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FOWLER CHAPTER 13 LECTURE 13 RCL CIRCUITS CHAPTER 13 COMBINED RESISTANCE, INDUCTANCE AND CAPACITANCE(RCL) CIRCUITS IMPEDANCE (Z): COMBINED OPPOSITION TO RESISTANCE AND REACTANCE. MEASURED IN OHMS. ALL 3 RCL COMPOENTS ARE IN SERIES ELI THE ICEMAN FOR CAPACITORS: CURRENT LEADS VOLTAGE FOR INDUCTORS: VOLTAGE LEADS CURRENT FOR SERIES RCL CIRCUIT CURRENT IS THE SAME IN EACH COMPOENT VOLTAGE IS ALWAYS OUT OF PHASE IN EACH COMPOENT. ANOTHER WAY TO REPRESENT COMPLEX WAVEFORMS IS BY THE USE OF VECTORS SERIES RCL CIRCUITS 1. CURRENT IN RCL CIRCUITS. CURRENT FLOW IN ALL PARTS OF THIS CIRCUIT ARE THE SAME AND IN PHASE. IT I R I L I C 2. VOLTAGE IN RCL CIRCUITS R VR 0 I VL 90 VS FOR INDUCTORS: VOLTAGE LEADS CURRENT L 0 C VR I R 0 VC 90 VL I X L I X L 6.28 fL I FOR CAPACITORS: CURRENT LEADS VOLTAGE VC I X C XC 1 6.28 fC THESE V/I PHASE DIAGRAMS ARE DIFFICULT TO FOLLOW, LETS LOOK AT THIS IN ANOTHER LIGHT. 33 R FIND TOTAL Z FIRST FIND X L AND X C VS X L 6.28 fL 6.28 60 Hz 20mH 7.5 115V 60 Hz XC THINK IN TERMS OF VECTORS OR PHASORS. L 20mH C 10 F 1 1 265.3 6.28 fC 6.28 60 Hz 10 F 7.5 XL 33 R 33 R 33 R X X L XC X 7.5 265.3 X 257.8 X 257.8 X C 265.3 X 257.8 Z Z USE PYTHAGOREAN THEOREM TO FIND THE INPEDANCE Z OF THIS CIRCUIT R 33 Z 2 R2 X 2 OR Z R X R X L X C 2 X 257.8 2 2 2 FOR OUR EXAMPLE Z Z 332 257.8 2 X X L XC X 7.5 265.3 X 257.8 Z 260 SINCE I T I R I C I L USING OHM ' S LAW FIND VT, SINCE R 33 115V 60 Hz L 20mH C 10 F VS 115V 0.44 A 440mA Z 260 VT VR VL VC NONE OF THE VOLTAGES ARE IN PHASE. MUST BE ADDED AS VECTORS. VR IT R VS IT VC I T X C VL IT X L VR IT R 0.44 A 33 14.52V VL IT X L .44 A 7.5 3.3V VC IT X C .44 A 265.3 116.73V MUST USE PHASORS TO FINE VT FOR THIS SERIES CIRCUIT VL 3.3V VR 14.52 VR 14.52 VR 14.52 VC 116.73 V VL VC V 3.3V 116.73V V 113.43V V VT VT V 2 R VT VS 113.43 VT 2 VT V 2 R VL VC 2 VT 14.522 113.43 2 VT 225 12,996 VT 115V PARALLEL RCL CIRCUITS VOLTAGE ACROSS ANY PARALLEL CIRCUIT ELEMENT WILL BE THE SAME AND IN PHASE. SO; VS VR VL VC R VS 33 L 115V 60 Hz 20mH I L LAGS VL BY 90 TO FIND IT , SOLVE FOR I R , I L, I C FIND X L AND X C FIRST X L 6.28 fL 6.28 60 Hz 20mH 7.5 XC C 1 1 265.3 6.28 fC 6.28 60 Hz 10F 10 F I C LEADS VC BY 90 FIND I R , I L , I C USING OHM ' S LAW V 115V IR 3.5 A R 33 V 115V IL 15.3 A X L 7.5 V 115V IC 0.43 A X C 265.3 FIND IT, AGAIN SINCE IR,IL, IC ARE ALL OUT OF PHASE MUST USE VECTORS TO FIND A SOLUTION. I C 0.43 A I R 3.5 A I R 3.5 A IT I L 15.3 A I R 3.5 A I 14.9 A IT I 14.9 A CAPACITIVE CURRENT I IC I L RESISTIVE CURRENT I T I 2 R I C I L INDUCTIVE CURRENT I T 3.5 A2 14.9 A2 COMBINED INDUCTIVE AND CAPACITIVE CURRENT 2 I T 15.3 A TO FIND THE TOTAL IMPEDANCE FOR THIS CIRCUIT USING OHM’S LAW Z VS 115V 7.5 I T 15.3 A FOR A CIRCUIT WITH INDUCTANCE FOR A CIRCUIT WITH CAPACITANCE RESONANCE P.347 RESONANT OCCURS WHEN X L XC CAN OCCUR IN SERIES OR PARALLEL CIRCUITS WITH RCL OR LC COMPOENTS. FOR ANY VALVE OF L AND C THERE IS ONLY ONE FREQUENCY WHERE, X L XC fR THIS IS CALLED THE RESONANT FREQUENCY: 1 6.28 LC DO EX. 13-11 p.348 fR SERIES