Chapter 24
Alternating Current Circuits
Unlike a battery, which provides a steady, constant
emf, an ac source provides an emf, or voltage,
which varies with time (sine function repeats
when argument wt increases by 2p):
V = Vmaxsin wt
Vmax
w =2pf
w = angular frequency
voltage
T = period = 1/f = 2p/w
time
The ac voltage in your house
has a frequency f of 60 Hz
= 60 oscillations/sec.
Resistor in an ac circuit
The voltage across the resistor is
changing with time, so the current
through the resistor must also be
time-varying.
I = V/R = Imax sin wt
Notice that as the voltage changes sign,
the current changes direction.
The voltage and the
current are in
phase. They peak
at the same time.
rms Values
It is useful to compare ac circuits and dc circuits. We
cannot use, however, the average ac current or
voltage which are both zero. Instead we must use the
root mean square or rms values.
I rm s
I m ax

,
2
V rm s
V m ax

2
The rms value of current or voltage in an ac circuit can
be compared to the equivalent quantities in a simple
dc circuit.
Vrms = IrmsR
Pav = Irms2R = Vrms2/R
Root Mean Square Voltage and Current
V (t )  Vmax sin(wt )
I (t )  V (t ) / R  (Vmax / R) sin(wt )
V2
Power  P 
 I 2R
R
2
Vmax
P
sin 2 (wt )
R
2
Vmax
1
2
P 
 Vrms
/ R,
R 2
Vrms  VMax / 2
Vrms = Square root of the mean
(average) of V-squared.
Average AC Power consumed by a resistor
2
2
independent of form of V vs t curve
P  Vrms
/ R  I rms
R
Home Circuits
• US standard: 120 V, 60 Hz AC
• Vrms = 120 V, Vmax = (2) 120 V =170 V
• Circuit Breakers trip at Irms = 15 A
– Imax = (2) 15 A =21.2 Amp
• Maximum power in typical home circuit:
• P = Irms Vrms < (15A) (120 V) = 1800 W
• Clothes Dryers, AC units frequently require higher
power:
– Higher current circuits (20 A), and higher voltage:
• Vrms = 240 V
Walker Problem 5 pg. 812
A 3.33-kW resistor is connected to a generator with a
maximum voltage of 101 V. Find (a) the average
and (b) the maximum power delivered to this circuit.
Vrms  Vmax / 2
 101V / 1.414  71.4V
2
P  Vrms
/ R  (71.4V ) 2 /(3330 W)  1.53W
2
PMax  Vmax
/ R  (101V ) 2 /(3330 W)  3.06W
Capacitors in ac circuits
The “resistance” of the capacitor to
current in the circuit is known as
capacitive reactance
XC = 1/(wC)
SI unit of reactance: Ohm (W) = s/F
Vrms = IrmsXC OR Vmax = ImaxXC
The voltage lags behind the current by 90°.
V=Q/C: When I>0, capacitor is charging,
When I<0, capacitor is dis-charging.
The average power consumed by a
capacitor in an ac circuit is zero.
Every ½ cycle capacitor stockpiles energy;
During next ½ cycle capacitor disharges energy to circuit.
Walker Problem 14, pg. 813
An rms voltage of 10.0 V with a frequency of 1.00 kHz is applied to a 0.395-mF
capacitor. (a) What is the rms current in this circuit? (b) By what factor
does the current change if the frequency of the voltage is doubled? (c)
Calculate the current for a frequency of 2.00 kHz.
X C  1 /(wC ) 
1
 403
s
F
(2p )(1000 Hz)(0.395  10 6 F )
s
V
 403
 403  403W
C /V
A
I rms  Vrms / X C  10.0V /( 403W)  0.0248 A  24.8mA
If frequency is doubled, XC drops by factor of 2,
Current is doubled: Irms = 49.6 mA @2.00 kHz
Inductors in AC circuits
The “resistance” of the inductor to
current in the circuit is known as
inductive reactance
XL = wL
SI unit of reactance: Ohm (W) = H/s
Vrms = IrmsXL OR Vmax = ImaxXL
The voltage leads the current by
90°.
The voltage required to drive the current
is maximized when the current is
changing the fastest.
The average power consumed by an
inductor in an ac circuit is zero.
Walker Problem 29, pg. 813
What is the rms current in a 125-mH inductor when it is
connected to a 60.0-Hz generator with an rms voltage
of 115 Hz?
Ground Fault Interupter
• The principle of magnetic flux is exploited in an important electrical
safety circuit
• As long as the input and output currents are equal and opposite, there
is zero flux in through the secondary.
• If current flows through an alternate path to ground (e.g. through your
body!!) the imbalance produces a magnetic flux, which induces an
EMF in the sensing coil and trips the Circuit Breaker.
• This can save your life.
RC circuits:
Filters &
AC-coupling
I
I
• In a series RC circuit, the same current flows from the generator,
through the resistor, onto the top plate of the capacitor, and out of the
bottom plate of the capacitor.
– This is true for both AC and DC power.
• The voltage drop across the resistor is in phase with the current.
• The voltage drop across the capacitor lags the current by ¼ oscillation.
• The resistance and the capacitive reactance combine “in quadrature” to
yield the effective AC impedance
 1 
Z  R 2  X C2  R 2  
wC 
2
RC Circuit:
Equivalent Circuit
=
Vrms
I rms  Vrms / Z
Z
2
R  X C2 
 1 
R 
wC 
2
2
Irms
Z
RC Circuit:
Filter
Irms
Vrms
Vout
• Analyze output voltage, as a function of frequency w of
source.
I rms  Vrms / Z
Vout ,rms  I rms X C  Vrms
Vout ,rms  Vrms
1 /(wC )
XC
Z
 1 
R 
wC 
2
2
 Vrms
1
wRC 2  1
• For w >> 1/(RC), Vout  0
• For w << 1/(RC), Vout  Vrms
RC Circuit:
Filter
VDC
Irms
Vout
Vrms
• RC Circuit filters out unwanted noise (Vrms at
frequency w/2p) from a DC power supply (VDC).
• Treat the high frequency noise and DC supply
independently
– For w >> 1/(RC), Vout(w)  0
– For w << 1/(RC), Vout  VDC
RC Circuit:
AC Signal Coupling
C
Vout
Vrms
Irms
• By looking at the signal across the resistor (instead of the capacitor) an
AC circuit removes any DC bias from the input, and transmits the high
frequency signal to the output
I rms  Vrms / Z
Z  X C2  R 2
Vout ,rms  I rms R  Vrms
Vout ,rms  Vrms
R
Z
R
2
 1 
2

