1. ELE130 Electrical Engineering 1 Week 5 Module 3 AC (Alternating Current) Circuits

2. Administration u Quiz - next week in lecture l material from weeks 4, 5 & 6 u Laboratory Test l postponed from next week to the week following Easter break (from week 6 to week 7) l will include work from laboratory 1, 2 & 3 l laboratory no. 4 will be done in week 6 instead of week 7 u Help Desk

3. Why use ac? u In power applications l easier to generate (no brushes or rectifiers) l easier to transform voltages (reduced losses) l easier to distribute (easier switching) l easy utilization (brushless motors, switching) u In communications l ac required to reproduce audio signals, to transmit over distance (radio, microwave, etc.) u DC circuits are special case of ac circuits (when frequency is zero)

4. Parameters of an ac wave u v s = V m sin(  t +  ) l V m : amplitude of the wave l  : radian frequency  = 2  f, f in Hz,  in rad/sec l t : time in seconds u  t :  (in radians) l  : relative angle (in degrees or radians, but don’t MIX them up!)

5. Generation of an ac wave

6. Note u f (cycles / sec) and  (radians/sec) l 1 cycle = 2  radians = 360 o u angle can be specified in degrees or radians l 90° =  /2 radians u generally, frequencies will be the same (if not big hassles!) u generally waves will have different amplitudes u  is relative to reference. Of more importance however is relative phase, or phase difference

7. Consider two waves u Phase difference (  1 -  2 ) u (if frequency different, then relative phase changes with time) u Leading and lagging waves l using time line, the wave whose peak is occurs first is leading u Phase ‘wraps’ around i.e. 20° = 380 ° l usually work in range -180 ° to +180 °

8. Lagging and Leading Waveforms

9. Example - what is the phase difference? u x 1 (t) = X 1 sin(  t - 30°) & x 2 (t) = X 2 cos (  t - 40°) l convert sin to cos : sin(x) = cos (x - 90 o ) u x 1 (t) = X 1 cos (  t - 30 o - 90 o ) = X 1 cos (  t - 120°) u Phase difference is |-120° - (-40°)|= 80° u x 2 (t) leads x 1 (t)

10. Inductor response to an ac source u The current i L is: u as v L = v s : u by Ohm’s Law: |Z L |=  L u current lags voltage by 90 o u therefore angle of Z L = +90 o u  multiplying by j u  Z L = j  L

11. Capacitor response to an ac source u The current i C is: u by Ohm’s Law:  Z C  u  current leads voltage by 90 o u  angle of Z C = -90 o u  multiplying by -j

12. Voltage - Current Relationships

13. Example - (solving circuits with an ac source) u RL circuit l solve in dc. - steady state + transient response ( revision) l solve in time domain l solve in frequency domain (need complex numbers) u real (time domain ) circuits can be solved by adding an imaginary part, do the calculations in the complex domain, and then simply take the magnitude part of the complex answer as the solution to the real circuit!

14. Revision of DC Transients -RC u V s is given u Time constant: u i C may change instantaneously u v C can not change instantaneously u also: u Which implies that capacitor “appears” like an open circuit to DC

15. Revision of DC Transients-RL u V s is given u Time constant u i L can not change instantaneously u v L may change instantaneously u also: u Which implies that an inductor “appears” like a short circuit to DC

16. Time Domain Analysis Assume: i(t) = I m cos(  t -  ) i(t) = ? v s = V m cos(  t)

17. Solution u KVL : v s = v R + v = i(t) R + L di/dt V m cos(  t) = RI m cos(  t -  ) + L d(I m cos(  t -  ))/dt = RI m cos(  t -  ) - LI m  sin(  t -  ) = RI m [ cos(  t)cos  + sin(  t)sin  ] - LI m  [sin(  t) cos  - sin  cos(  t)] = cos(  t)[RI m cos  + LI m  sin  ] + sin(  t) [RI m sin  - LI m  cos  ] u cos (  t) & sin (  t) are othoganal functions ( ie no multiples of cos (  t) will ever equate to sin (  t) ) u Hence, the 2 terms may be considered separately l V m = RI m cos  + LI m  sin  ………………... (1) l 0 = RI m sin  - LI m  cos  …………………(2)

18. Solution (cont.) u From (2) l sin  / cos  = LI m  / RI m l tan  = L  / R l  = tan -1  L/R l Followes that sin  =  L /  R 2 + (  L) 2 & cos  = R /  R 2 + (  L) 2 u From (1) l V m = RI m cos  + LI m  sin  = RI m R /  R 2 + (  L) 2 + LI m   L /  R 2 + (  L) 2 = I m [R 2 + (  L) 2 /  R 2 + (  L) 2 ] = I m [  R 2 + (  L) 2 ] l I m = V m /  R 2 + (  L) 2 u i(t) = V m /  R 2 + (  L) 2 cos (  t - tan -1  L/R )

19. Complex number revision u Electrical Engineers already use i therefore use ‘j’ u What is j 2 ? l if multiplication by j is a rotation by 90°, then j 2 must be a rotation by 180° l j 2 = -1 j =  -1 (which cannot really exist so must be imaginary) u Complex numbers can be expressed in l cartesian co-ordinates (addition & subtraction) or l polar co-ordinates (multiplication & division) u Euler’s Identity : z=a+jb = M[cos  t  jsin  t] = Me  j  t u where M = magnitude

20. Phasor notation u X = X m  l where  is assumed or stated as a side note u Phasor notation - omits the time depend part and allows us to write the magnitude and phase only l Polar: X = X m  Cartesian: X = X R + jX I l X R = X m cos  X I = X m sin  l X m =  (X 2 R + X 2 I )  = tan -1 (X I / X R )  XRXR jX I XmXm

21. Complex Impedance's u the ratio of voltage to current across and element is known as IMPEDANCE u Impedance is frequency complex, containing both real and imaginary components u Symbol is Z (which is complex) : Z = R + jX u R & X are both real u R is always 0 or positive u X is reactive component, can be positive or negative

22. Impedance u Z R = R u Inductance gives a positive reactance  Z L = j  L u Capacitance gives negative reactance

23. Steps in solve AC Problems 1. Confirm sources are of the same frequency 2. Convert to Phasor notation 3. Treat as a DC problem (with complex numbers) 4. Convert Back to Time domain

24. Example (cont. in phasor notation) u Total impedance : Z T = R + j  L =  R 2 + (  L) 2 tan -1 (  L/R) u Ohms Law : V s = I T. Z T u  I T = V m  0° /  R 2 + (  L) 2 tan -1 (  L/R) = [V m /  R 2 + (  L) 2 ]  0° - tan -1 (  L/R) = V m /  R 2 + (  L) 2 cos (  t - tan -1 (  L/R)) i(t) = ? v s = V m cos(  t) V m  0° jLjL