Mutual inductance ppt

inductance and mutual inductance ppt and self and mutual inductance ppt
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Published Date:23-07-2017
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AC Circuits II Physics 2415 Lecture 23 Michael Fowler, UVa Today’s Topics • Review self and mutual induction • LR Circuits • LC Circuits Definition of Self Inductance • For any shape conductor, when the current changes there is an induced emf E opposing the change, and E is proportional to the rate of change of current. • The self inductance L is defined by: dI E L dt • and symbolized by: • Unit: for E in volts, I in amps L is in henrys (H). Mutual Inductance • We’ve already met mutual Coil 2: Coil 1: • . inductance: when the current N loops N loops 2 1 I in coil 1 changes, it gives 1 rise to an emf E in coil 2. 2 • The mutual inductance M is 21 M  N / I defined by: 21 2 21 1 where is the magnetic  21 flux through a single loop of Coil 2 coil 2 from current I in coil 1. 1 d dI 21 1 ENM 2 2 21 Coil 1 dt dtMutual Inductance Symmetry • Suppose we have two coils close to each other. A changing current in coil 1 gives an emf in coil 2: E M dI / dt 2 21 1 • Evidently we will also find: E M dI / dt 1 12 2 • Remarkably, it turns out that M = M 12 21 • This is by no means obvious, and in fact quite difficult to prove. Mutual Inductance and Self Inductance • For a system of two coils, such as a transformer, the mutual inductance is written as M. • Remember that for such a system, emf in one coil will be generated by changing currents in both coils: dI dI 12 ELM 11 dt dt dI dI 12 EML 22 dt dtEnergy Stored in an Inductance • If an increasing current I is flowing through an inductance L, the emf LdI/dt is opposing the current, so the source supplying the current is doing work at a rate ILdI/dt, so to raise the current from zero to I takes total work I 2 1 U LIdI LI 2  0 • This energy is stored in the inductor exactly 2 1 as U  C V is stored in a capacitor. 2Energy is Stored in Fields • When a capacitor is charged, an electric field is created. • The capacitor’s energy is stored in the field 2 1  E with energy density . 0 2 • When a current flows through an inductor, a magnetic field is created. • The inductor’s energy is stored in the field 2 1 B / with energy density . 2 0LR Circuits • . L • Suppose we have a R steady current flowing I from the battery through the LR circuit Switch shown. • Then at t = 0 we flip the V 0 switch… • This just takes the battery out of the circuit. LR Circuits • The decaying current • . L R generates an emf E LdI / dt I and this drives the Switch current through the resistance: V 0 LdI / dt IR dx / dt ax • This is our old friend at x x e . which has solution 0LR Circuits L R A B C • The equation • . I LdI / dt IR has solution  (R/L)t t/ I I e I e 00 I 0 I(t) so the decay time: 0.37I 0 LR / t 0 L/R 3L/R 2L/R LR Circuits continued… • . L • Suppose with no initial R A B C current we now reconnect I(t) to the battery. • How fast does the current S build up? • Remember that now the V 0 inductance is opposing the battery: V LdI / dt IR 0LR Circuits continued… • . L • Suppose with no initial R A B C current we now reconnect I(t) to the battery. • How fast does the current S build up? • Remember that now the V 0 inductance is opposing the battery: V LdI / dt IR 0LR Circuits continued… • We must solve the equation • . L R A B C V LdI / dt IR 0 or I(t) dI / dt(R / L)IV / L 0 S This differs from the earlier equation by having a V constant term added on the 0 dy / dxayb right. It’s like which you can easily check ax has solution y  Ae b / a . LR Circuits continued… • We’re solving dI / dt(R / L)IV / L 0 • . dy / dxayb • We know the solution to ax is y  A e b / , a where A is a constant to be fixed by the initial conditions. I y, t x, R / L a, V / L b • Equating 0 (R/L)t gives I Ae V / R 0 and A is fixed by the requirement that the current is zero initially, so V t/ 0 I 1 e , L / R  RLR Circuits continued… L R A B C • We’ve solved • . I(t) LdI / dtRIV 0 V 0 and found V V /R t/ 0 0 I 1 e , L / R  R I(t) • Initially the current is zero but changing rapidly—the inductance emf is equal and opposite to the 0 L/R 3L/R 2L/R battery. Clicker Question • The switch S is closed… • . L R R S V 0 Clicker Question • The switch S is closed and • . L R current flows. • The initial current, R immediately after the switch is closed, is: S VR / • A 0 I(t) • B 2/ VR V 0 0 • C VR /2 0 Clicker Answer • The switch S is closed and • . L R current flows. • The initial current, R immediately after the switch is closed, is: S VR / • A 0 I(t) • B 2/ VR V 0 0 • C VR /2 0 The current through the inductance takes time to build up—it begins at zero. But the current through the other R starts immediately, so at t = 0 there is current around the lower loop only. Clicker Question • The switch S is closed and • . L R current flows. • What is the current a long R time later? S VR / • A 0 I(t) V • B 2/ VR 0 0 VR /2 • C 0