R
wC 
 Vrms
1
2
 1 
wRC   1
AC Coupling
C
Vout
Vrms
Irms
• At high frequency, w>>1/(RC), the capacitor acts as a
short circuit, Vout = Vrms
• At low frequency w << 1/(RC), the capacitor acts as an
open circuit, Vout  0
Vout ,rms  Vrms
1
2
 1 

1
wRC 
Phasors
• V=V0sin(wt) can be thought of as the yprojection of a vector of length V0 rotating
in a imaginary x-y plane with angular
frequency w. Why bother?
• For a Resistor, I = V/R,
– I in phase with V.
• For a Capacitor IRMS = VRMS (wC),
– I leads V by ¼ oscillation = 90° on phasor plot
• For an Inductor IRMS = VRMS / (wL)
– I lags V by ¼ oscillation = 90° on phasor plot
• Also, eip = -1 is a really cool theorem
Phasors
We can represent the current and the voltage across the inductor
(VL), capacitor (VC) and resistor (VR) by vector diagrams called
phasors. The current I is always parallel to VR. In this graph,
the phasor is frozen at the moment I(t) is parallel to x-axis. The
voltage vectors indicate their phase relative to the current:
VL
VR
VC
I
V
f
VR
VL- VC
f is the phase angle between the
current and the voltage in the circuit
VL  VC I  X L  X C  , cos f  VR  R
tan f 

V
Z
VR
IR
Walker Problem 22, pg. 813
A 65.0-Hz generator with an rms voltage of 115 V is connected in series to a
3.35-kW resistor and a 1.50-mF capacitor. Find (a) the rms current in the
circuit and (b) the phase angle, f, between the current and the voltage.
VR= IR
VC= I XC
= I/(wC)
I
115 V
115V  ( IR ) 2  ( IX C ) 2  I R 2  X C2  IZ
I  115V / Z  115V / (3350 W) 2  1 /[ 2p (65 / s )(1.50  10 6 F )]2
I  115V /(3726 W)  30.9  10 3 A
tan f  ( IX C ) /( IR )   X C / R  1 /(wRC ),
f  64.0
RC Circuit, Alternating Current
I(t)
erms
•
•
•
•
•
•
VR = I(t) R
VC = Q(t)/C
VR,rms= Irms R
VC,rms = Irms /(wC)
erms = Irms Z
Z = [R2+XC2]1/2
RC Circuit
Current in resistor R
equals current
flowing onto
capacitor.
Positive current means
the capacitor charge
is increasing.
2.50
2.00
1.50
1.00
Volts
0.50
0.00
0.00
-0.50
1.57
3.14
4.71
6.28
7.85
9.42
11.00
12.57
14.14
15.71
-1.00
-1.50
-2.00
-2.50
w t
V_R = IR
V_C=I X_C
V_R+V_C
ref pts
a)
b)
c)
d)
e)
f)
When is the capacitor
charge
increasing?
From wt = 0 to 1.57
From wt = 1.57 to 3.14
a) and b)
From wt = 3.14 to 4.71
From wt = 4.71 to 6.28
a) and e)
RC Circuit
Current in resistor R equals
current flowing onto
capacitor.
Positive current means the
capacitor charge is
increasing.
2.50
2.00
1.50
1.00
Volts
0.50
0.00
0.00
-0.50
1.57
3.14
4.71
6.28
7.85
9.42
11.00
12.57
-1.00
-1.50
-2.00
-2.50
w t
V_R = IR
V_C=I X_C
V_R+V_C
ref pts
14.14
15.71
When the capacitor
charge is
maximum:
a) The current is also
maximum
b) The current is at its
negative maximum
c) The current is
increasing
d) The current is zero.
RC Circuit
Phase shift
3.00
Volts
2.00
1.00
0.00
-1.000.00
1.57
3.14
4.71
6.28
7.85
9.42
11.00
12.57
14.14
-2.00
-3.00
w t
V_R = IR
V_C=I X_C
I (t )  2I rms cos(wt )
I
Q (t )  2 rms sin(wt )
w
Q (t )
C
1


