# 20. Alternating Currents - [email protected] ```University of Rhode Island
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PHY 204: Elementary Physics II
Physics Course Materials
2015
20. Alternating Currents
Gerhard M&uuml;ller
University of Rhode Island, [email protected]
Abstract
Part twenty of course materials for Elementary Physics II (PHY 204), taught by Gerhard M&uuml;ller at
the University of Rhode Island. Documents will be updated periodically as more entries become
presentable.
Some of the slides contain figures from the textbook, Paul A. Tipler and Gene Mosca. Physics for
Scientists and Engineers, 5th/6th editions. The copyright to these figures is owned by W.H. Freeman.
We acknowledge permission from W.H. Freeman to use them on this course web page. The textbook
figures are not to be used or copied for any purpose outside this class without direct permission from
W.H. Freeman.
Recommended Citation
M&uuml;ller, Gerhard, &quot;20. Alternating Currents&quot; (2015). PHY 204: Elementary Physics II. Paper 4.
http://digitalcommons.uri.edu/elementary_physics_2/4
This Course Material is brought to you for free and open access by the Physics Course Materials at [email protected] It has been accepted for
[email protected]
Alternating Current Generator
Coil of N turns and cross-sectional area A rotating with angular frequency ω
~
in uniform magnetic field B.
• Angle between area vector and magnetic field vector: θ = ωt.
• Flux through coil: ΦB = N BA cos(ωt).
dΦB
= Emax sin(ωt) with amplitude Emax = N BAω.
dt
• U.S. household outlet values:
√
◦ Emax = 120V 2 ≃ 170V
◦ f = 60Hz, ω = 2πf ≃ 377rad/s.
• Induced EMF: E = −
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Single Device in AC Circuit: Resistor
Voltage of ac source : E = Emax cos ωt
Current through device: I = Imax cos(ωt − δ)
Resistor
VR = RI = Emax cos ωt ⇒ I =
Emax
cos ωt
R
Emax
, phase angle: δ = 0
R
Emax
≡
= R (resistance)
Imax
amplitude: Imax =
impedance: XR
I
VR
ωt
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Single Device in AC Circuit: Inductor
Voltage of ac source : E = Emax cos ωt
Current through device: I = Imax cos(ωt − δ)
Inductor
Emax
dI
= Emax cos ωt ⇒ I =
sin(ωt)
dt
ωL
Emax
π
amplitude: Imax =
, phase angle: δ =
ωL
2
Emax
impedance: XL ≡
= ωL (inductive reactance)
Imax
VL = L
VL
ωt
I
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Single Device in AC Circuit: Capacitor
Voltage of ac source : E = Emax cos ωt
Current through device: I = Imax cos(ωt − δ)
Capacitor
dQ
Q
= Emax cos ωt ⇒ I =
= −ωCEmax sin(ωt)
C
dt
π
amplitude: Imax = ωCEmax , phase angle: δ = −
2
1
Emax
(capacitive reactance)
=
impedance: XC ≡
Imax
ωC
VC =
I
VC
ωt
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Single Device in AC Circuit: Application (1)
The ac voltage source E = Emax sin ωt has an amplitude of Emax = 24V
and an angular frequency of ω = 10rad/s.
In each of the three circuits, find
(a) the current amplitude Imax ,
(b) the current I at time t = 1s.
(1)
~
(2)
~
(3)
~
8Ω
25mF
0.6H
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Single Device in AC Circuit: Application (2)
Consider an ac generator E(t) = Emax cos(ωt), Emax = 25V, ω = 377rad/s
connected to an inductor with inductance L = 12.7H.
(a) Find the maximum value of the current.
(b) Find the current when the emf is zero and decreasing.
(c) Find the current when the emf is −12.5V and decreasing.
(d) Find the power supplied by the generator at the instant described in (c).
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Single Device in AC Circuit: Application (3)
Consider an ac generator E(t) = Emax cos(ωt), Emax = 25V, ω = 377rad/s
connected to a capacitor with capacitance C = 4.15&micro;F.
(a) Find the maximum value of the current.
(b) Find the current when the emf is zero and decreasing.
(c) Find the current when the emf is −12.5V and increasing.
(d) Find the power supplied by the generator at the instant described in (c).
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RLC Series Circuit (1)
Applied alternating voltage: E = Emax cos ωt
Resulting alternating current: I = Imax cos(ωt − δ)
Goals:
• Find Imax , δ for given Emax , ω.
