Course Updates. Reminders: 1) Assignment #10 due Today. 2) Quiz # 5 Friday (Chap 29, 30) 3) Start AC Circuits

Transcription

1 ourse Updates eminders: 1) Assignment #10 due Today 2) Quiz # 5 Friday (hap 29, 30) 3) Start A ircuits

2 Alternating urrents (hap 31) In this chapter we study circuits where the battery is replaced by a sinusoidal voltage or current source. The circuit symbol is, cos( ωt) i t) I cos( ωt) v( t) 0 An example of an circuit connected to sinusoidal source is, or ( 0 Important: I(t) is same throughout just like D case.

3 Alternating urrents Since the currents & voltages are sinusoidal, their values change over time and the averages are zero. A more useful description of sinusoidal currents and voltages are given by considering the average of the square of this quantities. We define the MS (root mean square), which is the square root of the average of, i 2 ( I cos( ω )) 2 ( t) t 0 ( ( )) i ( t) I cos ( 1 cos( 2 ) ) 0 ωt I 0 + ωt 2 I I MS i ( t) 2 I 0 2 2

4 Alternating urrents ; Phasors A convenient method to describe currents and voltages in A circuits is Phasors. Since currents and voltages in circuits with capacitors & inductors have different phase relations, we introduce a phasor diagram. For a current, We can represented this by a vector rotating about the origin. The angle of the vector is given by ωt and the magnitude of the current is its projection on the X-axis. If we plot simultaneously currents & voltages of different components we can display different phases. i I cos ωt ( ) Note this method is equivalent to imaginary numbers approach where we take the real part (x-axis projection) for the magnitude

5 Beam me up Scotty It ate my phasor!

6 Alternating urrents: esistor in A circuit A resistor connected to an A source will have the voltage, v, and the current across the resistor has the same phase. We can draw the current phasor and the voltage phasor with the same angle. and v I ( ωt) i I cos( ωt) cos (just like D case) Phasors are rotating 2 dimensional vectors

7 esistor in A circuit; I & versus ω t I(t)Icos(ωt) (t)icos(ωt) ωt a b c d e f Note: oltage is in phase with current

8 Alternating urrents: apacitor in A circuit A capacitor connected to an A current source will have the voltage lagging behind the current by 90 degrees. We can draw the current phasor and the voltage phasor behind the current by 90 degrees. i dq dt I cos Find voltage: ( ωt) q 1 I v idt ω sin MAX I 1 ω ( ωt) q I ω sin ( ωt)

9 Alternating urrents ; apacitor in A circuit We define the capacitive reactance, X, as 1 X ω MAX I 1 cap I ω ω I X ike: I We stated that voltage lags by 90 deg., so equivalent solution is I v cos ω I sinωt ω I ω ( ωt 90) [ cosωt cos90 + sinωt sin90]

10 apacitor in A circuit; I & versus ω t i(t)icos(ωt) v(t)(i/ω)sin(ωt) a b c d e f ωt Note voltage lags 90 deg. Behind current (t)(i/ω)sin(ωt) (I/ω)cos(ωt-π/2)

11 Alternating urrents: Inductor in A circuit An inductor connected to an A current source will have the voltage leading (ahead of) the current by 90 degrees. We can draw the current phasor and the voltage phasor ahead the current by 90 degrees. i I cos MAX Iω di dt ( ωt) and Iω sin( ωt) Define inductive reactance, X, as MAX ind ( ) I X Iω I ω X ω ike: I

12 Inductor in A circuit; I & versus ω t i(t)icos(ωt) v(t) - Iωsin(ωt) a b c d e f ωt Draw phasor diagram for each point Note voltage leads 90 deg. ahead current v(t)iω sin(ωt) Iω cos(ωt + π/2)

13 What is reactance? You can think of it as a frequency-dependent resistance. X 1 ω For high ω, χ ~0 - apacitor looks like a wire ( short ) For low ω, χ - apacitor looks like a break X ω For low ω, χ ~0 - Inductor looks like a wire ( short ) For high ω, χ - Inductor looks like a break (inductors resist change in current) (" X " )

14 Frequency Filtering with inductor circuit The voltage across the inductor is, ab di dt Iω sin ωt ( ) and the magnitude depends on ω. So OW frequencies are reduced or FITEED out. Frequency Filtering with capacitor circuit The voltage across the capacitor is, I ab ω sin ωt ( ) and the magnitude depends on (1/ω). So HIGH frequencies are reduced or FITEED out.

