!
A
CHAPTER 29
ALTERNATING CURRENT CIRCUITS
φ
!
B
• Introduction to phasors
• Inductors in ac circuits
• Capacitors in ac circuits
• R-C-L circuits with an ac generator
slip
rings
b
• ac generators
• Resistance in ac circuits and rms values
φ = ωt
ω is the angular velocity
b
ε
brushes
The flux through coil is
! !
Φ = B • A = BA cosφ = BA cos(ωt).
dΦ
So
= −ωBA sin(ωt).
dt
∴ε = −
dΦ
= ωBA sin(ωt).
dt
• Resonance and tuned circuits
• Transformers
ε
t
Φ
Time for one revolution
T
For one turn we have ε = ωBA sin(ωt). So for N turns,
Question 29.1: A coil with area 2.0 m2 rotates in a 0.01 T
magnetic field at a frequency of 60 Hz. How many turns
ε = NωBAsin(ωt).
Peak voltage occurs when sin(ωt) = 1,
i.e., ε peak = NωBA.
are needed to generate a peak voltage of 160 V?
∴N =
ε peak
ωBA
=
160
= 21 turns.
2π × 60 × 0.01 × 2
instantaneous
value
Alternating p.d. ⇒ alternating current:
~
i(t) = i = Icos(ωt)
v(t) = V cos(ωt)
i(t)
I
amplitude
v(t)
T
V
T
2
T
3T
2
2T
t
−V
+
+
t
v(t) = Vsin(ωt)
Frequency f = ω 2π
If w ⇒ rad/s then f ⇒ Hertz (Hz)
With f = 60 Hz, ω = 376 rad/s
Periodic time T = 1 f = 16.7ms
2
2T
t
+
t
V
−V
2
3T
−I
+
v(t)
T
t + T2
With household a.c. (60Hz) the current reverses direction
every 1120 s, i.e., 120 times every second!
∴T = 1 f = 1.67 × 10−2 s.
Distance travelled in one-half cycle:
~ v D × T 2 ⇒ 10 −4 × 0.83 × 10−2 = 8.3 ×10 −7 m.
(~ 3000 atom diameters)
Not very far!!
Resistor in an ac circuit
Voltage phasor: v R = IR cos(ωt) = VR cos(ωt)
~
i
angular
velocity
!
VR
vR
ωt
vR
R
Instantaneous current i(t) = i = Icos(ωt)
Instantaneous p.d. (voltage) across R is
v R = iR = IR cos(ωt) = VR cos(ωt),
which is the same as the emf of the generator.
v R (t)
v R (t) & i(t)
VR = IR
vR
Max v R
v R (t)
I=
VR
R
Max i
i(t)
t
VR cos(ωt)
t
A “Phasor” is a rotating vector. Phasors may be added
The voltage and current are in-phase.
just like ordinary vectors.
Power dissipated in the resistor:
Current and voltage phasors:
i = I cos(ωt)
v R = IR cos(ωt) = VR cos(ωt)
The instantaneous power p R = v R i.
But i = I cos(ωt) and v R = IR cos(ωt)
∴p R = I2R cos2 (ωt).
!
VR
!
I
i
I 2R
ωt
vR
pR
P av
i
0
T
T
2
t
vR
v R (t) & i(t)
i
vR
I cos(ωt)
The average power (over one complete period):
Energy dissipated in one period
P av =
time of one period
=
VR cos(ωt)
t
Note: the current and voltage are in-phase.
1T 2
I2R T 2
2
I
R
cos
(ωt)dt
=
∫
∫ cos (ωt)dt
T0
T 0
=
I2R 1 ⎡ ωt sin(2ωt) ⎤ T 1 2
+
= I R
T ω ⎢⎣ 2
4 ⎥⎦ 0 2
i.e., half the maximum power.
Let’s look for a moment at: i2 = I2 cos2 (ωt)
i 2 (t)
I2
hatched area
= I2T
i2
t
0
T
T
2
The average “squared” current (over one period) is:
i2 =
Area under curve of i2 (t) over one period
Time of one period
2
1T 2
= ∫ I cos2 (ωt) dt = I 2 .
T0
We define the root mean square (rms) current as:
2
i2 = I 2 = I
( = 0.707I)
2
I rms =
... and similarly:
Vrms = V
2
( = 0.707V).