 2I rms  R cos(wt ) 
sin(wt )
wC


1
R

 2I rms Z  cos(wt ) 
sin(wt )
wCZ
Z

e (t )  I (t ) R 
V_R+V_C
ref pts
Z  R2  X C 2
cosf  R / Z
sin f  X C / Z
e (t )  2I rms Z cos(wt  f )
e rms  I rms Z
15.71
RC Circuit
3.00
Volts
2.00
1.00
0.00
-1.000.00
1.57
3.14
4.71
6.28
7.85
9.42
11.00
12.57
14.14
15.71
-2.00
-3.00
w t
V_R = IR
V_C=I X_C
e
ref pts
VR
VR
VC
V_R+V_C
e
VC
VC
RL Circuit
& Phasors
• The source
determines e and
w,
• The circuit
determines the
magnitude and
phase of the
current.
e
e
LR Circuit, Oscillating EMF
LR Circuit
2.50
2.00
1.50
1.00
Volt
0.50
0.00
0.00
-0.50
1.57
3.14
4.71
6.28
7.85
9.42
11.00
12.57
-1.00
-1.50
-2.00
-2.50
wt
V_R = IR
V_L = I X_LI
V_R+V_L
When is the current decreasing most rapidly?
a)
At wt=0
c) At wt = p = 3.14
b)
At wt = p/2 = 1.57
d) At wt = 3 p /2 = 4.71
14.14
15.71
LC Circuit
Total Voltage drop around loop is zero,
I XL – I X C = 0
wL – 1/(wC) = 0
w2=1/(LC)
a) Initially, Capacitor is charged, current starts to flow, carrying
charge from top plate to bottom plate, through inductor.
b) Capacitor is completely discharged, but inductance opposes
any change in current. Current keeps flowing, building up
c) Positive charge on Bottom plate. This completes one half
cycle in time p/w.
LC Circuit (no dissipation)
A LC circuit is like a mass on a spring:
mass  Inductance
Kinetic Energy Magnetic Energy
Spring constant  1/Capacitance Potential Energy Electric Energy
The RLC Series Circuit
The effective
resistance of the
circuit is given by
the impedance Z:
Z  R 2  ( X L  X C )2
VIZ
Imax = Vmax / Z
Irms = Vrms / Z
SI unit of impedance: ohm
The RLC Series Circuit
Z  R 2  ( X L  X C )2
I V / Z
VR  IR  V ( R / Z )
VC  IX C  V ( X C / Z )
VL  IX L  V ( X L / Z )
w2=1/(LC)
• Resonance
XL=XC
VL=VC = VR XC /R
If XC >>R, at resonance, then
VC >>VR
VL
VR
VC
Walker Problem 40, pg. 814
Find the rms voltage across each element in an RLC
circuit with R = 9.9 kW, C = 0.15 mF, and L = 25
mH. The generator supplies an rms voltage of 115 V
at a frequency of 60.0 Hz. (b) What is the phase
angle between the voltage and the current in the
circuit?
Resonance in a Series RLC Circuit
The current in a series RLC circuit is
Vmax
I max 

Z
Vmax
R2  ( X L  X C )2
The current will be
maximum when XL = XC
at a resonance frequency
w of
1
w0 
LC
Z (w0 )  R
Energy Storage at Resonance
• Energy in Inductor
 EL= (1/2) L(Irms)2 = (1/2) L(Irms)2
• Energy lost in one cycle
 ER= PT = (Irms)2R(2p/w) = (1/2) L(Irms)2
• [Energy Stored]/[Energy dissipated in one cycle]






Oscillator ‘Q-value’ = EL / ER= Lw/4pR)
At Resonance w = 1/(LC)1/2
‘Q-Value’ = (L/C)1/2 /(4pR)
Jefferson Lab `Q-Value’=109
RLC oscillator in lab, L = 0.010H, R=15 W, C = 1.00e-4F
‘Q-Value’ = (0.01H/0.0001F) 1/2 /(4p 15W) = (10W)/(15W 4p) = 0.4
 Resonance frequency = 1/[2p(LC)1/2] = 160 Hz
Jefferson Lab Accelerator
• RLC Circuit: f = 1.5 GHz
– Free oscillation decays by factor e=2.718 in 109 oscillations
– “Like a church bell that rings for a month”: (109)/(440/sec) =26 days