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UF PHY 2049 - Lecture notes

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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11HitthittCircuit elementAverage PowerReactance Phase of current Voltage amplitudeResistorR Current is in phase with the voltage CapacitorC Current leads voltage by a quarter of a periodInductor L Current lags behind voltage by a quarter of a period1CXCw=LX Lw=R RV I R=CC C CIV I XCw= =L L L LV I X I Lw= =RSUMMARY22mRPR� �=E0CP� �=0LP� �=(31 - 17)An ac generator with emf is connected toan in series combination of a resistor , a capacitor and an inductor , as shown in the figure. The phasorfor the ac genera sinmR CLtw=The series RCL circuitE E( )tor is given in fig.c. The current inthis circuit is described by the equation: sini I tw f= -( )sini I tw f= -The current is for the resistor, the capacitor and the inductorThe phasor for the current is shown in fig.a. In fig.c we show the phasors for thevoltage across , the voltage across R Civ R v Ccommon, and the voltage across .The voltage is in phase with the current . The voltage lags behindthe current by 90 . The voltage leads ahead of the current by 90 . LR CLv Lv i vi v i� �(31 - 18)OABKirchhoff's loop rule (KLR) for the RCL circuit: . This equationis represented in phasor form in fig.d. Because and have opposite directionswe combine the two in a single phasor R C LL CLv v vV VV= + +E( ) ( ) ( ) ( )( )( )2 2 2 22 2 2 22222. From triangle OAB we have: The denominator is known as the " " of the circuit. The current amplitude Cm R L C L C L CmL CmL CVV V V IR IX IX I R X XI ZR X XZ R X X I-� �= + - = + - = + - �� �=+ -= + - � =EEEimpedance22 1mZIR LCww=� �+ -� �� �E( )22 L CZ R X X= + - mIZ=E( )sini I tw f= -(31 - 19)OABFrom triangle OAB we have: tanWe distinguish the following three cases depending on the relative valuesof and . 0 The current phasor lags behind the generatL C L C L CRL LL CV V IX IX X XV IR RX XX Xff- - -= = => � >1. or phasor.The circuit is more inductive than capacitive 0 The current phasor leads ahead of the generator phasorThe circuit is more capacitive than inductive 0 The current phasoC LC LX XX Xff> � <= � =2.3. r and the generator phasor are in phase( )sini I tw f= -( )22 L CZ R X X= + - LX Lw=tanL CX XRf-=1 CXCw=(31 - 20)Fig.a and b: 0 The current phasor lags behind the generator phasor. The circuit is moreinductive than capacitiveL CX X f> � >1. Fig.c and d: 0 The current phasor leads ahead of the generatorphasor. The circuit is more capacitive than inductive Fig.e and f: 0 The current phasor and the generator phC LC LX XX Xff> � <= � =2.3. asor are in phase(31 - 21)In the RCL circuit shown in the figure assume thatthe angular frequency of the ac generator can be varied continuously. The current amplitude in the circuit is given by the equation: mIw=ResonanceE22 The current amplitude11has a maximum when the term 01This occurs when R LCLCLCwwwww� �+ -� �� �- == The equation above is the condition for resonance. When its is satisfied A plot of the current amplitude as function of is shown in the lower figure.This plot is known as a "mresIRI w=Eresonance c "urve mresIR=E1LCw =(31 - 22)2We already have seen that the average power used bya capacitor and an inductor is equal to zero. The power on the average is consumed by the resistor.The instantaneous power P i=Power in an RCL ciruit( )( )20222 20sin1The average power 1sin2cosThe term cos in the equation above is known asthe "TavgTavg rmsrmsavg rms rms rms rms rms rms rmsR I t RP PdtTI RP I R t dt I RTRP I RI I R I IZ Zw fw fff= -� �� �=� �= - = =� �� �= = = =��EE Epower fac " of the circuit. The averagepower consumed by the circuit is maximum when 0f =tor2avg rmsP I R=cosavg rms rmsP I f= E(31 - 23)Power Station Transmission lines rms=735 kV , I rms = 500 A home110 VT1T2Step-up transformerStep-down transformerR = 220Ω1000 km=l2The resistance of the power line . is fixed (220 in our example) Heating of power lines This parameter is also fixed ( 55 MW in our exaheat rmsR RAP I Rr= W=Energy Transmission Requirementslmple)Power transmitted (368 MW in our example)In our example is almost 15 % of and is acceptableTo keep we must keep as low as possible. The only watrans rms rmsheat transheat rmsP IP PP I=Ey to accomplish thisis by . In our example 735 kV. To do that we need a device that can change the amplitude of any ac voltage (either increase or decrease)rms rms=increasing E E(31 - 24)The transformer is a device that can change the voltage amplitude of any ac signal. It consists of two coils with different number of turns wound around a common iron core. The transformerThe coil on which we apply the voltage to be changed is called the " " andit has turns. The transformer output appears on the second coils which is knownas the "secondary" and has turnsPSNNprimary. The role of the iron core is to insure that the magnetic field lines from one coil also pass through the second. We assume that if voltage equal to is applied across the primary then a voltagPV e appears on the secondary coil. We also assume that the magnetic field through both coilsis equal to and that the iron core has cross sectional area A. The magnetic fluxthrough the primary SPVBF ( )The flux through the secondary ( )PP P PSS S S Sd dBN BA V N Adt dtddBN BA V N Adt dtF= � =- =-FF = � =- =-eqs.1eqs.2(31 - 25)( ) ( )If we divide equation 2 by equation 1 we get: PSPP P P PSS S S SSS SP P S PPd dBN BA V N Adt dtddBN BA V N Adt dtdBN AV NdtdBV NN AVNdtVNFF = � =- =-FF = � =- =--= =-=�eqs.1eqs.2 The voltage on the secondary If 1 We have what is known a " " transformerIf 1 We have what is known a " " transformerBoth types of tranSS PPSS P S PPSS P S PPNV VNNN N V VNNN N V VN=> � > � >< � < � <step upstep downsformers are used in the transport of electric power over large distances. S PS PV VN N=(31 - 26)PISIIf we close switch S in the figure we have in addition to the primary current a current in the secondary coil. We assume that the transformer is " " i.e. it suffers no losses due to heating PSII idealthen we have: (eqs.2) If we divide eqs.2 with eqs.1 we get: In a step-up transformer ( ) we have that In a step-down transformer ( ) P …


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UF PHY 2049 - Lecture notes

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