• Find voltages VR , VL , VC across devices.
~
ε
A
Loop rule: E − VR − VC − VL = 0
Note:
• All voltages are time-dependent.
R
C
L
V
V
V
VR
VC
VL
I
• In general, all voltages have a different phase.
• VR has the same phase as I.
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RLC Series Circuit (2)
Phasor diagram (for ωt = δ):
VL
Voltage amplitudes:
• VR,max = Imax XR = Imax R
ε
VL−VC
• VL,max = Imax XL = Imax ωL
• VC,max = Imax XC
δ
Imax
=
ωC
VC
I
VR
Relation between Emax and Imax from geometry:
2
Emax
=
=
2
+ (VL,max − VC,max )2
VR,max
&quot;
&laquo;2 #
„
1
2
R2 + ωL −
Imax
ωC
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RLC Series Circuit (3)
Emax
Impedance: Z ≡
=
Imax
s
&laquo;2
„
1
R2 + ωL −
ωC
VL
ε
Current amplitude and phase angle:
• Imax
Emax
Emax
= q
=
`
Z
2
R + ωL −
• tan δ =
VL−VC
&acute;
1 2
ωC
δ
VL,max − VC,max
ωL − 1/ωC
=
VR,max
R
VC
I
VR
Voltages across devices:
• VR = RI = RImax cos(ωt − δ) = VR,max cos(ωt − δ)
“
dI
= −ωLImax sin(ωt − δ) = VL,max cos ωt − δ +
• VL = L
dt
Z
“
Imax
1
sin(ωt − δ) = VC,max cos ωt − δ −
Idt =
• VC =
C
ωC
π”
2
π”
2
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AC Circuit Application (1)
In this RLC circuit, the voltage amplitude is Emax = 100V.
Find the impedance Z, the current amplitude Imax ,
and the voltage amplitudes VR , VC , VL , VLC
(a) for angular frequency is ω = 1000rad/s,
(b) for angular frequency is ω = 500rad/s.
~
100V
A
500 Ω
0.5&micro;F
2H
V
V
V
VR
VC
VLC
VL
V
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AC Circuit Application (2)
In this RLC circuit, we know the voltage amplitudes VR , VC , VL across each device, the current
amplitude Imax = 5A, and the angular frequency ω = 2rad/s.
• Find the device properties R, C, L and the voltage amplitude Emax of the ac source.
~
ε max
A
R
C
L
V
V
V
50V
25V
25V
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Impedances: RLC in Series (1)
Z=
s
R2 +
„
1
ωL −
ωC
resonance at ω0 = √
Ω0
&laquo;2
limit R → 0
˛
˛
˛
1 ˛˛
Z = ˛˛ωL −
ωC ˛
1
LC
ΩL
ΩL
Z
Z
1ΩC
1ΩC
R
Ω
Ω0
Ω
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Impedances: RLC in Series (2)
limit L → 0
limit C → ∞
q
Z = R2 + (ωL)2
Z=
s
R2 +
1
(ωC)2
Z
ΩL
Z
R
R
Ω
1ΩC
Ω
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Filters
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RLC Series Resonance (1)
impedance
Z=
s
R2
+
„
current
1
ωL −
ωC
&laquo;2
Vmax
Imax = q
`
R2 + ωL −
Imax
phase angle
&acute;
1 2
ωC
δ=
ωL − 1/ωC
R
∆
Π2
VmaxR
ΩL
Z
Ω
Ω0
1ΩC
Ω0
R
Ω
Ω0
Ω
-Π2
resonance angular frequency:
ω0 = √
1
LC
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RLC Series Resonance (2)
resistor
inductor
capacitor
VRmax = Imax R
VLmax = Imax ωL
VCmax = Imax /ωC
VRmax
VLmax
VC max
V0max
V0max
Vmax
Vmax
Vmax
Ω0
Ω
• relaxation times: τRC = RC,
• angular frequencies: ωL = p
• voltages:
V0max
Ω
Ω0
Ω
Ω0
τRL = L/R
ω0
1 − (ω0 τRC
= Vmax ω0 τRL ,
)2 /2
VLmax (ωL )
q
= ω0 1 − (ω0 τRC )2 /2
,
ωC
=
VCmax (ωC )
V0max
= p
1 − (ω0 τRC )2 /4
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RLC Parallel Circuit (1)
Applied alternating voltage: E = Emax cos ωt
Resulting alternating current: I = Imax cos(ωt − δ)
~
Goals:
• Find Imax , δ for given Emax , ω.
