Chapter 28: Alternating Current

Size: px

Start display at page:

Download "Chapter 28: Alternating Current"

Oswald Parrish
6 years ago
Views:

1 hapter 8: Alternating urrent Phasors and Alternating urrents Alternating current (A current) urrent which varies sinusoidally in tie is called alternating current (A) as opposed to direct current (D). One exaple of A current source is a coil of wire rotating with constant angular velocity in a agnetic field. The sybol ~ is used to denote an A source. n general a source eans either a source of alternating current or voltage. υ sinωt for alternating voltage, voltage aplitude i sinωt for alternating current, current aplitude n the U.S. and anada, coercial electric-power distribution syste uses a frequency of f 6 Hz, corresponding to ω 377 rad/s. n uch of the rest of the world uses f 5 Hz. n Japan, however, the country is divided in two regions with f 5 Hz and 6 Hz. Note : υ sinωt, i sinωt sinωt, P P sinωt in textbook

2 Phasors and Alternating urrents Phasors A convenient way to express a quantity varying sinusoidally with tie is by a phasor in phasor diagra as shown. ectifier and rectified current P sin ωt O ω phasor ωt P

3 Phasors and Alternating urrents ectifier and rectified current (cont d)

4 Phasors and Alternating urrents oot-ean-square current and voltage oot-ean-square current of a sinusoidal current tie averaged P P sinωt P sin ωt P ( cos ωt) rs P oot-ean-square voltage of a sinusoidal voltage P rs For -volt A, 7.

5 eluctance esistance, inductance, capacitance and reactance esistor in an A circuit Given: sinωt sin ωt oltage across in phase with current through sinωt ωt t t υ At tie t /

6 eluctance esistance, inductance, capacitance and reactance nductor in an A circuit Given: sinωt d sinωt d sin ωtdt dt d cosωt ω ω sin ( ωt π / ) oltage across leads current through by one-quarter cycle (9 ). ω ωt t ω t /(ω) At tie t

7 eluctance esistance, inductance, capacitance and reactance apacitor in an A circuit Given: sinωt Q sin ωt Q sin ωt dq π ω cosωt ω sin( ωt + ) dt oltage across lags current through by one-quarter cycle (9 ). t ω ω t ωt ωt ω At tie t

8 eluctance series circuit and reluctance circuit suary sinωt Given: Assue the solution for current: ( t) sin( ωt φ) sin( ωt φ) cos( ωt φ) ω ω cos( ωt φ) (See derivation later) aplitude ω ω X X reactance

9 eluctance series circuit and reluctance (cont d) What is reactance? fω/π You can think of it as a frequency-dependent resistance. X ω For high ω, χ ~ - apacitor looks like a wire ( short ) For low ω, χ - apacitor looks like a break X ω For low ω, χ ~ - nductor looks like a wire ( short ) For high ω, χ - nductor looks like a break (inductors resist change in current) (" X " )

10 ircuits series circuit (cont d) Given: sin t Assue: sin( ωt φ ) ω aplitude Q d dt cos( ωt φ ) ω ω cos( ωt φ ) sin( ω t φ ) Q cos( ωt φ ) ω d ω cos( ωt φ ) dt This picture corresponds to a snapshot at t. Note :sin( ωt φ + π / ) cos( ωt φ) The projections of these phasors along the vertical axis are the actual values of the voltages at the given tie. ω ω φ φ φ ω

11 ircuits series circuit (cont d) Proble: Given drive sin ωt, find,,,,, Strategy:. Draw drive phasor at t. Guess i phasor i i i sin( ωt φ) sin( φ ) at t -φ φ 3. Since i, this is also the direction for the phasor. φ (No or f ) 4. ealize that due to Kirchhoff s current law, i i i (i.e., the sae current flows through each).

12 ircuits series circuit (cont d) 5. The inductor current always lags draw 9 further counterclockwise. 6. The capacitor voltage always lags draw 9 further clockwise. X X -φ φ The lengths of the phasors depend on,,, and ω. The relative orientation of the,, and phasors is always the way we have drawn it. φ is deterined such that + + (Kirchhoff s voltage rule) These are added like vectors.

13 ircuits Phasor diagras for circuits: Exaple y ~ out x y x ( ) + ( X ) ( + X ) + X aplitude of current

14 Filters : Exaple ircuits ~ Ex.: µf, Ω out out + X out ( ) ( ω ) ω ω + + "transission" High-pass filter e+.e+6.e+6 3.E+6 4.E+6 5.E+6 6.E+6 (Angular) frequency, oega ω High-pass filter Note: this is ω, f ω π

15 ircuits ~ Filters out ω No current out ω apacitor ~ wire out out Highpass filter ω ω ~ out ω No current out ω nductor ~ wire out out owpass filter ω ω ~ ω No current because of capacitor ω No current because of inductor out (onceptual sketch only) ω Band-pass filter

16 ircuits Phasor diagras for circuits: Exaple X φ φ φ (X -X ) φ X eluctance for inductor X ω eluctance for capacitor X ω pedance Z Z + ( ) X X tanφ aplitude X X ( ( ) ) X X + + ( X X ) Z

17 ircuits Phasor diagras for circuits: Tips This phasor diagra was drawn as a snapshot of tie t with the voltages being given as the projections along the y-axis. Soeties, in working probles, it is easier to draw the diagra at a tie when the current is along the x-axis (when ). X X f φ X y f φ f φ x f φ X Full Phasor Diagra Fro this diagra, we can also create a triangle which allows us to calculate the ipedance Z: Z fφ X pedance Triangle X

18 esonance in Alternating urrent ircuits esonance For fixed,, the current will be a axiu at the resonant frequency w which akes the ipedance Z purely resistive. i.e.: Z + ( X X ) reaches a axiu when: X X This condition is obtained when: ω o ω o ω o ; f resonance frequency Note that this resonant frequency is identical to the natural frequency of the circuit by itself! At this frequency, the current and the driving voltage are in phase! X X tan φ π ~

19 esonance in Alternating urrent ircuits esonance (cont d) X Z tan φ X Z cos φ X cosφ X φ Z X - X o Plot the current versus ω, the frequency of the voltage source: ω o ω o

20 esonance in Alternating urrent ircuits esonance (cont d) On esonance: φ and Z ~ X X X Q X Q On resonance, the voltage across the reactive eleents is aplified by Q! Necessary to pick up weak radio signals, cell phone transissions, etc.

