Physics 227: Lecture 23
AC Circuits II - RLC circuits
(Lots of Algebra)
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Lecture 22 review:
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Vrms = Vmax/√2. Pave = Pmax/2.
Diodes.
Phasors: rotating vectors, I, V are x-components of vectors
Reactance: X = Vmax/Imax, no phase information.
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Thursday, December 1, 2011
XR = R, XL = ωL, XC = 1/ωC
Now AC voltage divider just like DC case for resistors (for
magnitudes): V1/Vsupply = X1/(x1+X2), etc. (not quite right)
Voltage iclicker
You are told that electrical power in your house is at 120 V
and 60 Hz. The 120 V is the:
A. Voltage amplitude or maximum.
B. A meaningless number.
C. The average voltage.
D. The rms voltage.
E. The voltage squared.
Thursday, December 1, 2011
The peak / maximum / amplitude
of the voltage is √2 x 120 ≈ 170 V.
Monday we did the following...
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Consider an AC voltage applied over two circuit elements
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X1
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VAC
X2
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With two resistors:
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V1 = R1/(R1+R2) VAC.
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V1 = R1/(R1+ωL) VAC.
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V1 = R1/(R1+1/ωC) VAC = ωCR1/(1+ωCR1) VAC.
V2 = R2/(R1+R2) VAC.
What about, e.g. X2 = ωL?
V2 = ωL/(R1+ωL) VAC.
What about, e.g. X2 = 1/ωC?
V2 = (1/ωC)/(R1+1/ωC) VAC = 1/(1+ωCR1) VAC.
It is not quite right - it ignores the phases. So
today let’s do it right, with impedance!
We are going to find that the “a+b” in the denominator → √(a2+b2)
Thursday, December 1, 2011
Impedance
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ZR = R, ZL = jωL = jXL, ZC = 1/jωC = -j/ωC = -jXC.
The different ± signs for L & C reflect the different phases.
For any element, V = IZ.
For multiple elements, impedances add like resistances, but
with complex number algebra.
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Z1
VAC
Z2
Thursday, December 1, 2011
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What is j?
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j = √(-1).
Mathematicians and physicists like i = √(-1).
Electrical engineers think i is already reserved
for something more important, current, so they
use j.
Now we are nearly ready to ``turn the crank’’
following voltage divider algebra, with complex
arithmetic.
Review - complex numbers
Imaginary # axis
C = |A+jB|
B
φ
A
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magnitude C = √(A2+B2)
tanφ = B/A = Imaginary part / Real part
cosφ = A/C
sinφ = B/C
Real # axis
We can write the complex number as:
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C = Ccosφ + j Csinφ, or
recalling ejφ = cosφ + jsinφ: C = Cejφ
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Thursday, December 1, 2011
You can prove this from Taylor series expansions
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sin(x) = ∑n odd (-1)(n-1)/2xn/n!
cos(x) = ∑n even (-1)n/2xn/n!
exp(x) = ∑n xn/n!
Notes...
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I don’t know why the book does phasors and reactance - I like
going straight to impedance. ``Z’’ = impedance.
I read through the chapter, and it seems to me unclear from the
text whether R is a reactance or not. Most texts I have used
clearly define XR = R, which I agree with. Others separate
impedance into resistance (real) and reactance (imaginary) parts.
The book defines the impedance of an AC circuit with V = IZ,
and gives Z = √[R2 + (XL-XC)2] for an RLC circuit, in Eq. 31.21. I
think impedance is a complex quantity, so that formally this is
wrong. This gives the magnitude of the impedance, not the
impedance, which is a complex quantity.
Our approach is that the magnitude of the complex quantities
gives the physical quantity, and the phase gives the phase angle
in the time dependence (cos(ωt+φ)). (There are other conventions that
give equivalent answers.)
Thursday, December 1, 2011
Faissler: “Introduction to Modern
Electronics”
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My favorite introductory electronics book for students.
