Physics 2102 Jonathan Dowling Lecture 21 Alternating Current Circuits Alternating Current To keep oscillations going we need to drive the circuit with an external emf that produces a current that goes back and forth Notice that there are two frequencies involved one at which the circuit would oscillate naturally The other is the frequency at which we drive the oscillation However the natural oscillation usually dies off quickly exponentially with time Therefore in the long run circuits actually oscillate with the frequency at which they are driven All this is true for the gentleman trying to make the lady swing back and forth in the picture too Alternating Current We have studied that a loop of wire spinning in a constant magnetic field will have an induced emf that oscillates with time E Em sin d t That is it is an AC generator AC s are very easy to generate they are also easy to amplify and makes them easy to send in distribution decrease in voltage This in turn grids like the ones that power our homes Because the interplay of AC and oscillating circuits can be quite complex we will start by steps studying how currents and voltages respond in various simple circuits to AC s AC Driven Circuits emf vR 0 1 A Resistor v R emf E m sin d t vR Em iR sin d t R R Resistors behave in AC very much as in DC current and voltage are proportional as functions of time in the case of AC that is they are in phase For time dependent periodic situations it is useful to represent magnitudes using phasors These are vectors that rotate at a frequency d their magnitude is equal to the amplitude of the quantity in question and their projection on the vertical axis represents the instantaneous value of the quantity under study AC Driven Circuits 2 Capacitors vC emf E m sin d t qC C emf CE m sin d t dqC iC d CE m cos d t dt iC d CE m sin d t 90 0 Em iC sin d t 90 0 X 1 where X reactance d C Em im X V looks like i R Capacitors oppose a resistance to AC reactance of frequency dependent magnitude 1 d C this idea is true only for maximum amplitudes the instantaneous story is more complex AC Driven Circuits v L emf E m sin d t 3 Inductors d iL v L dt v L L iL dt L Em Em iL cos d t sin d t 90 0 L d L d Em iL sin d t 90 0 X Em im where X L d X Inductors oppose a resistance to AC reactance of frequency dependent magnitude d L this idea is true only for maximum amplitudes the instantaneous story is more complex All elements in parallel Emsin dt VR VC VL IC IR C IL Once again VR is always in phase with IR VL leads IL by 900 IC leads VC by 900 ELI the ICE man L R Impedance or reactance Resistors Inductors Capacitors X R X d L X 1 d C Example Circuit shown is driven by a low frequency 5 Hz voltage source with an rms amplitude of 7 07 V Compute the current amplitudes in the resistor capacitor and inductor IR IC and IL respectively Rms peak and peak to peak If E Emsin dt Then the peak to peak emf is 2Em The amplitude of the emf is Em peak The rms emf is Em 2 Erms 7 07 V f 5 Hz 10 F 10mH 100 Example 2 solution The three components are in parallel across the driving emf Amplitude of voltage across the three components is the same So current amplitude for any given component is inversely proportional to X Im Em X R 100 IR V 100 0 1 A XC 1 C 1 10 IC 10V 3 14 mA XL L 10 IL 10V 0 31 85 A Erms 7 07 V f 5 Hz 10 F 10mH 100 Em 10 V rad s Driven RLC Circuit Phase differences between voltage and current Resistors 0 Capacitors 2 I leadsV Inductors 2 V leads I Driven RLC Circuit Series circuit current is the same in all devices Taking a walk we see that the emfs in the various devices should add up to that of the AC generator But the emfs are out of phase with each other How to add them up Use phasors Current in circuit Emf in devices Resulting emf Driven RLC Circuit Applying Pythagoras theorem to the picture 2 E 2 VR VL VC 2 E 2 i X R 2 i X L i X C 2 E i X R X L X C 2 E i Z 2 2 2 i E X R 2 X L X C 2 Which resembles i E R 2 Z is called impedance Z X R X L X C 2 Also for the phase VL VC i X L i X C X L X C tan VR i XR XR Summary E Em sin dt I Imsin dt 1 2 Em Z R d L Im d C Z d L VL VC d C tan VR R 2 We have used XR R XC 1 C XL L Example In a given series RLC circuit Em 125 V Im 3 2 A Current LEADS source emf by 0 982 rad Determine Z and R Z Em Im 125 3 2 39 1 How to find R Look at phasors Emcos VR ImR R Emcos Im 125V 0 555 3 2A 21 7 V I R R m VL Em ImZ VC Example In a given series RLC circuit VL 2VR 2VC Determine VL VC tan VR 2VR VR 1 VR Hence 450 VL Em VR VR VC VL VC Em
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