# Mutual Inductance and Self Inductance

###### Mutual Inductance and Self Inductance
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 Clicker Answer • 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 After the current has built up to a steady value, the inductance plays no further role as long as the current remains steady. Clicker Question • After this long time, the switch • . L R is suddenly opened • What are the currents R immediately after the switch is opened VR / • A round the upper loop 0 S VR /2 • B round the upper loop V 0 0 • C all currents zero Clicker Question • After this long time, the switch • . L R is suddenly opened • What are the currents R immediately after the switch is opened VR / • A round the upper loop 0 S VR /2 • B round the upper loop V 0 0 • C all currents zero Clicker Answer • After this long time, the switch • . L R is suddenly opened • What are the currents R immediately after the switch is opened VR / • A round the upper loop 0 VR /2 • B round the upper loop V 0 0 • C all currents zero The inductance will not allow sudden discontinuous change in current, so the current through it will be the same just after opening the switch as it was before. This current must now go back via the other resistance. Clicker Question • The two circuits shown have the • . same inductance and the same t = 0 current, no battery, and resistances R and 2R. • In which circuit does the current decay more quickly A. R B. 2R C. Both the same Clicker Answer • The two circuits shown have the • . same inductance and the same t = 0 current, no battery, and resistances R and 2R. • In which circuit does the current decay more quickly A. R B. 2R 2 The decay is by heat production I R. LC Circuits Question • . • Suppose at t = 0 the switch S is closed, and the resistance in this initial charge circuit is extremely small. Q Q 0 0 • What will happen C S A. Current will flow until the capacitor discharges, after which nothing further will happen. L B. Current will flow until the capacitor is fully charged the opposite way, then a reverse current will take it back to the original state, etc. LC Circuits Answer: B • This is an oscillator • . • The emf V = Q/C from the capacitor Q Q builds up a current through the C inductor, so when Q drops to zero I S there is substantial current. • As this current decays, the inductor L generates emf to keep it going—and with no resistance in the circuit, this is enough to fully charge the oscillator. • We’ll check this out with equations. LC Circuit Analysis • The current . I dQ / dt • . • With no resistance, the voltage across the capacitor is exactly balanced by Q Q the emf from the inductance: C I Q dI S  L C dt • From the two equations above, L 2 d Q Q S in the diagram is  2 the closed switch dt LCQuick review of simple harmonic motion from Physics 1425… Force of a Stretched Spring • If a spring is pulled to • A extend beyond its Natural length natural length by a distance x, it will pull back with a force F kx Spring’s force where k is called the F kx “spring constant”. Extension x The same linear force is also generated when the spring is compressed. Quick review of simple harmonic motion from Physics 1425… Mass on a Spring • Suppose we attach a • A Natural length mass m to the spring, free to slide backwards m and forwards on the frictionless frictionless surface, then Spring’s force pull it out to x and let go. F kx • F = ma is: m 22 md x / dt kx Extension x Quick review of simple harmonic motion from Physics 1425… Solving the Equation of Motion • For a mass oscillating on the end of a spring, 22 md x / dt kx • The most general solution is x Acos t  • Here A is the amplitude,  is the phase, and by 2 putting this x in the equation, mω = k, or  km / • Just as for circular motion, the time for a complete cycle T1/ f 2 / 2 m / k ( f in Hz.)Back to the LC Circuit… • The variation of charge with time is 2 Q Q d Q Q • .  2 dt LC C I S • We’ve just seen that 22 md x / dt kx L has solution x Acost , k / m  from which Q Q cos t, 1/ LC. 0Where’s the Energy in the LC Circuit • The variation of charge with time is Q Q cos t, 1/ LC 0 Q Q • . so the energy stored in the capacitor is C 2 2 2 I S U Q / 2C Q / 2C cost  E 0 • The current is the charge flowing out L IdQ / dt Q sin t 0 so the energy stored in the inductor is 2 2 2 2 2 2 2 11 U LI LQ sint Q / 2C sint1/ LC  B2200 Compare this with the energy stored in the capacitor Clicker Question • Suppose an LC circuit has a very large capacitor but a small inductor (and no resistance). • During the period of one oscillation, is the maximum energy stored in the inductor A. greater than B. less than C. equal to the maximum energy stored in the capacitor Clicker Answer • Suppose an LC circuit has a very large capacitor but a small inductor (and no resistance). • During the period of one oscillation, is the maximum energy stored in the inductor A. greater than B. less than C. equal to the maximum energy stored in the capacitor Energy in the LC Circuit • We’ve found the energy in the capacitor is 2 2 2 Q Q U Q / 2C Q / 2C cost  E 0 • . • The energy stored in the inductor is C I S 2 2 2 1 U LI Q / 2C sint  B 0 2 • So the total energy is L 2 2 2 2 U Q / 2C cos tsin t Q / 2C.  B00 • Total energy is of course constant: it is cyclically sloshed back and forth between the electric field and the magnetic field. Energy in the LC Circuit • . • Energy in the capacitor: electric field energy • Energy in the inductor: magnetic field energy
Website URL
Comment
Presentations
Free
Category:
Presentations
User Name:
Dr.GriffinWood
User Type:
Teacher
Country:
United Kingdom