2
Average power: P av = I R 2 = I2rmsR
V 2
= VrmsI rms = rms R .
i(t)
I
t
−I
Question 29.2: A current, i(t), varies with time as a
square wave, as shown above, what is its rms value?
i(t)
I
t
−I
Sketch i2 (t) versus t and we get ...
i2 (t)
I2
Question 29.3: A 100 W lightbulb is screwed into a
= I2 T
0
T
T
2
By definition I rms =
standard 120 V(rms) socket. Find
3T
2
2T
i2 , where the average square
value is
i
2
Area under the i2 (t) curve over one period
=
Time of one period
I2 T 2
=
=I .
T
∴I rms =
i2 = I2 = I.
(a) the rms current,
(b) the peak current, and
(c) the peak power.
100W
~
(a) Given : P av = 100W and Vrms = 120V.
But P av = VrmsI rms
∴I rms =
120V
P av 100
=
= 0.833A.
Vrms 120
I
V
(b) Also I rms = max and Vrms = max
2
2
∴I max = 2 × I rms = 1.18A
and Vmax = 2 × Vrms = 169.7V.
(c)
~
i
vL
L
Instantaneous current: i = I cos(ωt)
Instantaneous p.d. across inductor:
di
v L = L = −IωLsin(ωt),
dt
which is the same as the emf of the generator.
But − sin(ωt) = cos ωt + π 2 ... check it out!
∴v L = IωL cos ωt + π 2
(
Pmax
Vmax
I max
t
The peak (maximum) power Pmax = I maxVmax
= 1.18 × 169.7 = 200W,
i.e., maximum power = twice average power.
Note: the power pulses occur at 2ω .
Inductor in an ac circuit
(
)
)
X L = ωL“reactance” ( Ω)
“phase angle”
Reactance ( X L) is like “ac resistance” but it is not
constant; it depends on frequency.
(
)
∴v L = IX L cos ωt + π 2 .
Therefore, in terms of angle, v L is 90! ahead of the
current!
Progression of the voltage and current phasors for an
i = I cos(ωt)
inductor in series with an ac generator.
"
VL "
I
vL
i
v L = IX L cos(ωt + 90! ) = VL cos(ωt + 90! ),
i.e., VL = IX L ( = IωL).
I
i(t)
45!
The voltage ‘leads’ the
VL = IωL
t
v L (t)
90!
"
I
90!
90
135!
current by 90!. The plots
show how the projections
of the phasors on the xaxis, i.e., the
v L is 1 4 of a cycle ( 90!) ahead of i.
"
VL
!
ωt
vL
i = I cos(ωt)
= IX L cos(ωt + 90! )
Corresponding phasor diagram of the voltage and current.
instantaneous values of
the current and voltage,
vary with time.
Capacitor in an ac circuit
Power in an ideal inductor in an a.c. circuit:
The instantaneous power p L = v L i.
But i = I cos(ωt) and v L = IωLcos(ωt + 90! )
∴p L = I2ωL cos(ωt)cos(ωt + 90! ).
I
i
pL
VL
i
2
T
t
The average power (over one cycle):
1T
P av = ∫ I2ωLcos(ωt)cos(ωt + 90! ) dt = 0.
T0
... You can tell that from the graph! ...
The energy stored in inductor in one quarter-cycle is
released in the next quarter-cycle.
vC
Instantaneous charge on the capacitor:
I
q = ∫ idt = I ∫ cos(ωt)dt = sin(ωt).
ω
Therefore, the instantaneous p.d. across the capacitor is:
q
I
vC = =
sin(ωt),
C ωC
vL
0
Instantaneous current: i = I cos(ωt)
= dq dt .
C
t
T
~
which is the same as the emf of the generator.
But sin(ωt) = cos ωt − π 2 .
I
∴v C =
cos ωt − π 2 .
ωC
1
XC =
reactance ( Ω)
phase angle
ωC
(
(
)
)
(
)
i.e., v C = IX C cos ωt − π 2
Therefore, in terms of angle, v C is 90! behind the current.
i = I cos(ωt)
v C = IX C cos ωt − π 2 = VC cos ωt − π 2
⎛ I ⎞
i.e., VC = IX C ⎜ =
⎟.
