Lecture 17 with revised notes for HW 17

Lecture 17 with revised notes for HW 17 - Lecture 17...

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Lecture 17 Chapter 31 Electromagnetic Oscillations and Alternating Current (31 - 1)
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Outline of topics: -Electromagnetic oscillations in an LC circuit -Alternating current (AC) circuits with capacitors -Resonance in RCL circuits -Power in AC-circuits -Transformers -AC power transmission (31 - 1)
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L C The capacitor is initially chanrged with . Discharge of the capacity through the inductor resulting in a time dependent current . Q i LC Oscillations : 2 2 22 2 2 Conservation of the total energy in the circuit leads to 0 0. Both the charge on the capacitor p 1 0 EB q Li U U U C dU dt dU q dq di dq di d q Li i dt C dt dt d d t dt dt q q Lq dt C lates and the current in the inductor oscillate with constant amplitude at an angular frequency 1 i LC (31 - 2)
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L C 2 2 2 2 2 2 2 1 0 ( ) This is an equation for the simple har 0 monic os with solution cillator (SHO 1 ) ( ) : ( ) cos( ) dx dq q dt LC Lq dt x dt x t X t C eqs.1 eqs.2 is the phase angle. The current sin dq i Q t dt ( ) co s q t Q t 1 LC (31 - 3)
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L C 22 2 2 2 2 2 2 The energy stored in the electric field of the capacitor cos The energy stored in the magnetic field of the inductor sin sin 2 2 2 The total energy 2 E B EB qQ Ut CC Li L Q Q U t t C U U U Q U 2 cos sin is a constant. 2 Q tt When is maximum is zero, and vice versa UU Note: (31 - 4)
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0 t 1 2 /8 tT 3 /4 4 3 /8 5 5 /2 4 3 2 1 6 6 5 /8 3 /4 7 /8 7 8 7 8 (31 - 5)
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2 2 If we add a resistor in an RL cicuit (see figure) we must modify the energy equation because now energy is being dissipated on the resistor. 2 EB dU iR dt q U U U C Damped oscillations in an RCL circuit 2 2 2 Li dU q dq di Li i R dt C dt dt /2 2 2 22 1 0 For the damped RCL circuit the solution is: The angular frequenc ( ) cos y 1 4 Rt L dq di d q d q dq i L R q dt dt dt q t Qe dt dt C t R LC L (31 - 6)
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/2 Rt L Qe Rt L Qe () qt Q Q ( ) cos Rt L q t Qe t 2 2 1 4 R LC L (31 - 7)
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Three simple circuits: - A circuit with an ac generator with resistive load - A circuit with an ac generator with capacitive load - A circuit with an ac generator with inductive load
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From KLR we have: 0 sin Both the resistor current and the resistor voltage reach their maximum values at the same time. We say that voltage and current are m RR R R i R i t i v A resistive load in . phase (31 - 10) V I R
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phasors The resistor voltage and the resistor current are represented by rotating vectors known as phasors. Phasors rotate in the counterclockwise direction with angular speed The length of each p RR vi 1. 2. hasor is proportional to the of the ac quantity The projection of the phasor on the vertical axis gives the instantaneous value of the ac quantity. The rotation angle for each phasor is amplitude 3. 4. equal to the phase of the ac quantity ( in this example) t (31 - 11)
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2 We define the "root mean square" (rms) value of as follows: The equation looks the same as in the DC case. This power appears as he 2 at on rms m rms V V P R R V 2 0 2 2 0 2 2 0 2 1 ( ) sin 1 1 si 1 sin n 2 2 T m T m m T P P t dt P t TR td P tdt R t T T P R Average Power for R (31 - 12)
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From KLR we have: 0 sin cos sin 90 The voltage amplitude equal to The current amplitude 1/ The quantity is known a C Cm C C m m C CC
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This note was uploaded on 03/26/2010 for the course PHYS 021 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Lecture 17 with revised notes for HW 17 - Lecture 17...

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