PHYS 2421 - Fields and Waves
Idea: how to deal with alternating currents Edison wanted direct current, Tesla alternating current, who won? Mathematically: Instantaneous current: Instantaneous voltage: w is the angular frequency i I cost v V cost w=2pf, f = 60 Hertz (50 Hertz in Europe) V=120 volts (220 in Europe, South America) Symbol in circuits: Average I?: i 0 Ave How to measure a changing current? There are several ways.
How to characterize alternating currents? Method 1: Rectified average i I cost Use average of i Average of rectified I?: I rav t2 / t / 2 I cos dt 2I d t0 t / 2 t2 dt t cos 2 2I 0.637I t0
How to characterize alternating currents? Method 2: Root mean square average i I cost Use average of i I 2 Average of i?: 2 Ave Irms rms 1 2 i 2 Ave t2 t2 / 2 I dt dt 2 i t0 t0 t2 t0 i I cos t I 1cos2t 2 2 2 2 1 dt cos2t 2 1 2 I I Irms 2 2 0.7071 I ompare to I rav 0.637 I
How to characterize alternating currents? Method 3: Phasor i I cost Take instantaneous current as component of a vector Not a Mexican Wrestler! et vector rotate with angular speed. Useful for comparing times et us see some examples
Given: I rms 2.7 A a) I Ave 0 b) c) I i 2 i I 2 2.7 A 7.3 A 2 2 2 rms Ave Ave rms I I I I rms 2.7 A 2 2 2.7 rms A 3.8 A 2 Hmwk: Probls 11 th ed.: 31.1 and 2 (soln.: 2.97 A and 1.89 A) or 12 th Ed.: 31.3 and 2
Summary of Section 31.1 i I cost Symbol in circuits: v V cost Average I?: i 0 Ave 2 I rav I I rms I 2 Phasor Hmwk Sect. 31.1: Probls 11 th ed.: 31.1 and 2 or 12 th Ed.: 31.3 and 2
Resistor in an ac circuit i I cost Voltage drop across R? v R ir IRcost V R cost Phasor diagram
di dt v d I Isint Inductor in an ac circuit cos dt Icos t 90 t o Voltage drop across? i I cost V cos t90 Phasor diagram o V I is like a resistance, call it "Reactance" X Inductive reactance Reactance
a) Want X to have i=250 ma when V=3.6 V 6 V IX X V / I 3.6 V / 25010 A 14.4 k X 2 f 3 6 X / 2 f 14.4 / 2 Now want 10 1.6 10 1.43mH b) Want I for 16 MHz and 160 khz I V / X V / 2 f As f increases by 10, I decreases by 10 I=25mA As f decreases by 10, I increases by 10 I=2500mA Hmwk: Probls 31.3 and 5 (11 th Ed) or 5 and 9 (12 th Ed.)
i dq dt q apacitor in an ac circuit I I cost sint I q v v sin t cos t 90 o I i I cost Voltage across? Phasor diagram X 1 Reactance Take V I IX is capacitive reactance
v 1.2 Acos 2500t a) i 0.006cos 2500t A R 200 b) X 1 1 6 2500 rad/s 510 F 80 v V cos cos c) t / 2 0.48 V 2500 rad/s t / 2 Hmwk: Probls 11th Ed.: 31.7 & 10 Soln: a) 1736 W, vr=1.10cos [120t] 12 th Ed.: 31.11 & 12 Soln: a) 1736, v R =1.10cos [120t]
Summary of Section 31.2 R in an ac circuit i I cost v V cost R R in an ac circuit v V cos t 90 o Reactance X in an ac circuit v I cos Reactance t90 o X 1
Summary of Section 31.2 Homework for section 31.2: 11th Ed.: Probls 3, 5, 7 and 10 12 th Ed.: Probls 5, 9, 11 and 12
i and v R are in phase i I cost v V cost R R v I cos t90 o v is behind 90 o v V cos t 90 o v is ahead 90 o At this point phasors become useful
v R is in phase with i v is behind 90 o v across source? v v v v v is ahead 90 o R cos cos 90 I o cos o 90 R v V t V t t Horizontal projection Horizontal projection Horizontal projection
Sum of horizontal projections of vectors = horizontal projection of sum of vectors Magnitude of sum : Sum voltage vectors: V 2 2 V V V IR IX IX I R X X 2 2 2 2 R 2 2 2 2 IZ; " impedance" is Z R X X R 1/ V V IX IX X X Phase angle : tan In conclusion, if the current is : The source voltage is : V IR R R i I cost v IZ cost
Formalism is still valid if an element is missing If R is missing, set R=0 If is missing, set =0 If is missing, set 1/ 2 2 2 2 Z R X X R et us see an example!
