1. AC Circuits and Resonance Conclusion
April 20, 2005

2. Alternating Current Circuits
An “AC” circuit is one in which the driving voltage and
hence the current are sinusoidal in time. p fv 2p
V(t) wt Vp
-Vp
V = VP sin (wt - fv )
I = IP sin (wt - fI )
w is the angular frequency (angular speed) [radians per second].
Sometimes instead of w we use the frequency f [cycles per second]
Frequency  f [cycles per second, or Hertz (Hz)] w = 2p f

3. Phase Term
V = VP sin (wt - fv ) p fv 2p
V(t) wt Vp
-Vp

4. Resistors in AC Circuits~
EMF (and also voltage across resistor):
V = VP sin (wt)
Hence by Ohm’s law, I=V/R:
I = (VP /R) sin(wt) = IP sin(wt)
(with IP=VP/R) V I
V and I
“In-phase” p 2p wt

5. Capacitors in AC CircuitsE ~ C
Start from: q = C V [V=Vpsin(wt)]
Take derivative: dq/dt = C dV/dt
So I = C dV/dt = C VP w cos (wt)
I = C w VP sin (wt + p/2) V
This looks like IP=VP/R for a resistor
(except for the phase change).
So we call Xc = 1/(wC)
the Capacitive Reactance I p wt 2p
The reactance is sort of like resistance in that IP=VP/Xc. Also, the current leads the voltage by 90o (phase difference).
V and I “out of phase” by 90º. I leads V by 90º.

6. I Leads V??? What the **(&@ does that mean??2 V f 1 I
I = C w VP sin (wt + p/2)
Current reaches it’s maximum at
an earlier time than the voltage!

7. Inductors in AC Circuits
V = VP sin (wt)
Loop law: V +VL= 0 where VL = -L dI/dt
Hence: dI/dt = (VP/L) sin(wt).
Integrate: I = - (VP / Lw) cos (wt)
or I = [VP /(wL)] sin (wt - p/2) ~ L V
Again this looks like IP=VP/R for a
resistor (except for the phase change).
So we call XL = w L
the Inductive Reactance I p 2p wt
Here the current lags the voltage by 90o.
V and I “out of phase” by 90º. I lags V by 90º.

8. SUMMARY

9. 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 voltage or current.
Resistor Vp Ip
w t

10. 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 voltage or current.
Resistor Capacitor Vp Ip Ip
w t
w t Vp

11. 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 voltage or current.
Resistor Capacitor Inductor Vp Vp Ip Ip Ip
w t
w t
w t Vp

12. i i + + +
time i i
LC Circuit i i + + +

13. Analyzing the L-C Circuit
Total energy in the circuit:
Differentiate :
N o change in energy

14. Analyzing the L-C Circuit
Total energy in the circuit:
Differentiate :
N o change in energy

15. Analyzing the L-C Circuit
Total energy in the circuit:
Differentiate :
N o change in energy

16. Analyzing the L-C Circuit
Total energy in the circuit:
Differentiate :
N o change in energy
The charge sloshes back and
forth with frequency w = (LC)-1/2

17. LC Oscillations C L Work out equation for LC circuit (loop rule)
Rewrite using i = dq/dt
 (angular frequency) has dimensions of 1/t
Identical to equation of mass on spring C L

18. LC Oscillations (2) Solution is same as mass on spring  oscillations
qmax is the maximum charge on capacitor
 is an unknown phase (depends on initial conditions)
Calculate current: i = dq/dt
Thus both charge and current oscillate
Angular frequency , frequency f = /2
Period: T = 2/

19. Plot Charge and Current vs t

20. Energy Oscillations
Equation of LC circuit
Total energy in circuit is conserved. Let’s see why
(Multiply by i = dq/dt)
Use
UL + UC = const

21. Oscillation of Energies
Energies can be written as (using 2 = 1/LC)
Conservation of energy:
Energy oscillates between capacitor and inductor
Endless oscillation between electrical and magnetic energy
Just like oscillation between potential energy and kinetic energy for mass on spring

22. Plot Energies vs t
Sum

23. UNDRIVEN RLC Circuit Work out equation using loop rule
Rewrite using i = dq/dt
Solution slightly more complicated than LC case
This is a damped oscillator (similar to mechanical case)
Amplitude of oscillations falls exponentially

24. Charge and Current vs t in RLC Circuit

25. RLC Circuit (Energy) Basic RLC equation Multiply by i = dq/dt
Collect terms (similar to LC circuit)
Total energy in circuit
decreases at rate of i2R (dissipation of energy)

26. Energy in RLC Circuit
Sum

27. AC Circuits Enormous impact of AC circuits Basic components
Power delivery
Radio transmitters and receivers
Tuners
Filters
Transformers
Basic components R L C
Driving emf

