1. Physics Electricity and Magnetism Lecture 14 - AC Circuits, Resonance Y&F Chapter 31, Sec
Phasor Diagrams for Voltage and Current
The Series RLC Circuit. Amplitude and Phase Relations
Impedance and Phasors for Impedance
Resonance
Power in AC Circuits, Power Factor
Examples
Transformers

2. Apply Sinusoidal AC Source to Circuit
load
the driving frequency
resonant frequency
STEADY STATE RESPONSE (after transients die away):
. Current in load is sinusoidal, has same frequency as source ... but
. Current may be retarded or advanced relative to E by “phase angle” F for the
whole circuit (due to inertia L and stiffness 1/C).
KIRCHHOFF RULES APPLY:
. In a series branch, current has the same amplitude and phase everywhere.
. Across parallel branches, voltage drops have the same amplitudes and phases
. At nodes (junctions), instantaneous net currents in & out are conserved
peak applied voltage
peak current
DEFINE: IMPEDANCE IS THE RATIO OF PEAK (or RMS) EMF TO PEAK (or RMS) CURRENT.
Impedance for a circuit is a function of resistances, reactances, and how they are interconnected

3. Current/Voltage Phases in Resistors, Capacitors, and Inductors
Sinusoidal current i (t) = Imcos(wDt). Peak is Im
vR(t) ~ i(t) vL(t) ~ derivative of i(t) Vc(t) ~ integral of i(t)
Peak voltages across R, L, or C lead/lag current by 0, p/2, -p/2 radians
Ratios of peak voltages to peak currents are called Reactances in C’s and L’s
Phases of voltages in a series branch are relative to the current phasor.
VR& Im in phase
Resistance
currrent
Same
Phase
VL leads Im by p/2
Inductive Reactance
VC lags Im by p/2
Capacitive Reactance

4. Phasors applied to a Series LCR circuit
Applied EMF:
Current: R L C E vR vC vL
Same current including phase everywhere in the single branch
Refer voltage phasors to current phasor
Everything oscillates at the driving frequency wD
Same phase for the current in E, R, L, & C, but......
Current leads or lags E (t) by a constant phase angle F Im Em F
wDt+F
wDt
Phasors for VR, VL, & VC all rotate at wD :
VC lags Im by p/2
VR has same phase as Im
VL leads Im by p/2
Instantaneous voltages add normally (Loop Rule) Im Em F
wDt VL VC VR
along Im
perpendicular to Im
Phasor magnitudes add like vectors

5. Applies to a single series branch with L, C, R
Impedance Definition applied to the Series LCR Circuit
Voltage peak addition
rule in series circuit:
Reactances:
Same peak current flows in each component: Z Im Em F
wDt
VL-VC VR
XL-XC R
Divide each voltage in |Em| by (same) peak current:
Magnitude of Z:
Applies to a single series branch with L, C, R
Phase angle F:
See phasor diagram
R ~ 0  Im normal to Em  F ~ +/- p/2  tiny losses, no power absorbed
RESONANCE: XL=XC  Im parallel to Em  F = 0  Z = R  maximum current
F measures the power absorbed by the circuit:
F is zero  circuit acts purely resistively.
Otherwise: F >0 (current lags EMF) or F<0 (current leads EMF)

6. Summary: AC Series LCR CircuitZ Im Em F
wDt
VL-VC VR
XL-XC R
sketch shows
XL > XC L C E vR vC vL
VL = ImXL
+90º (p/2)
Lags VL by 90º
XL=wdL L
Inductor
VC = ImXC
-90º (-p/2)
Leads VC by 90º
XC=1/wdC C
Capacitor
VR = ImR
0º (0 rad)
In phase with VR R
Resistor
Amplitude Relation
Phase Angle
Phase of Current
Resistance or Reactance
Symbol
Circuit Element

7. Example 1: Analyzing a series RLC circuit
A series RLC circuit has R = 425 Ω, L = 1.25 H, C = 3.50 μF.
It is connected to an AC source with f = 60.0 Hz and εm= 150 V.
Determine the impedance of the circuit.
Find the amplitude of the current (peak value).
Find the phase angle between the current and voltage.
Find the instantaneous current across the RLC circuit.
Find the peak and instantaneous voltages across each circuit element.

8. Example 1: Analyzing a Series RLC circuit
A series RLC circuit has R = 425 Ω, L = 1.25 H, C = 3.50 μF.
It is connected to an AC source with f = 60.0 Hz and εm=150 V.
Determine the impedance of the circuit.
Angular frequency:
Resistance:
Inductive reactance:
Capacitive reactance:
(B) Find the peak current amplitude:
Current phasor Im leads the Voltage Em
Phase angle will be negative
XC > XL (Capacitive)
Find the phase angle between the current and voltage.

