Lecture25 - Lecture 25 Chapter 23. Faraday’s Law...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lecture 25 Chapter 23. Faraday’s Law Inductance Constant voltage – constant I, no curly electric field. Increase voltage: dB/dt is not zero emf For long solenoid: B = µ0 NI d Change current at rate dI/dt: emfbat R emfcoil Inductance ENC emfbat emf = Inductance ENC emfbat R L emfind emf ind = L dI dt R emfcoil EC EC µ0 N 2 2 dI πR d dt dI dt Increasing I increasing B emf ind = L emf = − dΦ mag dt emfbat R emfind L – inductance, or self-inductance Unit of inductance L: Henry = Volt.second/Ampere Increasing the current causes ENC to oppose this increase 1 Inductance: Decrease Current ENC EC emfbat R L emfind emf ind = L dI dt RL Circuits •  At t=0, the switch is closed and the current I starts to flow. a I b R I # L emf = − dΦ mag dt Conclusion: Inductance resists changes in current Orientation of emfind depends on sign of dI/dt 4/19/11 Initially, an inductor acts to oppose changes in current through it. A long time later, it acts like an ordinary connecting wire. 6 RL Circuits •  Find the current as a function of time. I= a I b R I RL Circuits ( on) ε 1 − e− Rt / L R Current /1 R L/R 2L/R ε ε 1 − e− Rt / L = 1 − e− t /τ RL R R ( ) ( ) # L I= ( ) Q f( x ) 0.5 I 0 0 •  What about potential differences? 0 1 2 x t/RC t 3 4 Voltage on L VL = L dI = ε e− Rt / L dt 1 1 # f( x V ) 0.5 L 0.0183156 4/19/11 7 4/19/11 0 0 0 1 2 x t 3 4 4 8 2 RL Circuits •  Why does RL increase for larger L? a I b R I RL Circuits After the switch has been in position for a long time, redefined to be t=0, it is moved to position b. a I b R I # L # L •  Why does RL decrease for larger R? 4/19/11 9 4/19/11 10 RL Circuits ( off) ε − Rt / L e R (on) /R 11 L/R 2L/R /1 R L/R 2L/R 1 (off) /R 1 L/R 2L/R Current I= f( x ) 0.5 I Q f( x ) 0.5 I I= ε 1 − e− Rt / L R ( ) 4 f( x ) 0.5 I I= ε − Rt / L e R 0.0183156 0 0 1 2 x t 3 4 4 1 0 0 0 1 2 x t/RC t 0.0183156 3 0 0 1 2 x t 3 4 4 Voltage on L VL = L dI = −ε e− Rt / L dt 1 00 #1 1 00 Q f( x ) V 0.5 L f( xV ) 0.5 L VL = L dI = ε e− Rt / L dt Q f( x ) 0.5 VL VL = L dI = −ε e− Rt / L dt -0 # 4/19/11 0.0183156 0 1 2 x t/RC t 3 4 0 11 0 1 2 x t 3 4 4 0 -# 4/19/11 0 0 1 2 x t/RC t 3 4 12 3 Inductor in Series i L1 i L2 Inductor in Parallel L1 i i1 i2 L2 4/19/11 13 4/19/11 14 LC Circuits •  Consider the LC and RC series circuits shown: •  Suppose that at t=0 the capacitor is charged to a value of Q. ++++ ---- LC Oscillations Kirchoff’s loop rule CR ++++ ---- I Q ++ - C L dI Q VL + VC = L + = 0 dt C C L Is there is a qualitative difference in the time development of the currents produced in these two cases. Why?? 15 4/19/11 4/19/11 16 4 x 0, r1 n .. r1 1.01 1 Q f( x ) 0 Q = Q0 cos(ω0 t ) x 0, r1 n .. r1 LC Oscillations LC Oscillations: Energy Check ω0 = 1 LC 0 1.01 VC 1 f( x ) 0 1.01 0 I 1 1.01 1 0 0 2 4 6 1.01 1 6.28 I = −ωx Q0 sin(ω 0 t ) 0 f( x ) 0 0 0 1.01 1 0 2 x 4 6 6.28 VL 1.01 1 0 1.01 1 0 f( x ) 0 2 x 4 6 6.28 0 •  The other unknowns ( Q0, ) are found from the initial conditions. eg in our original example we took as given, initial values for the charge (Qi) and current (0). For these values: Q0 = Qi, = 0. •  Question: Does this solution conserve energy? dI dt 1.01 1 f( x ) 0 0 0 0 2 tx 4 6 6.28 1.01 1 0 0 2 x 4 4/19/11 6.28 t6 17 4/19/11 18 Energy Check Energy in Capacitor 12 U E (t ) = Q0 cos 2 (ω 0t + φ ) 2C x 0, r1 n .. r1 1 UB versus UE UE f( x ) 0.5 Energy in Inductor 1 2 0 U B (t ) = Lω 0 Q02 sin 2 (ω 0tx+ φ0), r1 .. 0 0 r1 2 n 1 ω0 = LC 1 2 tx 4 6 U B (t ) = Therefore, 12 Q0 sin 2 (ω 0t + φ ) 2C UB f( x ) 0.5 U E (t ) + U B (t ) = Q02 2C 19 0 0 4/19/11 0 2 x t 4 6 20 4/19/11 5 LC Oscillations with Finite R •  If L has finite R, then energy will be dissipated in R and x 0, r1 n Driven Oscillations •  An LC circuit is a natural oscillator. n 100 the oscillations will become damped. 10 r1 r1 .. r1 x 0, n .. r1 ω resonance = 1 in absence of resistive loss LC + - + - R C r1 10 n 100 L Q 1 Q f( x ) 0 0 1 f( x ) 0 0 •  In a real LC circuit, we must account for thex 0, r1 .. r1 resistance of the inductor. This resistance will n Q damp out the oscillations. 1 f( x ) 0 1 1 0 t5 x R=0 10 1 0 5 x 10 0 R=0 t5 x 10 t 21 4/19/11 4/19/11 22 6 ...
View Full Document

This note was uploaded on 04/23/2011 for the course PHYS 272 taught by Professor K during the Spring '07 term at Purdue University-West Lafayette.

Ask a homework question - tutors are online