RESONANT CIRCUITS F.13-26 AT RESONANT X L X C ALSO VL VC PARALLEL RESONANT CIRCUITS P.348 X L X C ALSO I L IC For resonance to occur in any circuit it must have at least one inductor and one capacitor. Resonance is the result of oscillations in a circuit as stored energy is passed from the inductor to the capacitor. Resonance occurs when XL = XC At resonance the impedance of the circuit is equal to the resistance value as Z = R. At low frequencies the circuit is capacitive as XC > XL. At low frequencies the circuit is inductive as XL > XC. The high value of current at resonance produces very high values of voltage across the inductor and capacitor. Series resonance circuits are useful for constructing highly frequency selective filters. However, its high current and very high component voltage values can cause damage to the circuit. Resonant Circuits XL3 XL1 XC3 XL2 XC1 XC2 Frequency fr Resonance occurs when XL equals XC. There is only one resonant frequency for each LC combination. However, an infinite number of LC combinations have the same fr . PARALLEL RESONANT TANK CIRCUIT THIS CIRCUIT WOULD PRODUCE A SINE WAVE FOREVER IF L AND C WERE IDEAL COMPOENTS. WITH REAL WORLD L AND C THE WAVEFORM WILL DAMP OUT WITH TIME. YOU MUST FEED ENERGY INTO THE TANK CIRCUIT TO KEEP THE SINE WAVE PROPOGATING. BANDWIDTH : RANGE OF f OF A CIRCUIT WHICH PROVIDES 70.7% OR MORE OF THE MAX. RESPONSE. Bandwidth, (BW) is the range of frequencies over which at least half of the maximum power and current is provided The selectivity of a circuit is dependent upon the amount of resistance in the circuit. The variations on a series resonant circuit are drawn below. The smaller the resistance, the higher the "Q" for given values of L and C. The parallel resonant circuit is more commonly used in electronics, but the algebra necessary to characterize the resonance is much more involved. SERIES LC CIRCUIT RESPONSE CURVE FOR LC CIRCUIT ARE PLOTS OF EITHER VOLTAGE, CURRENT OR INPEDANCE vs. FREQUENCY ABOVE AND BELOW RESONANCE Series Resonance The resonance of a series RLC circuit occurs when the inductive and capacitive reactance are equal in magnitude but cancel each other because they are 180 degrees apart in phase. The sharp minimum in impedance which occurs is useful in tuning applications. The sharpness of the minimum depends on the value of R and is characterized by the "Q" of the circuit. XL X L XC 0 XC An example of the application of resonant circuits is the selection of AM radio stations by the radio receiver. The selectivity of the tuning must be high enough to discriminate strongly against stations above and below in carrier frequency. FILTERS: USE RC, RL, LC, AND RCL CIRCUITS TO FILTER ONE GROUP OF FREQUENCIES FROM ANOTHER GROUP OF FREQUENCIES. 4 CLASSES OF FILTERS 1.LOW PASS 2.HIGH PASS 3.BAND PASS 4.BAND-REJECT 0R BAND STOP, YOU TUBE: Passive RC low pass filters http://www.youtube.com/watch?v=OBM5T5_kgdI YOU TUBE: Passive RC high pass filters http://www.youtube.com/watch?v=4CcIFycCnxU LOW PASS FILTER THE HIGH PASS FILTER BAND PASS FILTER