15 Question 1 An circuit is driven by an A generator as shown in the figure. For what driving frequency ω of the generator, will the current through the resistor be largest a) ω large b) ω small c) ω doesn t matter

16 Question 1 An circuit is driven by an A generator as shown in the figure. For what driving frequency ω of the generator, will the current through the resistor be largest a) ω large b) ω small c) ω doesn t matter The current amplitude is inversely proportional to the frequency of the generator.

17 Alternating urrents: circuit Figure (b) has X >X and (c) has X <X. Using Phasors, we can construct the phasor diagram for an ircuit. This is similar to 2-D vector addition. We add the phasors of the resistor, the inductor, and the capacitor. The inductor phasor is +90 and the capacitor phasor is -90 relative to the resistor phasor. Adding the three phasors vectorially, yields the voltage sum of the resistor, inductor, and capacitor, which must be the same as the voltage of the A source. Kirchoff s voltage law holds for A circuits. Also and I are in phase.

18 Phasors Problem: Given drive ε m sin(ωt), find,,, I, I, I ε Strategy: We will use Kirchhoff s voltage law that the (phasor) sum of the voltages,, and must equal drive.

19 Phasors, cont. Problem: Given drive ε m sin(ωt), find,,, I, I, I ε 1. Draw phasor along x-axis (this direction is chosen for convenience). Note that since I, this is also the direction of the current phasor i. Because of Kirchhoff s current law, I I I I (i.e., the same current flows through each element)., I

20 Phasors, cont. Problem: Given drive ε m sin(ωt), find,,, I, I, I ε 2. Next draw the phasor for. Since the inductor current I always lags draw 90 further counterclockwise. The length of the phasor is I X I ω I X I

21 Phasors, cont. Problem: Given drive ε m sin(ωt), find,,, I, I, I ε 3. The capacitor voltage always lags I draw 90 further clockwise. The length of the phasor is I X I /ω IX I I X The lengths of the phasors depend on,,, and ω. The relative orientation of the,, and phasors is always the way we have drawn it. Memorize it!

22 Phasors, cont. Problem: Given drive ε m sin(ωt), find,,, I, I, I ε The phasors for,, and are added like vectors to give the drive voltage + + ε m : From this diagram we can now easily calculate quantities of interest, like the net current I, the maximum voltage across any of the elements, and the phase between the current the drive voltage (and thus the power). ε m

23 Question 2 A series circuit is driven by emf ε. Which of the following could be an appropriate phasor diagram? ε m ε m ~ (a) (b) (c) ε m

24 Question 2 A series circuit is driven by emf ε. Which of the following could be an appropriate phasor diagram? ~ ε m ε m (a) (b) (c) ε m The phasor diagram for the driven series circuit always has the voltage across the capacitor lagging the current by 90. The vector sum of the and phasors must equal the generator emf phasor ε m.

25 Question 3 A series circuit is driven by emf ε. ~ ε m For this circuit which of the following is true? (a) The drive voltage is in phase with the current. (b)the drive voltage lags the current. (c) The drive voltage leads the current.

26 Question 3 For this circuit which of the following is true? (a) The drive voltage is in phase with the current. (b)the drive voltage lags the current. (c) The drive voltage leads the current., I ε m φ First, remember that the current phasor I is always in the same orientation as the resistor voltage phasor (since the current and voltage are always in phase). From the diagram, we see that the drive phasor ε m is lagging (clockwise) I. Just as lags I by 90, in an A driven circuit, the drive voltage will also lag I by some angle less than 90. The precise phase lag φ depends on the values of, and ω. ~

27 () ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) cos cos - cos - cos X X Z t IZ t X X I t IX IX I t t v φ ω φ ω φ ω φ ω ω ω φ 1/ tan oltage (t) across A source Z is called impedance Also: ) cos( cos φ ω ω + t v t I i Z I Z I rms rms MAX ike: I IX IX I

28 Alternating urrents: circuit, Fig Y&F Example v ω10000rad/s 300ohm 60mH 0.5μ

29 series circuit; Summary of instantaneous urrent and voltages I IX IX () t I cos( ωt) () t I cos( ωt) i v v v v ad 1 ω () t IX cos( ωt 90) I cos( ωt 90) ( t) IX cos( ωt + 90) Iω cos( ωt + 90) 2 2 () t I ( X ) + ( X - X ) cos( ωt + φ) tanφ ω 1/ ω

30 For next time Homework #10 [due now] Homework #11 posted this afternoon, due next Wed. Quiz on Friday: Faraday s aw, Inductance and Inductors, circuits