Thursday, December 1, 2011
The Voltage Divider with Impedance
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ZR = R, ZL = jωL, ZC = 1/jωC = -j/ωC
V1/VAC = Z1/(Z1+Z2)
Z1
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VAC
Z2
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Resistors or inductors add in series, as before.
What about, e.g. Z1 = R, Z2 = jωL?
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VL/VAC = jωL/(R+jωL) =
ejπ/2-jatan(ωL/R) ωL/√(R2+ω2L2).
For ω very small, VR/VAC → R/R = 1, φR → 0,
VL/VAC → 0/R = 0, φL → π/2.
For ω very large, VR/VAC → R/∞ = 0, φR → π/2,
VL/VAC → ωL/ωL = 1, φL → 0.
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Thursday, December 1, 2011
VR/VAC = R/(R+jωL) = e-jatan(ωL/R)R/√(R2+ω2L2).
As VR, L or C → VAC, the phase has to go to 0. (In
phase with source.)
The Voltage Divider with Impedance
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ZR = R, ZL = jωL, ZC = 1/jωC = -j/ωC
V1/VAC = Z1/(Z1+Z2)
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Z1
What about, e.g. Z1 = R, Z2 = 1/jωC?
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VAC
Z2
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VC/VAC = (-j/ωC)/(R-j/ωC) = -j/(ωCR-j) =
e-jπ/2+jatan(-1/ωRC)/√(1+ω2R2C2).
For ω very small, VR/VAC → 0, φR → π/2,
VC/VAC → 1, φL →0.
For ω very large, VR/VAC → 1, φR → 0,
VC/VAC → 0, φL → π/2.
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Thursday, December 1, 2011
VR/VAC = R/(R-j/ωC) = ωCR/(ωCR-j) =
ejatan(1/ωRC)ωRC/√(1+ω2R2C2).
Again, as VR, L or C → VAC, the phase has to go to
0.
The RLC Circuit
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ZR = R, ZL = jωL, ZC = 1/jωC = -j/ωC
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VR/VAC = e-jatan[(ωL-1/ωC)/R] R/√(R2+(ωL-1/ωC)2).
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For ω → ∞: VR/VAC → 0, φR → -π/2, VC/VAC → 0, φC → -π,
VL/VAC → 1, φL → 0.
V1/VAC = Z1/(Z1+Z2) → Z1/(Z1+Z2+Z3)
ZTOTAL = Z1+Z2+Z3 = R+j(ωL-1/ωC) =
ejatan[(ωL-1/ωC)/R]√(R2+(ωL-1/ωC)2) ➭
phase of V wrt I is tan-1(ωL-1/ωC)/R.
VC/VAC = e-π/2-jatan[(ωL-1/ωC)/R] 1/√(ω2C2R2+(ω2LC-1)2).
VL/VAC = eπ/2-jatan[(ωL-1/ωC)/R] ωL/√(R2+(ωL-1/ωC)2).
For ω → 0: VR/VAC → 0, φR → π/2, VL/VAC → 0, φL → π,
VC/VAC → 1, φC → 0.
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Again, as VR, L or C → VAC, the phase has to go to 0.
Thursday, December 1, 2011
The RLC Circuit
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What happens when ω = 1/√(LC)?
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VR/VAC = 1.
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ZTOTAL = Z1+Z2+Z3 = R+j(ωL-1/ωC) = R
➭ phase of V wrt I is 0 - no imaginary
part to impedance
VC/VAC = -j/ωCR = -j√(L/C)/R.
VL/VAC = jωL/R = j√(L/C)/R.
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Note ωL = 1/ωC.
Full applied voltage over resistor,
capacitor and inductor have equal and
opposite voltages across them.
Minimum total impedance at this
frequency, so maximum current.
Resonance!
Thursday, December 1, 2011
Demo
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Thursday, December 1, 2011
We have had a lot of algebra - let’s have a
demo break.
Tesla Coil
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High frequency air-core
transformer. HV at the
torus discharged by arcing.