28. AC Circuits and Forced Oscillations
RLC + “driving” EMF with angular frequency d
General solution for current is sum of two terms
“Steady state”: Constant amplitude
“Transient”: Falls exponentially & disappears
Ignore

29. Steady State Solution Assume steady state solution of form
Im is current amplitude
 is phase by which current “lags” the driving EMF
Must determine Im and 
Plug in solution: differentiate & integrate sin(t-)
Substitute

30. Steady State Solution for AC Current (2)
Expand sin & cos expressions
Collect sindt & cosdt terms separately
These equations can be solved for Im and  (next slide)
High school trig!
cosdt terms
sindt terms

31. Steady State Solution for AC Current (3)
Solve for  and Im in terms of
R, XL, XC and Z have dimensions of resistance
Let’s try to understand this solution using “phasors”
Inductive “reactance”
Capacitive “reactance”
Total “impedance”

32. REMEMBER 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.
Resistor Capacitor Inductor Vp Vp Ip Ip Ip
w t
w t
w t Vp 8

33. Reactance - Phasor Diagrams
Resistor Capacitor Inductor Vp Vp Ip Ip Ip
w t
w t
w t Vp 8

34. “Impedance” of an AC CircuitL C ~
The impedance, Z, of a circuit relates peak
current to peak voltage:
(Units: OHMS) 14

35. “Impedance” of an AC CircuitL C ~
The impedance, Z, of a circuit relates peak
current to peak voltage:
(Units: OHMS)
(This is the AC equivalent of Ohm’s law.) 15

36. Impedance of an RLC Circuit~ E
As in DC circuits, we can use the loop method:
E - VR - VC - VL = 0
I is same through all components.

37. Impedance of an RLC Circuit~ E
As in DC circuits, we can use the loop method:
E - VR - VC - VL = 0
I is same through all components.
BUT: Voltages have different PHASES
 they add as PHASORS. 17

38. Phasors for a Series RLC CircuitIp
VRp
VLp f VP
(VCp- VLp)
VCp 18

39. Phasors for a Series RLC CircuitIp
VRp
VLp f VP
(VCp- VLp)
VCp
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ] 19

40. Phasors for a Series RLC CircuitIp
VRp
VLp f VP
(VCp- VLp)
VCp
By Pythagoras’ theorem:
(VP )2 = [ (VRp )2 + (VCp - VLp)2 ]
= Ip2 R2 + (Ip XC - Ip XL) 2 20

41. Impedance of an RLC Circuit~
Solve for the current: 21

42. Impedance of an RLC Circuit~
Solve for the current:
Impedance: 22

43. Impedance of an 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. IP 1 2 3 4 5 R = W
L=1mH
C=10mF
The current dies away
at both low and high
frequencies. wr w 23

44. Phase in an RLC Circuit Ip f VLp (VCp- VLp) VRp VP VCp
We can also find the phase:
tan f = (VCp - VLp)/ VRp
or;
tan f = (XC-XL)/R. or
tan f = (1/wC - wL) / R 24

45. cos f = R/Z Phase in an RLC Circuit Ip f VLp (VCp- VLp) VRp VP VCp
We can also find the phase:
tan f = (VCp - VLp)/ VRp
or;
tan f = (XC-XL)/R. or
tan f = (1/wC - wL) / R
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). 25

46. Power in an AC Circuit V f = 0 I P p wt 2p (This is for a purely
V(t) = VP sin (wt) I
I(t) = IP sin (wt) p 2p wt
(This is for a purely
resistive circuit.) P
P(t) = IV = IP VP sin 2(wt)
Note this oscillates
twice as fast. p 2p wt 26

47. Power in an AC Circuit
The power 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 ) 27

48. Power in an AC Circuit
Now: 28

49. Power in an AC Circuit
Now: 29

50. Power in an AC Circuit
Now:
Use:
and: So 30

51. Power in an AC Circuit
Now:
Use:
and: So
which we usually write as 31

52. 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. 32

53. Power in an AC Circuit What if f is not zero? I P V wt
Here I and V are 900
out of phase. (f= 900)
(It is purely reactive)
The time average of
P is zero. wt 33

54. STOP!

55. Transformers Iron Core Power is conserved, though:
Transformers use mutual inductance to change voltages:
N2 turns V1 V2
N1 turns
Iron Core
Primary
Secondary
Power is conserved, though:
(if 100% efficient.)

56. 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
Power
=I1 V1
V1=110V
V2=20kV
We use high voltage (e.g. 365 kV) to transmit electrical
power over long distances.
Why do we want to do this?

57. 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
Power
=I1 V1
V1=110V
V2=20kV
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)