductance vs. Resistance Inductor affects the current flow only if there is a change in current (di/dt β 0). The sign of di/dt determines the sign of the induced e.m.f. The higher the frequency of current oscillations, the larger the induced e.m.f. 3 Mutual Inductance Letβs consider two wire loops at rest. A time-dependent current in loop 1 produces a time-dependent magnetic field B1. The magnetic flux is linked to loop 1 as well as loop 2. Faradayβs law: the time dependent flux of B1 induces e.m.f. in both loops. loop 2 π©π πΌ1 loop 1 Primitive βtransformerβ The e.m.f. in loop 2 due to the time-dependent I1 in loop 1: πΞ¦1β2 β°2 = β ππ Ξ¦1β2 = π1β2 πΌ1 The flux of B1 in loop 2 is proportional to the current I1 in loop 1. The coefficient of proportionality π1β2 (the so-called mutual inductance) is, similar to L, a purely geometrical quantity; its calculation requires, in general, complicated integration. Also, itβs possible to show that π1β2 = π2β1 = π so there is just one mutual inductance π. π can be either positive or negative, depending on the choices made for the senses of transversal about loops 1 and 2. Units of the (mutual) inductance β Henry (H). β°2 = βπ ππΌ1 ππ π= Ξ¦1β2 πΌ1 4 Calculation of Mutual Inductance l π1β2 = π2β1 = π Calculate M for a pair of coaxial solenoids (π1 = π2 = π, l 1 = l 2 = l, π1 and π2 - the densities of turns). π΅1 = π0 π1 πΌ1 the flux per one turn 2 π2 Ξ¦ π π π ππ 0 1 2 π= = π2 π0 π1 ππ 2 = l πΌ1 2 ππ 2 π π 0 πΏ = π0 π2 l ππ 2 = l Experiment with two coaxial coils. Ξ¦ = π΅1 ππ 2 = π0 π1 πΌ1 ππ 2 π= πΏ1 πΏ2 The induced e.m.f. in the secondary coil is much greater if we use a ferromagnetic core: for a given current in the first coil the magnetic flux (and, thus the mutual inductance) will be ~πΎπ times greater. ππΌ1 ππΌ1 β°2 = πβ = πΎπ π ππ ππ no-core mutual inductance 5 βDischargingβ an Inductor π π π‘ π βπΏ Initially the switch is closed, the coil L conducts a certain current (depends on the coilβs resistance π ), and the corresponding magnetic field energy is stored in the coil. ππ π‘ β° = βπΏ >0 ππ ππ π‘ β π π‘ π = 0, ππ πΌ0 The switch is open at t = 0. As the current begins to decrease, the e.m.f. is induced in the coil that opposes the change by βpushingβ the current through the coil and the resistor π . - π π‘ is the instantaneous value of the current through the coil and through the resistor π . Letβs assume that π β« π: ππ π‘ π + π π‘ = 0, ππ πΏ π π‘ = πΌ0 πππ β π‘ πΏ/π πΌ0 = π/π - the steady current through the inductor (switch S is closed). πΏ π= π π‘=π ππ π‘ π = β ππ, π π‘ πΏ - the time constant of the circuit which consists of an inductor πΏ and resistor π . Letβs check that πΏ/π has units of time: πΏ πΏπΌ 2 π½ = 2β =π π π πΌ π½/π 6 Capacitors vs. Inductors π π‘ = πΌ0 πππ β π‘ πΏ/π βcurrent supplyβ πΌ0 was the current trough the inductor when the switch was βONβ. It could be unrelated to the net R of the circuit. The initial rate of energy dissipation: π β πΌ0 2 π (the larger R, the faster energy decreases) β thus, π β 1/π . πΌ0 = π0 /π π = π π βvoltage supplyβ Looks similar, right? BUT the current πΌ0 = π0 /π is inversely proportional to R! Thus, the initial rate of energy dissipation: π β π0 2 /π (the larger R, the slower energy decreases) β thus, π β π . 