21 Power Power in Alternating urrent ircuits The instantaneous power (for soe frequency, w) delivered at tie t is given by: ( sinωt)( sin( ω )) P( t) ( t) ( t) t φ The ost useful quantity to consider here is not the instantaneous power but rather the average power delivered in a cycle. P( t) sinωt sin( ωt φ) To evaluate the average on the right, we first expand the sin(ωt-φ) ter.

22 Power Power in Alternating urrent ircuits Expanding, sinωt sin( ωt φ) sinωt(sinωt cosφ cosωt sinφ) Taking the averages, sin ωt cos ωt (Product of even and odd function ) + sinωtcosωt Generally: / P( t) t sin x π sin xdx π Putting it all back together again, { cos φ sin ωt sin φ sin ωt cos ω } P ( t ) cos φ ω t π sin ωt ω t π

23 Power Power in Alternating urrent ircuits This result is often rewritten in ters of rs values: rs rs P ( t ) rs rscosφ Power delivered depends on the phase, f, the power factor Phase depends on the values of,,, and ω Therefore... P t ) cosφ ( rs rs

24 Power Power in Alternating urrent ircuits Power, as well as current, peaks at ω ω. The sharpness of the resonance depends on the values of the coponents. ecall: cos φ rs P() t cos φ rs We can write this in the following anner (which we won t try to prove): P( t) rs x x + Q ( x ) introducing the curious factors Q and x...

25 esonance in Alternating urrent ircuits Power and resonance A paraeter Q is often defined to describe the sharpness of resonance peaks in both echanical and electrical oscillating systes. Q is defined as Uax Q π U where U ax is ax energy stored in the syste and U is the energy dissipated in one cycle For circuit, U ax is osses only coe fro : This gives ω Q res U ax U ax ax ax T period π ωres And for copleteness, note x ω ω res

26 esonance in Alternating urrent ircuits Power and resonance FWHM For Q > few, Q ωres FWHM Full Width at Half Maxiu rs <P> FWHM Q3 o o Q Quality of the peak Higher Q sharper peak better quality ω ω o

27 Transforers A voltages can be stepped up or stepped down by the use of transforers. Transforers The A current in the priary circuit creates a tie-varying agnetic field in the iron ~ iron This induces an ef on the secondary windings due to the utual inductance of the two sets of coils. N (priary) N (secondary) The iron is used to axiize the utual inductance. We assue that the entire flux produced by each turn of the priary is trapped in the iron.

28 Transforers deal transforer without a load No resistance losses All flux contained in iron The priary circuit is just an A voltage source in series with an inductor. The change in flux produced in each turn is given by: dφ turn dt N Nothing connected on secondary iron N N (priary) (secondary) The change in flux per turn in the secondary coil is the sae as the change in flux per turn in the priary coil (ideal case). The induced voltage appearing across the secondary coil is given by: Therefore, dφ turn N N dt N N > N secondary is larger than priary (step-up) N > N secondary is saller than priary (step-down) Note: no load eans no current in secondary. The priary current, tered the agnetizing current is sall!

29 Transforers deal transforer with a load What happens when we connect a resistive load to the secondary coil? hanging flux produced by priary coil induces an ef in secondary which produces current This current produces a flux in the secondary coil µ N, which opposes the change in the original flux -- enz s law N (priary) iron N (secondary) This induced changing flux appears in the priary circuit as well; the sense of it is to reduce the ef in the priary, to fight the voltage source. However, is assued to be a voltage source. Therefore, there ust be an increased current (supplied by the voltage source) in the priary which produces a flux µ N which exactly cancels the flux produced by. N N

30 Transforers deal transforer with a load (cont d) Power is dissipated only in the load resistor. P dissipated Where did this power coe fro? t could coe only fro the voltage source in the priary: N (priary) iron N (secondary) P generated N N N N N N N N N N The priary circuit has to drive the resistance of the secondary.

31 Exercise Exercises Suppose volts, f Hz, Ohs, 4. H, Find X, Z,,, and l. X ω 6.8.4H 6.5Ω Z + ( ω) Z + (6.5) 8.3Ω Z X A

32 Exercises Exercise : alculate power lost in in Exercise P avg rs rs 3.53A.44.5A P avg (.5A) 6.5Watts To calculate power produced by the generator you need to take account of the phase difference between the voltage and the current. n general you can write: P avg rs rs cosφ For an inductor P because the phase difference between current through the inductor and voltage across the inductor is 9 degrees

PH 222-2C Fall Electromagnetic Oscillations and Alternating Current. Lectures 18-19

PH 222-2C Fall Electromagnetic Oscillations and Alternating Current. Lectures 18-19 H - Fall 0 Electroagnetic Oscillations and Alternating urrent ectures 8-9 hapter 3 (Halliday/esnick/Walker, Fundaentals of hysics 8 th edition) hapter 3 Electroagnetic Oscillations and Alternating urrent