X 10, 000 rad/s 60 mh 600 6 X 1/ 1/ 10, 000 rad/s 0.510 F 200 2 2 2 2 Z R X X 300 600 200 500 V 50 V I 0.1 A Z 500 V X X 600 200 tan tan 53 R 300 1 1 0 V IX V IX 0.1 A 200 20 V IR 0.1 A 300 30 V; 0.1 A 600 60 V R Hmwk: Problem 31.12 (11 th Ed.) or 14 (12 th Ed.)
A rads/s i I cos t 0.1 cos 10, 000 t vr ir IR cos t 30 V cos 10, 000 rads/s t v ix V cos t 90 V sin t 60 V sin 10, 000 rads/s t 0 v ix V cos t 90 V sin t 20 V sin 10, 000 rads/s t 0 Instantaneous source voltage v V cos t 50 cos 10, 000 t 53 0 180 0 V rads/s Hmwk: Problem 31.13 (11 th Ed.) or 16 (12 th Ed.)
Summary of Section 31.3 If the current is : The source voltage is : I mpedance is i I cost v IZ cost 2 2 2 2 1 1/, Z R X X R X X V V IX IX X X Phase angle : tan VR IR R Voltages across elements: vr IR cost v V cos t 90 IX cos t 90 0 0 0 0 v cos 90 cos 90 V t IX t Instantaneous source voltage v V cost Hmwk Sect. 31.3: Probls 31.12 & 31.13 (11 th Ed.) or 14 & 16 (12 th Ed.)
Idea: vary to increase current I increases as Z decreases I V V Z 2 R X X 2 Z varies with 1/ 2 2 Z R X & X vary with X X 1/ Minimum of Z when X =X 0 1 Max. I at 0 Resonanting frequency
1 1 a) 0 5,000,000 rad/s 6 3 b) X 510 rad/s 0.410 H 2000 X 3 12 0.410 H 10010 F 6 rad/s 12 F 1/ 1/ 510 10010 2000 2 2 2 2 Z R X X R 2000 2000 R 500 Vrms 1 V c) irms 2 ma Z 500 d) Vrms R IrmsR 0.002 A 500 1 V Vrms Irms X 4 V Vrms Irms X 4 V
Summary of Section 31.5 I output is max at resonance X X Z R 0 1 Max. I at 0 Vrms irms Z V I R rmsr rms V I X rms rms V I X rms rms Hwk for sect. 31.5: Probs 27 a)& b) and 32 (11 th Ed.) or 31a)& b) and 36 (12 th Ed.) Solution of 32/36: 945 rad/s, 70.6, 450 V
Idea: Use an ac circuit at a given v and i to induce a different i and v in another circuit Start with a circuit with i and v Add a magnetic core Add a secondary loop A changing current I 1 induces a changing current I 2 I B B I 1 1 1 2 2 2
N Relations between currents and voltages 1 1 d dt B B N 2 2 d dt where is the flux per turn (equal on both sides of core) d B 1 2 2 V2 V1 dt N1 N2 N1 Notice that the frequency is the same on both circuits Assuming no resistance in wires Power-in in circuit 1 = power-out in circuit 2 V N P I V P I V I I or I I 1 1 1 1 1 2 2 2 2 1 2 1 V2 N2 B N
N2 V2 240 V a) Use a tranformer with 2 N1 V1 120 V V1 b) urrent: cannot use I2 I1 as I1 is not known V2 Must use P I V I V 1 1 2 2 P 960 W P 960 W I2 4 A or I1 8 V2 240 V V1 120 V c) Resistance: take V RI 2 2 2 2 R V / I 240 V / 4 A 60 Hmwk: Probl 33 (11 th Ed.) or 37 (12 th Ed.) A
Summary of Section 31.6 V N V 2 2 1 N1 V N I I or I I 1 1 2 1 2 1 V2 N2 P1 I1V 1 P2 I2V2 Frequency is the same on both circuits Power-in in circuit 1 = power-out in circuit 2 Hmwk Section 31.6: Probls 32 and 33 (11 th Ed.) or 36 and 37 (12 th Ed.)
PHYS 2421 - Fields and Waves
You got what grade?
You got what grade?