Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29HitthittChapter 31 Electromagnetic Oscillations and Alternating CurrentIn this chapter we will cover the following 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)Suppose this page is perpendicular to a uniform magnetic field and the magnetic flux through it is 5Wb. If the page is turned by 30◦ around an edge the flux through it will be:A. 2.5WbB. 4.3WbC. 5WbD. 5.8WbE. 10WbA car travels northward at 75 km/h along a straight road in a region where Earth’s magnetic field has a vertical component of 0.50 × 10−4 T. The emf induced between the left and rightside, separated by 1.7m, is:A. 0B. 1.8mVC. 3.6mVD. 6.4mVE. 13mVLCThe circuit shown in the figure consists of a capacitor and an inductor . We give the capacitor an initialchanrge and then abserve what happens. The capacitorwill discharge thCLQLC Oscillationsrough the inductor resulting in a timedependent current . i We will show that the charge on the capacitor plates as well as the current 1 in the inductor oscillate with constant amplitude at an angular frequency The total energy in the circuit is tq iLCUw =2 2he sum of the energy stored in the electric field of the capacitor and the magnetic field of the inductor. . 2 2The total energy of the circuit does not change with time. Thus E Bq LiU U UCdU= + = +222200. 1 0dtdU q dq di dq di d qLi idt C dd qLt dt dtqdtdt dt C+== + = = � � ==(31 - 2)LC22221 0 ( )This is a homogeneous, second order, linear differential equationwhich we have encountered previously. We used it to d10 escribethe simple harmonic oscillat o d qL qdt Cd qqdt LC� �+ == ���+��eqs.1222 0 with solr (SHO)( ) ution: ( ) cos( )d xxdtx t X tww f+ == +eqs.2( )If we compare eqs.1 with eqs.2 we find that the solution to the differentialequation that describes the LC-circuit (eqs.1) is:1( ) cos where , and is the phase angle.The current q t Q tLCw f w f= + =( )sindqi Q tdtw w f= =- +( )( ) co sq t Q tw f= +1 LCw =(31 - 3)LC( )( ) ( )2 222 2 2 22 22The energy stored in the electric field of the capacitor cos2 2The energy stored in the magnetic field of the inductor sin sin2 2 2The total energy 2EBE Bq QU tC CLi L Q QU t tCU U UQUw fww f w f= = += = + = += +=( ) ( )22 2cos sin2The total energy is constant; Qt tC Cw f w f� �� �+ + + =� �� �� �energy is conserved223The energy of the has a value of at 0, , , ,...2 2 23 5The energy of the has a value of at , , ,...2 4 4 4 When is maximum is zeE BQ T Tt TCQ T T TtCU U==electric field maximummagnetic field maximumNote : ro, and vice versa(31 - 4)0t =12/ 8t T=3/ 4t T=43 / 8t T=55/ 2t T=4321665 / 8t T=3 / 4t T=7 / 8t T=7878(31 - 5)22If we add a resistor in an RL cicuit (see figure) we mustmodify the energy equation because now energy isbeing dissipated on the resistor. 2E BdUi RdtqU U UC=-= + = +Damped oscillations in an RCL circuit222Li dU q dq diLi i Rdt C dt dt� = + =-( )222/ 222 210 This is the same equation as thatof the damped harmonics o 0 which hscillator: The aas the solution ( ) co ngul r fs a :bt mmdq di d q d q dqi L R qdt dt dt dt dd x dxm b kxdt dtx t x e tt Cw f-+ + =�= += � = � + + =( )2222/ 21 ( )requency For the damped RCL circuit the solutcosion is:The angular fre que4ncy 4Rt LRqk bm mt Qe tLC Lww f w-� �= + = -�= -(31 - 6)/ 2Rt LQe-/ 2Rt LQe-( )q tQQ-( )q t( )/ 2( ) cosRt Lq t Qe tw f-�= +2214RLC Lw�= -/ 222The equations above describe a harmonic oscillator with an exponetially decayingamplitude . The angular frequency of the damped oscillator1 is always smaller than the angular f4Rt LQeRLC Lw-�= -221requency of the 1undamped oscillator. If the term we can use the approximation 4LCRL LCww w=��=(31 - 7)A battery for which the emf is constant generates a current that has a constant direction. This typeof current is known as " " or " "In chapter 30 we encountered a dAlternating Currentdcdirect current ifferent typeof sourse (see figure) whose emf is:sin sin where , is the area of the generatorwindings, is the number of the windings, is the angular frequency of therotation of the windings, and is the magnetic field. m mNAB t t NAB ANBw w w ww= = =E E E This type of generator is known as " " or " " because the emf as well as the currentchange direction with a frequency 2 . In the US 60 Hz. Almost all commercial electricalf fpw= =acalternating current power used today is ac even though the analysis of ac circuits is more complicated than that of dc circuits.The reasons why ac power was adapted will be discussed at the end of thischapter. (31 - 8)sin mtw=E ELCOur objective is to analyze the circuit shown in thefigure ( circuit). The discussion will be greatly simplified if we examine what happens if we connecteach of the three elemRCLThree Simple Circuitsents ( , , and ) separatelyto an ac generator.R C LFrom now on we will use the standard notation for ac circuitanalysis. Lower case letters will be used to indicate the values of ac quantities. Upper case letterswill be usedA conventioninstantaneous( ) to indicate the constant amplitudes of ac quantities.Example: The capacitor charge in an LC circuit was written as:cosThe symbol is used for the instantaneous value of the charge The symbolq Q tqw f= + is used for the constant amplitude of Q q(31 - 9)In fig.a we show an ac generator connected to a resistor From KLR we have: 0 sinThe current amplitude The voltage across is equal to sinThe voltagemR RmRR mRi R i tR RIRv R tww- = � = ==A resistive loadEEEEE amplitude is equal to The relation between the voltage and current amplitudes is: In fig.b we plot the resistor current and the resistor voltage as function of time t. Both quantimRRR RV I Riv=Eties reach their maximum values at the same time. We say that voltage andcurrent are .in phase(31 - 10) R RV I R=A convenient method for the representation of acquantities is that of phasorsThe resistor voltage and the resistor current are representedby rotating
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