Lecture 27 inductors. stored energy. lr circuits
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This document discusses inductors and mutual inductance. It defines mutual inductance as the induced electromotive force (emf) in one coil due to a change in current in another nearby coil. It provides equations for mutual inductance M relating the induced emf to the rate of current change. It also discusses self-inductance L, where an inductor opposes changes in its own current. RL circuits are analyzed, showing the current grows exponentially with time constant L/R. Inductors can store energy in their magnetic fields, which is released when the current changes.
Lecture 27 inductors. stored energy. lr circuits
- 1. Lecture 27 Inductors. Stored energy. LR circuits.
- 2. Induction between two coils Change the current in coil 1 changes the B-flux through coil 2 induces an emf in coil 2
- 3. Mutual inductance di1 d Φ2 µ − ε 2 = −N2 dt dt ε 2 = −M21 di1 dt M21 = N2 ΦB 2 i1 M21 mutual inductance indicates how large an emf in 2 due to current change in 1 Of course it works both ways: di ε 1 = −M12 2 dt N1 ΦB 1 M12 = i2
- 4. Mutual inductance depends on the geometry, orientation and materials of the coils. It can be shown that M21 = M12 ≡ M Therefore, we have: ε 1 = −M di2 dt ε 2 = −M N2 ΦB 2 N1 ΦB 1 M= = i1 i2 Units: SI Henry 1 H = 1 Wb/A di1 dt
- 5. ACT: Mutual inductance If the current in coil 1 is as shown, which of the graphs gives the correct emf in coil 2? i1 t ε2 ε2 ε2 t A t B t C
- 6. Example: M for two solenoids l N2 turns, length l N1 turns, length l l µ0N2i2 B2 = µ0n2i2 = l Φ12 = M= µ0N2i2 π R12 l N1 Φ12 i2 µ0N2i2 N1 ( π R12 ) l = i2 µ0N1N2π R12 M= l Craig Ogilvie
- 7. Self-inductance Ideal coil (no resistance) di/dt If current through coil changes, → flux through coil changes → in each loop there is an induced emf → loops “in series” ⇒ emf induced between two ends of coil Single coil is called an inductor The effect is called self-induction ε = −L di dt N ΦB L= i
- 8. ACT: Inductor In the circuit shown, the voltage in the power supply is turned up, so that the current increases. During this operation, which point, y or z, is at a higher potential? A. y B. z C. Both have the same potential By Lenz law, induced emf tries to oppose change. Current is increasing so emf tries to reduce it. Imagine a “battery” with positive terminal near y. εin
- 9. ACT: Inductor II In the circuit shown, the voltage in the power supply has been on for a long time. The switch is then opened, as shown. Right after this, which point, a or b, is at a higher potential? A. a B. b C. Both have the same potential εin By Lenz law, induced emf tries to oppose change. Current is decreasing so emf tries to increase it. Imagine a “battery” with positive terminal near a.
- 10. In-class example: Inductance of a solenoid A solenoid is constructed out of 1000 loops of wire wound around a 1.0 cm 2 cross section tube that is 5 cm long. A 20-A current is run through the solenoid. What is the solenoid’s self inductance L? µ0Ni B = µ0ni = l A. 250 H B. 25 H C. 2.5 H D. 2.5×10−3 H E. 2.5×10−5 H length l, N turns µ0Ni π R 2 Φ = BA = l µ0Ni π R 2 N( ) NΦ l L= = i i 2 T m 4π × 10 −7 ( 1000 ) 1.0 × 10 −4 m2 ÷ A L= = 2.5 × 10 −3 H 5.0 × 10 −2 m ( ) µ0N 2π R 2 L= l Inductance of a solenoid
- 11. Inductors against di/dt Inductor always acts to oppose current change. → setting up a current in a circuit with an inductor will take some time. → used in circuit design to protect against rapid changes of currents or spikes
- 12. RL circuits: current growth Switch moved to position “a” at t = 0 induced emf Kirchhoff’s law ε − iR − L di =0 dt i (t = 0 ) = 0 Qualitatively: i ( t = 0 ) = 0 ε i (t → ∞ ) = R drop in potential from top of inductor to bottom Differential equation for i + initial condition (no inductor)
- 13. di ε − iR − L =0 dt di R = dt ε L −i R ε −i − ln R ε R ÷ R ÷= t ÷ L ÷ R R ln 1 − i ÷ = − t ε L i ε/R ε i = R t ε i (t ) = R R − t 1 − e L t − 1 − e τ If L is large, τ is large, i.e. current grows slowly. ÷ ÷ L ÷ τ = ÷ R
- 14. In-class example: RL circuit You have 1000 Ω resistor and you want to build an LR series circuit that will go from 0 to 0.9 V0/R in 1 ms once the switch is closed. What value should L have? R V0 A. 1 H B. 0.43 H L ε i (t ) = R 0.9 = 1 − e τ =− C. 0.1 H D. 0.043 H E. 0.01 H t − 1 − e τ τ = L R − 1 ms ÷ ÷ 1 ms τ ln ( 0.1 ) = 0.43 ms L = R τ = 0.43 H
- 15. ACT: Three circuits All batteries, inductors, resistors are identical. Rank the circuits according to the current through battery just after switch is closed 1 2 A. i1 > i2 > i3 initially B. i2 > i3 > i1 initially C. i3 > i2 > i1 initially 3 Initially i = 0 through inductors. ⇒ i1 = 0 R2 < R3 ⇒ i2 > i3
- 16. DEMO: ACT: Current decay RL circuit The switch has been in position “a” for a long time, then it is moved to position “b” at t = 0. What is the graph for current in the resistor? induced emf i i A B t t −iR − L di =0 dt di < 0÷ dt i → di R =− i dt L C D i t i (t ) = i0e t t − τ τ = L R
- 17. Inductors store energy i t After the switch is opened, the inductor drives the current through the resistor. The energy dissipated in the resistor must have been stored in the inductor.
- 18. Energy stored in an inductor To establish current (from i = 0 to i = I ): di Voltage across inductor: V = L dt (ignoring signs) 1) rate of energy supplied to inductor P = Vi = Li di dt 2) increase of energy within inductor dU = Pdt = Li di I 3) energy stored within inductor U = ∫ dU = L ∫ idi 0 1 U = LI 2 2
- 19. Using the stored energy • • • • DEMO: Spark Energy is stored in L after switch is moved to “a” position Switching to “b” releases energy stored in inductor Time it takes for i = I to i = 0 can be very short Energy released can cause an arc across switch contacts – spark plug, current was stored in ignition coil (an inductor) – pulling plug from wall-socket, spark (wires as the inductor)