⎝ ωC ⎠
(
)
(
Power in a capacitor in an a.c. circuit:
)
The instantaneous power pC = v Ci.
I
But i = I cos(ωt) and v C =
cos(ωt − 90! )
ωC
I
∴pC =
i(t)
VC =
I
ωC
t
v C (t)
I2
cos(ωt)cos(ωt − 90! )
ωC
I
pC
VC
t
90!
!
v C is 1 4 of a cycle ( 90 ) behind i.
"
I
vC
i
0
T
2
T
t
The average power (over one cycle):
1 T I2
P av = ∫
cos(ωt)cos(ωt − 90! ) dt = 0.
T 0 ωC
ωt
vC
i
"
VC
Corresponding phasor diagram of the voltage and current.
... look at the graph!! ...
The energy stored in the capacitor in one quarter-cycle is
released in the next quarter-cycle.
Comparison of resistance and reactance ...
R
Resistance does not change
with frequency
ω (= 2πf )
XL
Reactance of an inductor
X L (= ωL)
0.01µF
10Ω
10Ω
20Ω
20Ω
~
increases with frequency
Question 29.4: Four resistors and a capacitor are
ω (= 2πf )
connected as shown. What is the effective resistance
of the circuit at very low frequency (ω → 0), and very
XC
Reactance of a capacitor
1
XC (=
)
ωC
decreases with frequency
ω (= 2πf )
high frequency (ω → ∞)?
(a) At low frequencies ( ω → 0), XC =
10Ω
10Ω
1
→ ∞.
ωC
~
20Ω
~
1
(b) At high frequencies (ω → ∞), XC =
→0
ωC
10Ω
10Ω
13.3Ω
⇒
20Ω
20Ω
~
~
R, v R
L, v L
!
V
!
VL
!
!
δ
( VL − VC )
40Ω
= 13.3Ω
3
C, v C
i
!
I
!
VR
X L > XC
p.d. LEADS
current by δ
ωt
!
VC
By Kirchoff’s 2nd rule: the resultant p.d. supplied the
! !
!
!
generator:
V = VR + ( VL + VC ).
!
I
!
VR
!
VL
1
1
1
3
=
+
=
.
R eq 20Ω 40Ω 40Ω
∴R eq =
~
40Ω
⇒
20Ω
V
!
δ
!
( VL − VC )
!
VC
! ωt
V
X L < XC
p.d. LAGS
current by δ
The angle δ is called the phase angle.
!
V
!
VL
!
!
( VL − VC )
δ
!
I
!
VR
X L > XC
ωt
!
V
!
VL
!
!
( VL − VC )
δ
!
VC
!
V= V =
( VR )2 + ( VL − VC )2
= (IR)2 + (IX L − IXC )2 = I R 2 + (X L − XC )2 .
i.e., V = IZ
⇒ “ac Ohms’s law”
where Z is the impedance, given by:
(
)2
Z = R 2 + ωL − 1ωC
⇒ (units Ω).
Z ⇒ Z(ω) is like the “ac resistance” of the circuit.
The instantaneous p.d. across the generator is
v = V cos(ωt + δ) = IZ cos(ωt + δ),
⎛ V − VC ⎞
⎛ X − XC ⎞
where δ = tan −1 ⎜ L
⎟
⎟ = tan −1 ⎜ L
⎝
⎠
R
⎝ VR ⎠
⎛ ωL − 1 ⎞
ωC ⎟ .
= tan −1 ⎜⎜
⎟
R
⎝
⎠
!
I
!
VR
ωt
X L > XC
!
VC
The instantaneous power is:
P = vi = Vcos(ωt + δ) × Icos(ωt)
= VIcos 2 (ωt)cosδ + cos(ωt)sin(ωt)sin δ
1
= VIcos 2 (ωt)cosδ + sin(2ωt)sin δ.
2
If we average over one cycle we have, from earlier,
1
cos2 (ωt) = and sin(2ωt) = 0,
2
so second term → 0.
1
∴ P av = VI cosδ
2
= VrmsI rms cosδ.
power factor
Note: VrmsI rms is the maximum possible power.
(
)2
Since Z = R 2 + ωL − 1ωC , then Z is a minimum at
In summary ...
