LC circuit
C
L
Charge capacitor, connect to
inductor and a switch, then close
switch
what happens?
+
V
I
–
+
+
V
V
–
–
(a)
I
(b)
(h)
I
(g)
(c)
I
(f)
(d)
I
(e)
–
–
V
V
+
+
–
I
V
+
Analysis
dI
1
L
+ Q=0
dt
C
⇒
Q(t) = Qp cos(ωt)
C
L
d2 Q
1
L 2 + Q=0
dt
C
solve by comparing to spring
I
C
1
ω=√
LC
Energy is conser ved:
1 2
1 2
U = UB + UE = LI +
Q
2
2C
L
v
k
m
dU
=0
dt
Utotal
UE
UB
π
2π
ωt
More realistic, add resistance
R
C
Dissipation:
L
d2 Q
dQ
1
L 2 +R
+ Q=0
dt
dt
C
dU
= −I 2 R "= 0
dt
energy in L and C decreases with time, lost to the
environment as heat
a. under-damped
1
>
LC
b. critically damped
Displacement, x
c. over-damped
–A
!
R
2L
"2
!
R
2L
A
0
Time, t
–A
"2
"2
Heuristic argument:
The time bet ween peaks in the√
oscillation is T1 ≈ 2π/ω ≈ 2π LC
(c)
0
(a)
R
2L
1
=
LC
1
<
LC
A
(b)
!
Displacement, x
Three cases
Time, t
The time to die out is T2 ≈ L/R
To see oscillations we need T1 < T2
Time permitting, show how to use complex numbers
to easily solve:
d2 Q
dQ
1
L 2 +R
+ Q=0
dt
dt
C
Under-damped solution:
Q(t) = Q0 e
−t/τ
L
τ=
2R
ω=
cos(ωt + φ)
!
1
−
LC
"
L
2R
#2
AC circuits
Vp
Vrms
V (t) = Vp sin(ωt + φV )
–φ
π
_
2
π
3_π
2
2π
ωt
Vp–p
–Vp
Similarly
I(t) = Ip sin(ωt + φI )
and expect this functional form when V(t) applied to circuit
So we will want to find peak current and phase shift.
Will take V (t) = Vp sin(ωt) and determine current
phase shift relative to this unshifted emf (choice of zero time)
A resistor is connected across an ac source as
shown. Which graph correctly shows the
instantaneous current through the resistor
and the instantaneous voltage v = va-vb
across the resistor?
(current in purple, voltage in blue)
1.
2.
3.
A resistor is connected across an ac source as
shown. Which graph correctly shows the
instantaneous current through the resistor
and the instantaneous voltage v = va-vb
across the resistor?
(current in purple, voltage in blue)
1.
2.
3.
An inductor is connected across an ac source
as shown. Which graph correctly shows the
instantaneous current through the inductor and
the instantaneous voltage v = va-vb across the
inductor?
(current in purple, voltage in blue)
1.
2.
3.
An inductor is connected across an ac source
as shown. Which graph correctly shows the
instantaneous current through the inductor and
the instantaneous voltage v = va-vb across the
inductor?
(current in purple, voltage in blue)
1.
2.
3.
dI
v = va − vb = L
dt
(COB)
Vp sinω t
Vp sinω t
Vp sinω t
R
Vp
I(t) =
sin(ωt)
R
C
I(t) = ωCVp cos(ωt)
L
Vp
I(t) = −
cos(ωt)
ωL
Can write all these as
I(t) = Ip sin(ωt + φ)
where
Vp
Ip =
X
that is
Vp
I(t) =
sin(ωt + φ)
X
X : “reactance”
(copy to board)
(COB)
Vp sinω t
Vp sinω t
X=R
R
C
I(t) =
Vp
sin(ωt)
R
I(t) = ωCVp cos(ωt)
φ=0
1
X=
ωC
π
φ=
2
X = ωL
Vp sinω t
L
I(t) = −
Vp
cos(ωt)
ωL
π
φ=−
2
Capacitor
Vp
Ip = ω CVp
Inductor
Vp
Ip = Vp /ω L
V
I
Time
Time
I
V
Current in a capacitor “leads” the
voltage by 90 degrees
Voltage in an inductor “leads” the
current by 90 degrees
(current peaks 1/4 cycle
before voltage)
(voltage peaks 1/4 cycle
before current)
or voltage “lags” current
or current “lags” voltage
Capacitor
Vp
Ip = ω CVp
Inductor
Vp
Ip = Vp /ω L
V
I
Time
Time
I
V
High frequency = short circuit
High frequency = open circuit
Low frequency = open circuit
Low frequency = short circuit
1
XC =
ωC
XL = ωL
A capacitor is connected across an ac source
as shown. Which graph correctly shows the
instantaneous current through the capacitor
and the instantaneous voltage v = va-vb across
the capacitor?