Thursday, December 1, 2011
RLC iClicker
A series RLC circuit is operated at a
frequency where the applied voltage is
nearly all over the inductor. What is the
relative phase of the voltage over the
inductor with respect to the applied voltage?
A. 0.
B. π
C. ±π/2.
D. It depends on the order of the elements.
E. One cannot have the applied voltage nearly all over
the inductor - this is a trick question.
Thursday, December 1, 2011
See final comment
on slides 8, 9, 10
Power
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For a resistor, we have P = IV = IrmsVrms
What happens for a capacitor or inductor?
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We related current to voltage through the impedance: V = IZ.
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P(t) = Vmaxcos(ωt) (Vmax/ωL)cos(ωt+π/2) = (Vmax2/ωL)(cos2(ωt)
cos(π/2) - cos(ωt)sin(ωt)sin(π/2)) = -(Vmax2/ωL)cos(ωt)sin(ωt)
= -(Vmax2/2ωL)sin(2ωt) → 0, averaged over a cycle.
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For voltage over an inductor only, with ZL = jωL, there is a
phase difference of 90o between the inductor current and
voltage.
We have the same result for a capacitor.
Thursday, December 1, 2011
Power II
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Consider an arbitrary impedance Z = Z0ejφ.
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The instantaneous power is P(t) = [Vmax2/ωL][cos2(ωt) cos(φ) cos(ωt)sin(ωt)sin(φ)].
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If the applied voltage is Vmaxcos(ωt), the resulting current is
(Vmax/Z0)cos(ωt+φ).
For the average power, the second term averages to 0 over a
cycle, so Pave = (Vmax2/2Z0)cos(φ).
Equivalently, Pave = VrmsIrmscos(φ). The ``power factor’’ is cos(φ),
which ranges from 0 (no resistance) to 1 (pure resistance).
Thursday, December 1, 2011
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Transformers I
Transformers can be used to
step up or step down the
voltage.
Assume an iron core contains
all the magnetic flux, so there
is the same flux ΦB through
each loop.
Since ξ = NdΦB/dt, we have
ξ1/N1 = ξ2/N2 = dΦB/dt.
Neglecting resistance, we have
V1/N1 = V2/N2 - higher voltage
for the coil with more loops.
Power is conserved, so P1 = I1V1
= P2 = I2V2. The current is
reduced by the same factor
the voltage is increased.
Thursday, December 1, 2011
Transformers II
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For the secondary winding (on
the right) we have V2 = I2R.
We can replace V2 = N2V1/N1
and I2 = N1I1/N2, so ...
N2V1/N1 = I1(N1/N2) R, or ...
V1/I1 = (N1/N2)2 R.
Transformer transforms voltage
by a factor of N2/N1, current
by a factor of N1/N2, and
resistance seen by the source
by a factor (N1/N2)2 from the
actual ``load’’.
Thursday, December 1, 2011
Transformer
Pictures
circuit symbol
for iron core
transformer
Thursday, December 1, 2011
Transformers and
Eddy Currents /
Power Loss
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Changing flux in the iron core
of a transformer will cause
eddy currents, leading to larger
I2R power losses and heating.
This can be reduced by making
a core of several separate
pieces to reduce eddy
currents. Recall our eddy
current demos.
There are also energy losses
from wire resistance and
magnetostriction. Typically
efficiency > 90%.
Thursday, December 1, 2011
RLC iClicker
A series RLC circuit is operated at its
resonant frequency. Which of the following is
true at this frequency?
It might help to recall XR = R, XL = ωL, XC = 1/
ωC.
A. XL = XC.
B. XL = XR.
C. XR = XC.
D. XLXC = ω2.
E. None of the others is correct.
Thursday, December 1, 2011
See slide 11.
Monday: Light
Some summer job opportunites - internships
with Federal government:
www.orau.gov/dhsinternships
www.orau.org/ornl
http://science.energy.gov/wdts/suli/
http://www.fossil.energy.gov/education/lelandfellowships/
Thursday, December 1, 2011