V
~
R, v R
some ω = ω! when ω!L =
L, v L
C, v C
Then ω! =
i
1
⇒ resonant frequency.
LC
1
ωC
XC =
The amplitude of the potential difference across generator
(emf) is:
R
V = IZ
X L = ωL
where V is the magnitude of the phasor:
! !
!
!
V = VR + VL + VC
ω!
(not an algebraic sum!) and Z is the impedance, given by:
(
1
, i.e., when X L = XC.
ω!C
)2
Z = R 2 + ωL − 1ωC .
So, we have relationships between V, I and Z:
V
V
V = IZ, I = and Z =
Z
I
ω (= 2πf )
What is the sigificance of resonance? At resonance,
ω = ω! ,
∴Z = R,
i.e., the impedance is purely resistive.
... that are the “ac equivalent” of Ohm’s law.
Also ...
What’s the power factor and the average power at
resonance?
How does the current vary with frequency? Take a
generator with a constant voltage amplitude.
V
~
We know: P av = VrmsI rms cosδ
power factor
C
R
where δ is phase angle between current and p.d. of the
I
L
generator. But, at resonance, since X L = XC, δ = 0, i.e.,
the current and p.d. are in-phase, i.e., cosδ = 1.
"
I
Phasor diagram
at resonance
"
VL
"
"
VL − VC = 0
"
V
X L = XC
ωt
"
VC
So, P(ω ! ) av = VrmsI rms , i.e., a maximum.
Also, at resonance, Vrms = I rms R.
V 2
∴ P(ω ! ) av = rms R = I rms2R.
We have: I =
=
V
V
=
2
Z
R 2 + ωL − 1ωC
(
V
(
)
R 2 + L2 ω − 1ωLC
i.e., I =
)
V
=
2
2 ⎞2
ω
!
R + L ⎜ω −
ω ⎟⎠
⎝
2
2⎛
Vω
2 2
2
(
2
ω R + L ω − ω!
Note, when ω = ω! , I =
)
2 2
,
.
V
⇒ pure resistance, i.e., at
R
resonance, the current is determined only by R (and it is
a maximum ).
What does this function look like?
Vω
I=
2
ω 2R 2 + L2 ω 2 − ω !2
(
)
!
100Ω
L = 2.0 H
0.6
C = 0.5 µF
200Ω
V = 100 V
500Ω
1000Ω
ω! =
δ
ω (= 2πf )
1500 rad/s
1
1
=
= 1000 rad/s,
LC
2.0 × 0.5 × 10 −6
and independent of R.
ω
∴f ! = ! = 159.2Hz.
2π
V
Note: Imax = , when ω = ω! .
R
!
I
!
VR
X L > XC
ωt
!
VC
δ (rad)
π
2
π
ω!
1000
500
!
( VL − VC )
0.8
0.2
!
V
!
VL
ω!
I(A)
1.0
0.4
How does the phase angle vary with frequency?
ω"
500Ω
1000Ω
4
⎛ ωL − 1 ⎞
ωC⎟
δ = tan −1 ⎜⎜
⎟
R
⎝
⎠
L = 2.0 H : C = 0.5 µF
0
− π4
− π2
500
Current leads
p.d.
X L < XC
100Ω
ω (= 2πf )
1000
ω"
1500 rad/s
p.d. leads
current
X L > XC
How does the power vary with frequency?
Vrms2ω2R
P av = I rms2R =
2.
2 2
2 2
2
ω R + L ω − ω!
(
)
V 2
Maximum power: Pav (ω = ω ! ) ⇒ rms R .
P av
500W
100Ω
Δω
L = 2.0 H
C = 0.5 µF
Question 29.5: If the frequency difference between
V = 100 V
radio stations on the old AM band is 10kHz, what are
Vrms = 70.7 V
suitable Q values at the lowest AM frequency (500kHz)
and the highest AM frequency (1600kHz)?
100W
500Ω
500
ω (= 2πf )
ω!
1000
1500 rad/s
Q-value of a resonant circuit is defined as
ωL ω
f
Q= ! ≈ ! = ! ’
R
Δω Δf
where Δω ( Δf ) is the width of the power peak at halfmaximum.
