Alternating Current Circuits
Electromagnetism 1, PHYS1090
Dr Jamie Holder
School of Physics and Astronomy
University of Leeds
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Overview
• Time averaged power
- Root-mean-square current
• Inductors and capacitors in AC circuits
- Inductive reactance
- Capacitive reactance
• RLC circuit with a generator
- Frequency response, resonance
• Summary
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Time averaged power
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Resistor in an AC circuit: max. current
Kirchhoff’s loop rule:
E − VR = 0
For a sine-generator:
Emax sin (ωt + δ) − I R = 0
The phase δ is arbitrarily chosen as δ = π/2:
Emax cos (ωt) − I R = 0
I=
Emax
cos (ωt)
R
The maximum current Imax is
Imax =
Average Power
Emax
R
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Average power delivered by generator
Power dissipated by resistor equals that delivered by generator
P = I2 · R
2
= Imax
cos2 (ωt) · R
Pavg
h
i
2
2
= Imax · R · cos (ωt)
avg
2
Pavg = Imax
·R·
1
2
1 2
Pavg =
·I
·R
2 max
Average Power
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Current RMS value
Definition of root-mean-square current
q
hIirms =
I 2 avg
For a sinusoidal current
2
2
2
I avg = Imax · cos ωt avg
2
2
= Imax · cos ωt avg
1
2
= Imax
·
2
q
1
I 2 avg = √ · Imax
2
1
hIirms = √ · Imax
2
For the average power Pavg this means
Pavg
Pavg
1 2
·I
·R
2 max
= hIi2rms · R
=
AC current meters measure RMS values!
Average Power
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Average power delivered by a generator
Average power delivered by generator
Pavg
= hE · Iiavg
= h(Emax cos ωt) · (Imax cos ωt)iavg
2
= Emax · Imax · cos ωt avg
=
1
· Emax · Imax
2
Use RMS values to replace maximum values
√
hIirms = Imax / 2
√
hEirms = Emax / 2
then the average power Pavg is
Pavg = hEirms · hIirms
Average Power
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Example: average power
What is the average current for a 3 kW electric heater in the
UK? The mains voltage in the UK is 230 V.
Almost a trick question! The average current is actually zero.
We calculate the measured RMS current instead.
Note that 230 V is the RMS value of the mains voltage.
hIirms =
Pavg
3000W
= 13A
=
hEirms
230V
A single circuit in a normal house may only deliver up to 20 A
before the fuse trips.
Beware of electric equipment with heaters!
Average Power
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L and C in AC circuits
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Current for an AC circuit with inductor
Kirchhoff’s loop rule
E − VL = 0
dI
Emax · cos (ωt) − L ·
= 0
dt
Z
Emax ·
cos (ωt)dt − L · I
= 0
Emax
· sin (ωt) = L · I
ω
Emax
· cos (ωt − π/2)
ωL
Or with XL = ωL and δ = −π/2
I=
I=
Emax
· cos (ωt + δ)
XL
Inductors and capacitors in AC circuits
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Phase and inductive reactance
The phase δ = −π/2 means that the voltage leads the current by a one-fourth
of a period.
The maximum current is
Emax
XL
where the inductive reactance XL is defined as
Imax =
XL = ωL
The inductive reactance XL and therefore the maximum current Imax do depend on the frequency ω!
Inductors and capacitors in AC circuits
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Current for an AC circuit with a capacitor
Kirchhoff’s loop rule
E − VC = 0
Q
= 0
Emax · cos (ωt) −
C
1 dQ
−ω · Emax · sin (ωt) − ·
= 0
C dt
I = −ω · Emax · C · sin (ωt)
I = ω · Emax · C · cos (ωt + π/2)
Or with XC = 1/ωC and δ = π/2
I=
Emax
· cos (ωt + δ)
XC
Inductors and capacitors in AC circuits
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Phase and capacitive reactance
The phase δ = +π/2 means that the voltage lags the current by a one-fourth
of a period.
The maximum current is
Emax
XC
with the capacitive reactance XL defined as
Imax =
XC =
1
ωC
The capacitive reactance XC and therefore the maximum current Imax do
depend on the frequency ω!
Inductors and capacitors in AC circuits
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RLC circuit with a generator
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RLC circuit with a generator
Start with Kirchhoff’s loop rule
Emax · cos (ωt) − L
dI
Q
− − IR = 0
dt
C
then use I = dQ/dt and rearrange
dQ
1
d2 Q
L·
+R·
+ · Q = Emax · cos (ωt)
dt
dt
C
Solution of the differential equation, with Z called the impedance:
I
δ
Imax
RLC circuit with a generator
= Imax · cos (ωt − δ)
XL − XC
=
R
Emax
Emax
= p
=
Z
R2 + (XL − XC )2
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Resonance
The current reaches a maximum for
XL = XC
1
ωL =
ωC
1
ω = √
= ω0
LC
where ω0 is the resonance frequency.
Average power provided by the generator
Pavg = hIi2rms · R = hEi2rms ·
R
Z2
Z 2 = R2 + (XL − XC )2
1 2
)
= R2 + (ωL −
ωC
2 2
L
1
= R2 +
· ω2 −
ω
LC
2
2
L
= R2 +
· ω 2 − ω02
ω
RLC circuit with a generator
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Resonance (continued)
Inserting the impedance Z into the average power Pavg provided by the generator
hEi2rms R ω 2
Pavg =
2
ω 2 R2 + L2 · ω 2 − ω02
The power provided reaches a maximum at
ω0 = √
1
LC
Maximum power and width of the curve also depends on the resistance R.
RLC circuit with a generator
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Summary
• Average power
Pavg = hEirms · hIirms
• Inductive reactance
XL = ω · L
• Capacitive reactance
XC =
1
ω·C
√
• RCL circuit resonance at ω0 = 1/ LC
Pavg =
Summary
hEi2rms R ω 2
ω 2 R2 + L2 · ω 2 − ω02
2
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