(current in purple, voltage in blue)
1.
2.
3.
A capacitor is connected across an ac source
as shown. Which graph correctly shows the
instantaneous current through the capacitor
and the instantaneous voltage v = va-vb across
the capacitor?
(current in purple, voltage in blue)
1.
2.
dQ
d
dv
i=
= Cv = C
dt
dt
dt
3.
LR AC-circuit and Phasors
Note: this is somewhere bet ween “not explained” and “pathetically poorly explained” in the text... pay attention!
Choose the zero of time so that the phase of I(t) is zero, and
assume you know peak current Ip
I(t) = Ip sin(ωt)
Compute V(t)
V (t) = Vp sin(ωt + φV )
Then, if needed invert relation (since usually it is V(t) that is given)
V (t) = L
⇒
dI(t)
+ RI(t)
dt
V (t) = Ip XL cos(ωt) + Ip R sin(ωt)
Question: how do we write this in the form we want?
V (t) = Vp sin(ωt + φV )
Ans 1 (mostly left to student):
We have to put
V (t) = Ip XL cos(ωt) + Ip R sin(ωt)
into this form:
V (t)
= Vp sin(ωt + φV )
= Vp [cos(ωt) sin(φV ) + sin(ωt) cos(φV )]
= [Vp sin(φV )] cos(ωt) + [Vp cos(φV )] sin(ωt)
Ans 1 (mostly left to student):
We have to put
V (t) = Ip XL cos(ωt) + Ip R sin(ωt)
into this form:
V (t)
= Vp sin(ωt + φV )
= Vp [cos(ωt) sin(φV ) + sin(ωt) cos(φV )]
= [Vp sin(φV )] cos(ωt) + [Vp cos(φV )] sin(ωt)
Comparing
Vp sin φV = Ip XL
and
Vp cos φV = Ip R
or, solving:
Vp = Ip
!
XL2
+
R2
and
XL
φV = arctan
R
Ans 2: Use phasors
Recall relation bet ween circular
motion and oscillations.
Vp sin(ωt + φ)
! (t)
V
ωt + φ
These oscillating functions with
angular frequency ω can
be though of as vectors rotating
with angular velocity ω.
Then the projection on the vertical
axis gives oscillating function.
Circle of radius Vp
! (t) = V
!L (t) + V
!R (t)
The vertical component of V
is precisely the equation we need to solve
(COB+fig with I’s )
Vp = Ip
!
XL2
+
R2
XL
φV = arctan
R
Now we can invert: let t → t − φV /ω
I(t) = Ip sin(ωt)
V (t) = Vp sin(ωt + φV )
⇔
I(t) = Ip sin(ωt − φV )
V (t) = Vp sin(ωt)
with
Ip = !
Vp
R2 + XL2
XL
φI = −φV = − arctan
R
Exercise for student: find VR and VC
We will do the LRC case next
Driven Series LRC circuit
Note: Consider this an example of the application of phasors
R
V = Vp sinω dt
L
C
Solve using phasors:
! (t) = V
!L (t) + V
!R (t) + V
!C (t)
V
Vp = VRp2 + (VLp – VCp)2
VLp
Ip
Ip
VLp – VCp
φ
VRp
VCp
(COB)
Vp
Ip =
Zp
and
recall:
where
Zp =
(b)
!
R2 + (XL − XC )2
XL − XC
tan φ =
R
1
XC =
ωC
and
XL = ωL
“impedance”
Resonance
Ip = !
Vp
R2 + (ωL − 1/ωC)2
1
ω0 = √
LC
(demo)
Application:
Antenna
C
L
Amplifier
Detector
Audio
Amplifier
Speaker
Peak current, Ip
90°
Phase,φ
Voltage leads
0
ω0
Frequency,ω
ω0
Frequency,ω
Current leads
–90°
XL − XC
ωL − 1/ωC
tan φ =
=
R
R
An L-R-C series circuit as shown is operating
at its resonant frequency. At this frequency,
how are the values of the capacitive reactance
XC, the inductive reactance XL, and the
resistance R related to each other?
