Alternating Current
Circuits
Electromagnetism 1, PHYS1090
Dr Joachim Rose
Department of Physics and Astronomy
University of Leeds
EM1-L19-1
Overview
The plan for todays lecture
• 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
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
Emax
I=
cos (ωt)
R
The maximum current Imax is
Imax =
Average Power
Emax
R
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Average power delivered by generator
Power P dissipated by resistor R equals the
power delivered by generator
P = I2 · R
2
= Imax
cos2h(ωt) · R i
2
Pavg = Imax
· R · cos2 (ωt)
2
Pavg = Imax
·R·
avg
1
2
1 2
Pavg =
· Imax · R
2
Average Power
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Current RMS value
Definition of root-mean-square current
hIirms =
rD
I2
E
avg
For a sinusoidal current
D
I2
E
D
avg
E
2
2
= Imax · cos ωt
avg
D
E
2
= Imax
· cos2 ωt
avg
= Imax ·
rD
1
2
1
= √ · Imax
avg
2
1
√
hIirms =
· Imax
2
I2
E
For the average power Pavg this means
1 2
· Imax · R
2
Pavg = hIi2
rms · R
Pavg =
AC currents 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
D
E
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.
Pavg
3000W
hIirms =
= 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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Inductors and
capacitors 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
I=
· cos (ωt − π/2)
ωL
Or with XL = ωL and δ = −π/2
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
Imax =
XL
where the inductive reactance XL is defined as
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
= 0
−ω · Emax · sin (ωt) + ·
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
Imax =
XC
with the capacitive reactance XL defined as
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
dI Q
− − IR = 0
Emax · cos (ωt) − L
dt C
then use I = dQ/dt and rearrange
dQ
1
d2Q
+R·
+ · Q = Emax · cos (ωt)
L·
dt
dt
C
The solution of the differential equation is
I = Imax · cos (ωt − δ)
XL − XC
δ =
R
Emax
Emax
=
Imax = q
Z
R2 + (XL − XC )2
where Z is called the impedance.
RLC circuit with a generator
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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
R
2
2
Pavg = hIirms · R = hEirms · 2
Z
Z 2 = R2 + (XL − XC )2
1 2
= R2 + (ωL −
)
ωC
2 2
L
1
= R2 +
· ω2 −
ω
LC
2 2
L
2
2
2
= R +
· ω − ω0
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
Pavg =
2
hEi2
rms R ω
2
2
ω 2 R2 + L2 · ω 2 − ω0
The power provided reaches a maximum at
ω0 = √
1
LC
The maximum power and width of the curve
also depend of of 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
Pavg =
2
hEi2
rms R ω
2
2
2
2
2
2
ω R + L · ω − ω0
Reading: Tipler, chapter 31
Summary
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