R
L
Junction rule: I = IR + IL + IC
• All currents are time-dependent.
IR
A
I
A
• Find currents IR , IL , IC through devices.
Note:
ε
IL
A
C
IC
A
• In general, each current has a different phase
• IR has the same phase as E.
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RLC Parallel Circuit (2)
Phasor diagram (for ωt = δ):
IC
Current amplitudes:
• IR,max
Emax
Emax
=
=
XR
R
• IL,max
Emax
Emax
=
=
XL
ωL
• IC,max =
IR
ε
δ
I L − IC
I
Emax
= Emax ωC
XC
IL
Relation between Emax and Imax from geometry:
2
Imax
=
=
2
IR,max
+ (IL,max − IC,max )2
&quot;
&laquo;2 #
„
1
1
2
− ωC
+
Emax
2
R
ωL
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RLC Parallel Circuit (3)
1
Imax
Impedance:
≡
=
Z
Emax
s
1
+
R2
„
1
− ωC
ωL
&laquo;2
Current amplitude and phase angle:
s
&laquo;2
„
1
1
Emax
= Emax
− ωC
+
• Imax =
Z
R2
ωL
• tan δ =
IL,max − IC,max
1/ωL − ωC
=
IR,max
1/R
Currents through devices:
IC
IR
ε
δ
I L − IC
I
IL
E
Emax
=
cos(ωt) = IR,max cos(ωt)
R
R
Z
“
1
Emax
π”
• IL =
Edt =
sin(ωt) = IL,max cos ωt −
L
ωL
2
“
π”
dE
= −ωCEmax sin(ωt) = IC,max cos ωt +
• IC = C
dt
2
• IR =
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AC Circuit Application (3)
Find the current amplitudes I1 , I2 , I3
(a) for angular frequency ω = 2rad/s,
(b) for angular frequency ω = 4rad/s.
~
εmax =50V
R=10Ω
L=2.5H
C=0.1F
A
I3
A
I2
A
I1
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AC Circuit Application (4)
Given the current amplitudes I1 , I2 , I3 through the three branches of this RLC circuit, and given
the amplitude Emax = 100V and angular frequency ω = 500rad/s of the ac source, find the device
properties R, L, C.
ε
~
R
L
C
A
I 3 =213.6mA
A
I 2 =75mA
A
I 1 =25mA
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Impedances: RLC in Parallel (1)
1
=
Z
s
1
+
R2
„
&laquo;
1 2
ωC −
ωL
resonance at ω0 = √
Ω0
limit R → ∞
˛
˛
˛
1 ˛˛
1
= ˛˛ωC −
Z
ωL ˛
1
LC
ΩC
ΩC
1Z
1Z
1ΩL
1ΩL
Ω0
1R
Ω
Ω
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Impedances: RLC in Parallel (2)
limit C → 0
1
=
Z
s
limit L → ∞
1
1
+
R2
(ωL)2
1
=
Z
r
1
+ (ωC)2
2
R
1Z
ΩC
1Z
1R
1R
1ΩL
Ω
Ω
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[tsl505 – 24/34]
RLC Parallel Resonance (1)
current
impedance
1
=
Z
s
1
+
R2
„
1
ωC −
ωL
&laquo;2
Imax =
phase angle
Vmax
Z
δ=
1/ωL − ωC
1/R
∆
Imax
Π2
ΩC
1Z
Ω0
1ΩL
Ω0
Ω
VmaxR
1R
Ω
Ω0
Ω
-Π2
resonance angular frequency:
ω0 = √
1
LC
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RLC Parallel Resonance (2)
resistor
inductor
capacitor
max
IR
= Vmax /R
max
IL
= Vmax /ωL
max
IC
= Vmax ωC
ILmax
IRmax
IC max
I0max
I0max
VmaxR
Ω
Ω0
Ω0
Ω
Ω
Ω0
currents at resonance:
max
IR
Vmax
=
,
R
max
max
IL
= IC
= I0max = Vmax
r
C
.