Larger Q ⇒ sharper peak.
The power curves for two neighboring stations ( f i and f j)
should be well separated so that only one of them is
“detected” at a time.
If f j − f i = 10kHz then
Δf << 10kHz
Δf
for “good” separation.
Question 29.6: A series RLC circuit, with L = 10 mH,
R = 5.0 Ω and C = 2.0 µF, is driven by an ac voltage
source that has a peak emf of 100 V. Find
fi
fj
Make the width of the power functions ( Δf ) about 1 5 th
(i.e., 20%) of the channel separation,
i.e., Δf ~ 0.2(f j − f i ) = 2kHz .
At the low frequency end ( f = 500kHz):
f
500kHz
Q=
=
= 250.
Δf
2kHz
At the high frequency end ( f = 1600kHz):
f
1600kHz
Q=
=
= 800.
Δf
2kHz
Therefore, we need a Q value > 800.
(a) the resonant frequency, and
(b) I rms at resonance.
When the frequency is 8000 rad/s, find
(c) the reactances of the capacitor and the inductor,
(d) the impedance of the circuit, and
(e) the value of I rms , and
(f) the phase angle.
V
V
~
Given : L = 10mH,
C
R
C = 2.0 µF, R = 5.0 Ω and
I
L
Vpeak = 100V.
(a) ω! = 1 LC = 1
(10 × 10−3 × 2.0 × 10 −6 )
~
R
Given : L = 10mH,
I
L
(
Z . But at resonance Z ⇒ R = 5Ω
V
and Vrms = peak
= 100
= 70.7V.
2
2
∴I rms = 70.7 5 = 14.1A.
(c) XC = 1ωC and X L = ωL, where ω = 8000 rad/s.
∴XC = 1
= 62.5Ω,
(8000 × 2.0 × 10−6 )
and X L = 8000 × 10 × 10
−3
= 80.0Ω.
)2 =
R 2 + (X L − XC )2
= 52 + (80 − 62.5)2 = 18.2Ω.
= 7.07 × 10 rad/s (1125 Hz).
(b) I rms =
Vpeak = 100V.
(d) Z = R 2 + ωL − 1ωC
3
Vrms
C = 2.0 µF, R = 5.0 Ω and
C
V
(e) I rms = rms Z = 70.7 18.2 = 3.88A.
⎛ X − XC ⎞
⎛ 80.0 − 62.5⎞
(f) δ = tan −1 ⎜ L
⎟ = tan −1 ⎜
⎟ = 74.1! .
⎝
⎠
⎝
⎠
R
5
"
V
"
VL
"
"
( VL − VC )
δ
"
I
"
VR
ωt
"
VC
X L > XC
(a) The Q factor is defined as Q =
=
ω !L
R
ω !L 7.07 × 103 × 10 ×10 −3
=
= 14.1,
R
5.0
using ω! from the previous question.
Question 29.7: Find
(a) the Q factor, and
(b) From earlier, Q =
(b) the resonance width (in Hz),
for the circuit described in the previous question (29.5).
(c) What is the power factor when ω = 8000 rad/s?
=
ω!
f
f
ω
= ! , i.e., Δf = ! = !
Δω Δf
Q 2πQ
7.07 × 103
= 79.8 Hz.
2π × 14.1
(c) The power factor is cosδ. In the previous problem
1
−1 ⎛ ωL − ωC ⎞
!
δ = tan ⎜
⎟ = 74.1 .
R
⎝
⎠
∴cosδ = 0.27.
V
~
Question 29.8: A certain electrical device draws a rms
I rms = 10 A: P av = 720 W
Vrms = 120 V.
(a) By definition: Vrms = I rms Z.
V
∴Z = rms I
= 120 10 = 12 Ω.
rms
current of 10 A at an average power of 720 W when
connected to a 120 V (rms), 60 Hz power source.
(a) What is the impedance of the device?
(b) What series combination of resistance and
reactance would have the same impedance as the
device?
(c) If the current leads the emf, is the reactance
(b) Also, P av = I rms2R, where R is the resistance of the
“device” ... why only R??
P
∴R = av
= 7.2 Ω.
I rms2
From earlier: Z = R 2 + (X L − XC )2 ,
∴Z2 = R 2 + (X L − XC )2 = R 2 + X2 ,
inductive or capacitive?
where X is the reactance of the ‘device’.
