Alternating Current Circuits

1. Alternating Current Circuits Chapter 33 (continued)

2. Phasor Diagrams • A phasor is an arrow whose length represents the amplitude of an AC voltage or current. • The phasor rotates counterclockwise about the origin with the angular frequency of the AC quantity. • Phasor diagrams are useful in solving complex AC circuits. • The “y component” is the actual current or voltage. Resistor Capacitor Inductor VRp VLp Ip Ip Ip w t w t w t VCp

3. R ~ V C L Impedance in AC Circuits The impedance Z of a circuit or circuit element relates peak current to peak voltage: (Units: Ohms) (This is the AC equivalent of Ohm’s law.)

4. Phasor Diagrams Resistor Capacitor Inductor VRp VLp Ip Ip Ip w t w t w t VCp

5. R ~ V C L RLC Circuit Use the loop method: V - VR - VC - VL = 0 I is same through all components. BUT: Voltages have different PHASES  they add as PHASORS.

6. Ip RLC Circuit VRp VLp f VP (VCp- VLp) VCp

7. Ip RLC Circuit VRp VLp f VP (VCp- VLp) VCp By Pythagoras’s theorem: (VP )2 = [ (VRp )2 + (VCp - VLp)2 ] = Ip2 R2 + (Ip XC - Ip XL)2

8. R ~ V C L RLC Circuit Solve for the current:

9. R ~ V C L RLC Circuit Solve for the current: Impedance:

10. IP W R = 1 0 R = 1 0 0 W 0 2 3 4 5 1 0 1 0 1 0 1 0 RLC Circuit The current’s magnitude depends on the driving frequency. When Z is a minimum, the current is a maximum. This happens at a resonance frequency: The circuit hits resonance when 1/wC-wL=0: w r=1/ When this happens the capacitor and inductor cancel each other and the circuit behaves purely resistively: IP=VP/R. L=1mH C=10mF The current dies away at both low and high frequencies. wr w

11. Ip VRp VLp f VP (VCp- VLp) VCp We can also find the phase: tan f = (VCp - VLp)/ VRp = (XC-XL)/R = (1/wC - wL) / R Phase in an RLC Circuit

12. Ip VRp VLp f VP (VCp- VLp) VCp We can also find the phase: tan f = (VCp - VLp)/ VRp = (XC-XL)/R = (1/wC - wL) / R Phase in an RLC Circuit More generally, in terms of impedance: cos f = R/Z At resonance the phase goes to zero (when the circuit becomes purely resistive, the current and voltage are in phase).

13. Power in an AC Circuit The power dissipated in an AC circuit is P=IV. Since both I and V vary in time, so does the power: P is a function of time. Use V = VP sin (wt) and I = IP sin (w t+f ) : P(t) = IpVpsin(wt) sin (w t+f ) This wiggles in time, usually very fast. What we usually care about is the time average of this: (T=1/f )

14. Power in an AC Circuit Now:

15. Power in an AC Circuit Now:

16. Power in an AC Circuit Now: Use: and: So

17. Power in an AC Circuit Now: Use: and: So which we usually write as

18. Power in an AC Circuit (f goes from -900 to 900, so the average power is positive) cos(f) is called the power factor. For a purely resistive circuit the power factor is 1. When R=0, cos(f)=0 (energy is traded but not dissipated). Usually the power factor depends on frequency, and usually 0<cos(f)<1.

19. Power in a purely resistive circuit V f = 0 V(t) = VP sin (wt) I I(t) = IP sin (wt) p wt 2p (This is for a purely resistivecircuit.) P P(t) = IV = IP VP sin 2(wt) Note this oscillates twice as fast. p wt 2p

20. Power in a purely reactive circuit The opposite limit is a purely reactive circuit, with R=0. I P This happens with an LC circuit. Then f= 900 The time average of P is zero. V wt

21. Iron Core Np turns Ns turns Vs Vp Transformers Transformers use mutual inductance to change voltages: Primary (applied voltage) Secondary (produced voltage) Faraday’s law on the left: If the flux per turn is f then Vp=Np(df/dt). Faraday’s law on the right: The flux per turn is also f, so Vs=Ns(df/dt). 

22. Iron Core Np turns Ns turns Vs Vp Transformers Transformers use mutual inductance to change voltages: Primary (applied voltage) Secondary (produced voltage) In the ideal case, no power is dissipated in the transformer itself. Then IpVp=IsVs

23. Transformers & Power Transmission Transformers can be used to “step up” and “step down” voltages for power transmission. 110 turns 20,000 turns Power =I2 V2 V2=20kV V1=110V Power =I1 V1 We use high voltage (e.g. 365 kV) to transmit electrical power over long distances. Why do we want to do this?

24. Transformers & Power Transmission Transformers can be used to “step up” and “step down” voltages, for power transmission and other applications. 110 turns 20,000 turns Power =I2 V2 V2=20kV V1=110V Power =I1 V1 We use high voltage (e.g. 365 kV) to transmit electrical power over long distances. Why do we want to do this? P = I2R (P = power dissipation in the line - I is smaller at high voltages)