9. Example 1: Analyzing a series RLC circuit - continued
A series RLC circuit has R = 425 Ω, L = 1.25 H, C = 3.50 μF.
It is connected to an AC source with f = 60.0 Hz and εm=150 V.
Find the instantaneous current in the RLC circuit.
(E) Find the peak and instantaneous voltages in each circuit element.
VR in phase with Im
VR leads Em by |F| = 0
VL leads VR by p/2
VC lags VR by p/2
Add voltages above:
What’s wrong?
Voltages add with proper phases:

10. Why should fD make a difference?
Example 2:
Resonance in a series LCR Circuit:
R = 3000 W
L = 0.33 H
C = 0.10 mF
Em = 100 V.
Find |Z| and F for fD = 200 Hertz, fD = 876 Hz, & fD = 2000 Hz R L C E vR vC vL
Why should fD make a difference?
200 Hz
Capacitive
Em lags Im º
8118 W
415 W
7957 W
3000 W
Frequency f
Circuit Behavior
Phase Angle F
Impedance |Z|
Reactance XL
Reactance XC
Resistance R
876 Hz
Resistive
Max current 0º
Resonance
1817 W
2000 Hz
Inductive
Em leads Im
+48.0º
4498 W
4147 W
796 W Im Em
F < 0
F=0
F > 0

11. Resonance in a series LCR circuit
Vary wD: At resonance maximum current, minimum impedance
inductance dominates
current lags voltage
capacitance dominates
current leads voltage
width of resonance (selectivity, “Q”) depends on R.
Large R  less selectivity, smaller current at peak
damped spring oscillator
near resonance

12. Power in AC Circuits
Resistors always dissipate power, but the instantaneous rate varies as i2(t)R
No power is lost in pure capacitors and pure inductors in an AC circuit
Capacitor stores energy during two 1/4 cycle segments. During two other segments energy is returned to the circuit
Inductor stores energy when it produces opposition to current growth during two ¼ cycle segments (the source does work). When the current in the circuit begins to decrease, the energy is returned to the circuit

13. AC Power Dissipation in a Resistor
Instantaneous power
Power is dissipated in R, not in L or C
cos2(x) is always positive, so Pinst is always
positive. But, it is not constant.
Power pattern repeats every p radians (t/2)
The average or RMS power is the AC equivalent to DC power
Integrate Pinst over t:
Integral = 1/2
RMS means “Root Mean Square”
Square a quantity (positive)
Average over a whole cycle
Compute square root.
COMPUTING RMS QUANTITIES:
For any RMS quantity divide peak
value such as Im or Em by sqrt(2)
For any R, L, or C
Household power example: 120 volts RMS  170 volts peak

14. Power factor for an AC LCR Circuit
The PHASE ANGLE F determines the average RMS power actually absorbed due to the RMS current and applied voltage in the circuit.
How to compute power (proven below):
|Z|
Irms
Erms F
wDt
XL-XC R
Proof: Start with instantaneous power (not very useful):
Change variables:
Average it over one full period t:
Use trig identity:

15. Power factor for AC Circuits - continued
Odd integrand
Even integrand
Recall: RMS value = Peak value divided by sqrt(2)
Alternate form:
If R=0 (pure LC circuit) F  +/- p/2 and Pav = Prms = 0
Also note:

16. Example 2 continued with RMS quantities:
R = 3000 W
L = 0.33 H
C = 0.10 mF
Em = 100 V. R L C E VR VC VL
fD = 200 Hz
Find Erms:
Find Irms at 200 Hz:
Find the power factor:
Find the phase angle F directly:
Find the average power: or
Recall: do not use arc-cos to find F

17. Example 3 – Use RMS values inSeries LCR circuit analysis
A 240 V (RMS), 60 Hz voltage source is applied to a series LCR circuit consisting of a 50-ohm resistor, a 0.5 H inductor and a 20 mF capacitor.
Find the capacitive reactance of the circuit:
Find the inductive reactance of the circuit:
The impedance of the circuit is:
The phase angle for the circuit is:
The RMS current in the circuit is:
The average power consumed in this circuit is:
If the inductance is changed to maximize the current through the circuit, what new inductance L’ is needed?
How much RMS current would flow in that case?
F is positive since XL>XC (inductive)

18. Transformers power transformer Devices used to change
AC voltages. They have:
Primary winding
Secondary winding
Power ratings
power transformer
iron core
circuit symbol

19. Transformers Ideal Transformer Assume zero internal resistances,
iron core
zero resistance in coils
no hysteresis losses in iron core
all field lines are inside core
Assume zero internal resistances,
EMFs Ep, Es = terminal voltages Vp, Vs
Faradays Law for primary and secondary:
Assume: The same amount of flux FB cuts each turn in both primary and secondary windings in ideal transformer (counting self- and mutual-induction)
Assuming no losses: energy and power are conserved
Turns ratio fixes
the step up or step down voltage ratio
Vp, Vs are instantaneous (time varying)
or RMS averages, as can be the power and current.

20. Example: A dimmer for lights using a variable inductance
Light bulb
R=50 W
Erms=30 V L
f =60 Hz w = 377 rad/sec
Without Inductor:
a) What value of the inductance would dim the lights to 5 Watts?
Recall:
b) What would be the change in the RMS current?
Without inductor:
P0,rms = 18 W.
With inductor:
Prms = 5 W.