1. XL = R; XC can have any value
2. XC = R; XL can have any value
3. XC = XL; R can have any value
4. XC = XL = R
5. none of the above
An L-R-C series circuit as shown is operating
at its resonant frequency. At this frequency,
how are the values of the capacitive reactance
XC, the inductive reactance XL, and the
resistance R related to each other?
1. XL = R; XC can have any value
2. XC = R; XL can have any value
3. XC = XL; R can have any value
4. XC = XL = R
5. none of the above
In an L-R-C series circuit as shown, the
current has a very small amplitude if the emf
oscillates at a very high frequency. Which
circuit element causes this behavior?
1. the resistor R
2. the inductor L
3. the capacitor C
4. misleading question — the current actually has a very
large amplitude if the frequency is very high
In an L-R-C series circuit as shown, the
current has a very small amplitude if the emf
oscillates at a very high frequency. Which
circuit element causes this behavior?
1. the resistor R
2. the inductor L
3. the capacitor C
4. misleading question — the current actually has a very
large amplitude if the frequency is very high
Power in AC circuits
P = IV
I
P = IV
V
I
V
P = IV
V
Time
Time
Time
I
Resistor
Capacitor
(a)
(b)
Inductor
(c)
For inductor and capacitor P=IV alternates in sign. P>0: work is done on
circuit element. P<0: work is done by circuit element (returns energy to circuit).
Average power over cycle: define
compute (COB)
!P " = !I(t)V (t)"
1
!P " = Ip Vp cos φ
2
Define rms quatities
Vrms
1
= √ Vp
2
Irms
1
= √ Ip
2
!P " = Irms Vrms cos φ
then
cos φ is called “Power Factor”
CAUTION:
2
2
!P " = Irms
Z cos φ = Irms
R
but
2
2
2
Vrms
Vrms
Vrms
!P " =
cos φ =
R #=
2
Z
Z
R
In an L-R-C series circuit as shown, there is a
phase angle between the instantaneous
current through the circuit and the
instantaneous voltage v = va-vd across the
entire circuit. For what value of the phase
angle is the greatest power delivered to the
resistor?
1. zero
2. 90°
3. 180°
4. 270°
5. none of the above
In an L-R-C series circuit as shown, there is a
phase angle between the instantaneous
current through the circuit and the
instantaneous voltage v = va-vd across the
entire circuit. For what value of the phase
angle is the greatest power delivered to the
resistor?
1. zero
2. 90°
3. 180°
4. 270°
5. none of the above
Primary
Secondary
Transformers
If B field lines are al in Iron core, then
(a)
φ1
φ2
B-flux through one winding =
=
N1
N2
V1p
V2p
⇒
=
N1
N2
Device allows to “step-up” or “step-down” peak voltage
Ideal transformer, no power loss:
I1 V1 = I2 V2
(b)
In the transformer
shown in the drawing,
there are more turns in
the secondary than in
the primary. In this
situation,
1. the current amplitude is greater in the primary than in the
secondary
2. the current amplitude is smaller in the primary than in the
secondary
3. the current amplitude is the same in the primary and in
the secondary
4. not enough information given to decide
In the transformer
shown in the drawing,
there are more turns in
the secondary than in
the primary. In this
situation,
1. the current amplitude is greater in the primary than in the
secondary
2. the current amplitude is smaller in the primary than in the
secondary
3. the current amplitude is the same in the primary and in
the secondary
4. not enough information given to decide
Electric Power Distribution
step-up
Power
plant
20 kV
step-down
step-down
365 kV
4 kV
Transmission
line
Distribution
line in city
240 kV
Home
Use of a device at home requires using energy, at a rate of I2R
If transmission line has resistance r, without transformers,
energy is lost in lines at rate I2r
With transformer: for fixed current I in home device (we want to use our
device at fixed power, I2R) the power line has higher voltage,
lower current i, i << I. Power loss in lines is i2r << I2r
Diode:
DC Power Supplies
AC
power line
R
Current
Let’s current through
in one direction only
(a)
Time
(b)
Voltage across R
Voltage
Add low pass filter
(see homework)
C
(a)
I
Time
R
AC voltage from transformer
(b)