11 βDischargingβ an Inductor (contβd) π π Letβs check that the energy stored in the magnetic field is 100% transformed into Joule heat after switch S is open. The initial magnetic field energy stored in the inductor: π π‘ 1 ππ΅ = πΏ πΌ0 2 2 1 π = πΏ 2 π Switch S is open at t = 0. The back e.m.f.: β° π‘ Power dissipated in the resistors: πΌ0 = π/π - the steady current through the inductor (switch S is closed). ππ π‘ = πΏ ππ = π +π π π‘ π π‘ π π‘ = πππ β π πΏ/ π + π π = π + π π2 π‘ Net thermal energy released in the resistors: 2 β β π 2 οΏ½ π + π π π‘ ππ = π + π οΏ½ π 0 = π +π π π 2 0 2 βπ/2 π ββ/π β π β0/π π = πΏ/ π + π π β2π‘/π ππ 1 π = πΏ 2 π 2 14 Example π π π π‘ A. never Initially the switch is closed, the coil L conducts a certain current, and the corresponding magnetic field energy is stored in the coil. The switch is open at t = 0, and the back e.m.f. β° is induced in the coil. Can β° π‘ = 0 be greater than V? B. always C. only if R+r is sufficiently small D. only if R+r is sufficiently large β° = βπΏ ππ π‘ ππ π π‘ π0 πππ β ππ πΏ/ π + π π +π π‘ = πΏπ0 πππ β πΏ πΏ/ π + π π π‘ = π + π πππ β π πΏ/ π + π = βπΏ π0 = π/π 15 βChargingβ an Inductor As an inductor opposes any decrease in the current flowing through it, it also opposes any increase in that current. At t = 0 the switch is closed [π π‘ = 0 = 0]. As the current begins to increase, the e.m.f. is induced in the coil that opposes the change. β°βπΏ ππ π‘ βπ π‘ π =0 ππ ππ π‘ π β° + π π‘ = , ππ πΏ πΏ π‘ β« π: β° π‘ π π‘ = 1 β πππ β π πΏ/π π π‘ = β° π 16 Oscillations in L-C Circuits L-C circuits: the circuits with TWO elements that can store energy (ideally, without dissipation). The energy flow back and forth between L and C results in harmonic oscillations of π π‘ and π π‘ . π2 electric field energy ππΈ = in the capacitor 2πΆ π π‘ π0 2 = 2πΆ 2πΆ 2 πΏπΏ 2 magnetic field energy ππ΅ = in the inductor 2 Letβs say, at t = 0 the capacitor is fully charged, π=0. πΏ π π‘ + 2 2 19 Oscillations in L-C Circuits (contβd) ππ π‘ π2π π‘ β‘ , ππ ππ‘ 2 ππ π‘ π π‘ βπΏ β = 0, ππ πΆ π= 1 πΏπΏ 1 2π πβ‘ β‘ = 2π πΏπΏ π π Solution: π π‘ = π0 πππ ππ + π0 amplitude π 2 π0 2 π0 2 2 ππΈ = = πππ ππ = 1 + πππ 2ππ 2πΆ 2πΆ 4πΆ 2 2π πΏπΏ 2 πΏπ2 π0 πΏπ 0 ππ΅ = = π π π 2 ππ = 2 2 4 π/2 π2π π‘ 1 + π π‘ =0 ππ‘ 2 πΏπΏ 2 π π‘ = angl. frequency phase at t=0 ππ π‘ = βππ0 π π π ππ + π0 ππ 1 + π π π 2ππ π 20 L-C-R Circuits ππ π‘ π π‘ βπΏ β β ππ = 0, ππ πΆ Solution for weak damping π2π π‘ π ππ π‘ 1 + + π π‘ =0 ππ‘ 2 πΏ ππ πΏπΏ 1 πΏπΏ π 2 β« 4πΏ2: π π π π‘ = π0 πππ β πππ πβ π‘ + π0 2πΏ weak damping (βsmallβ R) πβ = 1 π 2 β πΏπΏ 4πΏ2 strong damping (βlargeβ R) 23 Next time: Lecture 22. AC circuits, Reactance. §§ 31.1 - 2 24