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Download AC Circuits - Part II
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PHY-2054 J. B. Bindell CHAPTER 22 – ALTERNATING CURRENT PART 2 THE FUTURE Monday – Continue with Chapter 22 Wednesday – Hope to return the exams. more of the same Friday – Quiz No 10:30 AM office hours (sorry) Next Monday – Probably a problem solving session at 7:00AM. Confirmation later in the week. RESULT - INDUCTOR vL LI sin(t ) sin( t ) cos(t ) 2 I is the MAXIMUM current in the circuit. sin(t ) cos(t vL LI cos(t 2 2 ) ) RESISTOR v iR IR cos(t ) vRmax I R COMPARING INDUCTOR vL LI cos(t 2 vLMax I L (L) looks like a resistance XL=L Reactance - OHMS ) FOR THE INDUCTOR vLMax I L IX L VL FOR THE RESISTOR vRMax IR VR SLIGHTLY CONFUSING POINT We will always use the CURRENT as the basis for calculations and express voltages with respect to the current. What that means? We describe thecurrent as varyingas : i I cos(t) and the voltageas v Vcos(t ) where is thephaseshift between the currentand the voltage. THE PHASOR vL LI cos(t ) 2 vL LI cos(t 2 ) cos(t 2 ) sin(t ) direction t t 2 t REMEMBER FOR AC SERIES CIRCUITS Compute the reactance of a 0.450 H inductor at frequencies of 60.0 Hz In the circuit below, R=30 W and L= 30 mH. If the angular frequency of the 60 volt AC source is is 3 K-Hz WHAT WE WANT TO DO: (a) calculate the maximum current in the circuit (b) calculate the voltage across the inductor (c) Does Kirchoff’s Law Work? R=30 W E=60V L= 30 mH =3 KHZ R=30 W E=60V R=30W XL=L=90W =3 KHZ L= 30 mH The instantaneous voltage across each element is the PROJECTION of the MAXIIMUM voltage onto the horizontal axis! This is the SAME as the sum of the maximum vectors projected onto the horizontal axis. I VL IX L VR IR t Source voltage leads the current by the angle . I V Vmax VL IX L Let V IZ VR IR t Z Impedance I Z I R I X L or 2 2 2 2 Z R 2 X L2 VL tan ( ) VR 2 2 R=30 W E=60V 1 L= 30 mH =3 KHZ The drawing is obviously NOT to scale. I R=30 W VL IX L Let V IZ VR IR t L= 30 mH Z Impedance I Z I R I X L or R 30W 2 2 2 2 Z R X 2 2 2 L =3 KHZ E=60V 2 V 60 I 0.632A Z 94.9 X L L 90W Z (30) 2 (90) 2 94.9W VL tan ( ) VR 1 90 0 tan 71 30 1 R 30W X L L 90W Z (30) 2 (90) 2 94.9W 90 0 tan 71 1.25 rad 30 1 V 60 I 0.632A Z 94.9 WHAT ABOUT THE CAPACITOR?? C: q vc c vc 1 q 1 1 i I cos(t ) t c t c c Without repeating what we did, the question is what function will have a f/t = cosine? Obviously, the sine! So, using the same process that we used for the inductor, 1 vc I sin(t ) C 1 Xc (ohms) C CAPACITOR PHASOR DIAGRAM 1 vc I sin(t ) C 1 Xc (ohms) C I vC cos(t ) C 2 NOTICE THAT The voltage lags the current by 90 deg I and V are represented on the same graph but are different quantities. SUMMARY An AC source with ΔVmax = 125 V and f = 25.0 Hz is connected between points a and d in the figure. Calculate the maximum voltages between the following points: (a) a and b 62.8 V (b) b and c 45.6 V (c) c and d 154 V (d) b and d 108 V AC CIRCUITS LOOK COMPLICATED http://www.ngsir.netfirms.com/englishhtm/RLC.htm