L
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Power in AC Circuits
Voltage of ac source: E = Emax cos ωt
Current through circuit: I = Imax cos(ωt − δ)
Instantaneous power supplied: P (t) = E(t)I(t) = [Emax cos ωt][Imax cos(ωt − δ)]
Use cos(ωt − δ) = cos ωt cos δ + sin ωt sin δ
⇒ P (t) = Emax Imax [cos2 ωt cos δ + cos ωt sin ωt sin δ]
Time averages: [cos2 ωt]AV =
1
,
2
[cos ωt sin ωt]AV = 0
Average power supplied by source: PAV =
Power factor: cos δ
1
Emax Imax cos δ = Erms Irms cos δ
2
ε
~
~
A
Z,δ
R
L
C
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Transformer
• Primary winding: N1 turns
(rms)
cos(ωt − δ1 )
• Secondary winding: N2 turns
(rms)
cos(ωt − δ2 )
(rms)
cos(ωt), I1 (t) = I1
V1 (t) = V1
(rms)
cos(ωt), I2 (t) = I2
V2 (t) = V2
(rms)
• Voltage amplitude ratio:
V1
(rms)
V2
(rms) (rms)
I1
• Power transfer: V1
=
N1
N2
(rms) (rms)
I2
cos δ1 = V2
cos δ2
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AC Circuit Application (5)
Find the current amplitudes I1 , I2 , I3 , I4 in the four RLC circuits shown.
εmax =1V
~
I1
1Ω
εmax =1V
~
1Ω
A
1F
1H
I2
A
1H
1F
εmax =1V
~
1H
1Ω
εmax =1V
1F
I3
~
1Ω
A
1H
I4
A
1F
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AC Circuit Application (6)
Consider an RLC series circuit with inductance L = 88mH, capacitance C = 0.94&micro;F, and
unknown resistance R.
The ac generator E = Emax sin(ωt) has amplitude Emax = 24V
and frequency f = 930Hz. The phase angle is δ = 75◦ .
(a) Find the resistance R.
(b) Find the current amplitude Imax .
max stored in the inductor.
(c) Find the maximum energy UL
max stored in the capacitor.
(d) Find the maximum energy UC
(e) Find the time t1 at which the current has its maximum value Imax .
(f) Find the time t2 at which the charge on the capacitor has its maximum value Qmax .
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AC Circuit Application (7)
Consider the two ac circuits shown.
(a) In the circuit on the left, determine the current amplitude I1
and the voltage amplitudes V1 and V2 .
(b) In the circuit on the right, determine the current amplitudes I2 , I3 , and I4 .
R = 3Ω
~
ε max = 50V
A
I1
~
C=0.1F
C = 0.1F
V
V1
V2
I3
A
I4
A
I2
A
L=2H
V
εmax = 50V
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AC Circuit Application (8)
Consider the two ac circuits shown.
(a) In the circuit on the left, determine the maximum value of current I1
and the maximum value of voltages V1 and V2 .
(b) In the circuit on the right, determine the maximum value of currents I2 , I3 , and I4 .
~
ε max = 50V
A
L=2H
C = 0.1F
I1
~
C=0.1F
εmax = 50V
I3
V
V1
V2
I4
A
I2
A
R = 3Ω
V
A
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AC Circuit Application (9)
In the two ac circuits shown the ammeter and voltmeter readings are rms values.
(a) In the circuit on the left, find the resistance R of the resistor, the capacitance C of the
capacitor, the impedance Z of the two devices combined, and the voltage Erms of the power
source.
(b) In the circuit on the right, find the capacitance C of the capacitor, the inductance L of the
inductor, the impedance Z of the two devices combined, and the rms value of the current I4 .
~
ε rms
A
R
C
I1= 9A
~
C
I 3 = 5A
V1 = 30V
A
I4
A
I 2 = 10A
A
L
V
εrms = 20V
V
V2 = 20V
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AC Circuit Application (10)
In the two ac circuits shown the ammeter and voltmeter readings are rms values.
(a) In the circuit on the left, find the capacitance C of the capacitor, the inductance L of the
inductor, the impedance Z of the two devices combined, and the voltage Erms of the power
source.
(b) In the circuit on the right, find the capacitance C of the capacitor, the resistance R of the
resistor, the impedance Z of the two devices combined, and the rms value of the current I4 .
L
~
ε rms
I 1 = 2A
C
~
A
C
I 3 = 5A
V1 = 40V
A
I4
A
I 2 = 2A
A
R
V
εrms = 50V
V
V2 = 25V
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```