∴X = Z2 − R 2 = 122 − 7.22 = 9.6 Ω.
(c) If the current leads the emf the reactance is
capacitive.
~
L
R
If the inductor has resistance, it is equivalent to an
‘ideal’ inductor with a resistance in series. Construct
the corresponding phasor diagram: the voltage across
Question 29.9: So far, we have considered only ideal
inductors and ideal capacitors, i.e., ones in which there is
no resistance. What happens to the voltage and current
with a non-ideal inductor, i.e., one with finite resistance?
the ‘ideal’ inductor (with R = 0) leads the current by
90! but in a resistor and inductor series circuit the
phase angle between voltage and the current is
0 < δ < 90! .
"
V
"
VL
"
I
δ "
VR
In fact,
⎛V ⎞
ωL
δ = tan −1 ⎜ L ⎟ = tan −1 ⎛ ⎞ .
⎝
⎝ VR ⎠
R⎠
Magnetic
(iron) core
(laminated)
Secondary coil
ε2
ε1
N 2 ⇒ Number
of turns on the
secondary coil
Primary coil
N1 ⇒ Number
of turns on the
primary coil
Power in
TRANSFORMER
Power out
POWER IN ~ POWER OUT
i.e., ε1 i1 ≈ ε2 i 2
By Faraday’s law, in the primary coil:
dΦ
dΦ
ε
ε1 = −N1
i.e.,
=− 1.
dt
dt
N1
Large, well designed transformers convert up to 99.5%
dΦ
dΦ
ε
In the secondary: ε2 = −N2
. But
⇒− 1.
dt
dt
N1
Losses occur through:
ε
N
∴ 1= 1
ε2 N2
•
•
If ε2 > ε1 STEP-UP TRANSFORMER
If ε2 < ε1 STEP-DOWN TRANSFORMER
of the input power into output power.
HEAT and VIBRATION (SOUND)
(a) We have
Question 29.10: A step-up transformer operates on a
110 V source and supplies a resistive load with 2.00 A.
V1,rms N1
N
=
, i.e., V2,rms = 2 V1,rms
V2,rms N2
N1
= 25 × 110 = 2750 V (rms).
The ratio of primary and seconday windings is 1:25.
(a) What is the secondary voltage?
(b) What is the current in the primary?
(c) What is the power output?
(b) With no power loss V1,rmsI1,rms = V2,rmsI2,rms ,
V2,rms I2,rms 2750 × 2
i.e., I1,rms =
=
= 50.0 A (rms).
V1,rms
110
Assume 100% efficiency. All voltages and currents are
rms values.
(c) With no losses P2 = P1 = V1,rmsI1,rms
= 110 × 50.0 = 5500 W.
Transmission lines
250,000 V
Home
120V
10,000 V
local distribution
5,000 V
Power Station
Power station
1 : 50
10’s to 100’s miles
250,000 V
Typical step-up transformers at a hydro-electric
power plant.
Area sub-station
25 : 1
Utility pole
83:1
10,000 V
M
Home
120 V
Question 29.11: You return from a trip to London with
an appliance advertized as the world’s greatest coffee
maker. Unfortunately, it was designed to operate from a
Typical step-down transformers used to produce
240 V source to obtain the 960 W of power that it
domestic p.d. (110V/120V); they are often located on
requires.
local neighborhood utility poles.
(a) What can you do to operate it at 120 V?
(b) What current will the coffee maker draw from
the 120 V line?
(c) What is the resistance of the coffee maker?
All voltages are rms values.
(a) To obtain V2 = 240 V with V1 = 120 V you need a
step-up transformer with a turns ratio
N2 V2 240
=
=
= 2,
N1 V1 120
i.e., the secondary of the transformer needs twice as
many turns as the primary.
(b) Assuming no losses P1 = P2 = 960 W.
But P1 = V1I1, where I1 is the rms current in the 120 V
primary.
P
960
∴I1 = 1 =
= 8.0 A.
V1 120
(c) The rms current in the secondary is
P
960
I2 = 2 =
= 4.0 A.
V2 240
V 240
∴R = 2 =
= 60 Ω
I2 4.0