Lesson 10 Calculation of Inductance LR circuits
Lesson 10: Inductance Lesson 10 Calculation of Inductance LR circuits Oscillations in LC circuits LRC circuits EMF due to mutual inductance 1 I E Field in Capacitor - - q By passing an electric current through
a capacitor we store energy as an electric E Field in Capacitor field between the plates which has potential energy 1 U = CV 2 2 and energy density 1 u = e E 2 2 I - + q - + 2 By passingan electric current through a solenoidwe form a magnetic Field Bwhich also stores energy B Field in Inductor I - + L INDUCTOR 3 Charging Inductor I Source of Back EMF 4 Charging an Inductor Charging Inductor II Current flows through coilproducing a magnetic field B As current builds up to itsequilibrium value it is changingthus B is changing and hence is changing Changing produces an EMF 5 Charging Inductor III This EMF ind will produce a changing current Iindin the opposite direction to the charging current Thus producing an induced magnetic fieldBind that opposes B After some time this results in equilibrium magnetic field Beq andcurrent Ieq 6 Ieq is actually equal to the steady state current R--why?
and Beq is the magnetic fieldcorresponding to this current The effect of the inductor is to cause a gradual increase of current from zero to the steady state value Charging Inductor IV 7 e Inductance = - N d F dt L dI changing flux in coil
B dt L dI induced emf changing flux in coil changing current For soleniod BA m nIA thus nA n 2 Al ; l length hence solenoid which depends on geometry of coil Inductance 8 SI units I e e [ ] [ ] S . I . units of Inductance L = - dI dt
V Vs [ ] L = = = = H ( Henry ) dI A A dt s Also d F N B dt L = - dI dt d F Wb N B Wb [ ] dt s L = = = = H ( Henry ) dI A A dt s 9 SI units II 10 RL Circuits: Charging I
- L + Kirchoffs Law e e ( ) ( ) - I + = t R t L 11 e e e e e Charging II [ ] ( ) ( ) ( ) ( ) Increasing current - I + = t
L e dI - I ( ) - = t R L dt e e t Rt - - ( ) = I t - = - 1 e L 1 e t L R R L t = Time constant for RL circuit : L R L H [ ] t = = = s L R W 12 Discharging 13 Energy Stored in the Magnetic Field
Energy in Inductor e dI - I ( ) t R - L = dt e dI ( ) ( ) 2 ( ) I t - I t R - L I t = dt Power suppliedJoule heat Rate at which by source emfdissipated energy is stored by load by Inductor dU dI ( ) = B = LI t dt dt U ( t ) I ( t ) 1 ( ) = = ( ) 2 U t dU L I d I = LI t B 2 14 Comparing Inductor and capacitor
15 Energy densities ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) energy
Energy Density = volume Energy densities For solenoid ( ) B t L = m n 2 = m 2 ( ) Al n Vol and I t = m n ( ) 2 1 1 B t ( ) ( ) 2 U t = LI t = m n 2 Vol B 2 2 m n ( ) ( ) 2 U t B t 1 ( ) = ( ) u t B = = 2 B t B Vol 2 m 2 m compare to energy density in capacitor e ( ) ( ) 2 u t = E t E 2 16 Oscillations in LC circuits
V(t) I(t) t) L 17 e 1 Start with fully charged capacitor Q Q U = ; I ( ) = ; V ( ) = 2 C
m ; I ( ) = ; V ( ) = m E 2 C C 2 . Close switch , charge flows from + plate to plate - e - = Kirchoffs Rule : V ( t ) ( t ) Q ( t ) dI Q ( t ) d 2 Q + L = + L = C dt C dt 2 d 2 Q 1 = - Q dt 2 LC 18 to equation for Harmonic Oscillator of mass on spring
d 2 Q 1 compare = - Q dt 2 LC to equation for Harmonic Oscillator of mass on spring for displacement X ( t ) from equilibrium position d 2 X k k = - X = - w 2 X ; w = = angular frequency m dt 2 m This has solution X ( t ) = A cos ( w t + d ) ; A = amplitude d = phase shift 19 Solution to Equations I
20 Solution to Equations II
+ +- +- Solution to Equations II +Qm Q(t) t -Qm +Im t I(t) -Im 21 Analogy to Harmonic Oscillator I
22 Analogy to Harmonic Oscillator II
+ t = 0 L Analogy to Harmonic Oscillator II - t = T/4 L - L t = T/2 + L t = 3T/4 + L t = T - + 23 - Analogy to Harmonic Oscillator III
Total Energy of Harmonic Oscillator Analogy to Harmonic Oscillator III 1 1 = U + K = kX 2 + mv 2 2 2 2 1 1 dX 1 = kX 2 + m = kA 2 ; A is amplitude 2 2 dt 2 Total Energy of LC Circuit 1 1 = U + U = Q 2 + LI 2 E B 2 C 2 1 1 dQ 2 1 = = 2 Q 2 + L Qmax 2 C 2 dt 2 C 24 PE / KE K.E. P.E. UB UE 25 In real circuits always have some resistance
RLC Circuit (dc) I V(t) C I(t) R t) L 26 RLC Circuit (dc) II In RLC circuit get energy lost as joule heat
and oscillations decay in amplitude and one has a damped harmonic oscillator . 27 Picture 28 Flux Current and Inductance Mutual Inductance Mutual Inductance e e e e Flux through coil 2
Nearby circuits effect each other define MUTUALINDUCTION M of coil 2 with 21 F respectto coil 1 as M = N 21 2 21 I 1 F Flux through coil 2 = 21 N = number of turns in coil 2 2 I = current in coil 1 1 Induced emf in coil 2 by coil 1 is e d F dI = - N 21 = - M 1 2 2 dt 21 dt if rate at which currents change are equal e e e dI = = = - M 2 1 dt 29