Exam 2 in two weeks!
• Lecture material
– Magnetism (Lect. 8) – AC circuits (Lect. 13)
– Will cover this weeks material!
• Discussion/HW material
– Discussion 4 – 7
– HW 4 – 7
• Review session Sunday, March 13, 3pm
– Will review HE2 from Fall ‘10
Physics 102: Lecture 12, Slide 1
Physics 102: Lecture 12
AC Circuits
L
R
Physics 102: Lecture 12, Slide 2
C
Review: Self-Inductance
Recall inductor
𝐿 = 𝜇0 𝑛2 𝑙𝐴
• Changing current
• Changing Bsol field
• Changing  through itself!
–  proportional to I:
Φ = 𝐿𝐼
• Induced EMF (voltage)
“Inductance”
Units: Henry (H)
– Recall Faraday’s law:
• Direction
𝐼𝑓 − 𝐼𝑖
∆Φ
∆𝐼
𝜀=−
= −𝐿 = −𝐿
Δ𝑡
Δ𝑡
𝑡𝑓 − 𝑡𝑖
– Given by Lenz’s Law
– Opposes change in current!
Physics 102: Lecture 12, Slide 3
Energy stored:
U = ½ LI2
Mutual Inductance
•
•
•
•
•
AC Generator
Changing current in P
Changing B-field thru P
Changing B-field thru S
Changing  thru S
– S proportional to IP: Φs = 𝑀𝐼𝑝
• Induced EMF (voltage) in S
– Recall Faraday’s law:
∆Φ
Φf − Φi
𝜀=−
=−
Δ𝑡
𝑡𝑓 − 𝑡𝑖
Physics 102: Lecture 12, Slide 4
Primary
Coil
Secondary
Coil
“Mutual
Inductance”
∆𝐼𝑝
𝜀𝑠 = −𝑀
Δ𝑡
Review: Generators and EMF
Voltage across generator:
 = w A B sin(q)
 = w A B sin(wt)
 = Vmax sin(wt)
1
•
w
v
2
x
q
v
r
Vmax 
Frequency = How fast its spinning
Amplitude = Maximum voltage
Physics 102: Lecture 12, Slide 5
-Vmax
t
AC Source
V(t) = Vmax sin(wt)=Vmax sin(2pf t)
Vmax = maximum voltage
f = frequency (cycles/second)
+24
V(t) = 24 sin(8p t)
2pf t = 8pt
-24
f = 4 Hz
T=(1/4)seconds/cycle
0.25
0.5
RMS: Root Mean Square Vrms=Vmax/√2
Physics 102: Lecture 12, Slide 6
RMS?
V(t) = Vmax sin(2pf t)
+Vmax
-Vmax
Square:
Mean:
Vmax2 / 2
Vmax2
square Root:
Vmax / √2
RMS: Root Mean Square Vrms=Vmax/√2
Physics 102: Lecture 12, Slide 7
Preflight 12.1, 12.2
L
R
I(t) = 10 sin(377 t)
C
Find Imax
Well… We know that the maximum value
sine is 1. So the maximum current is 10!
Imax = 10 A
Find Irms
78% correct
Just like Vrms=Vmax/√2 …
Irms=Imax/√2
=10/√2 A = 7.07 A
64% correct
Physics 102: Lecture 12, Slide 8
Resistors in AC circuit
VR = I R
always true – Ohm’s
Law
R
• VR,max = ImaxR
• Voltage across resistor is “IN PHASE” with current.
Resistance (R)
– VR goes up and down at the
same times as I does.
I
t
VR
Frequency does not
affect Resistance!
Frequency
Physics 102: Lecture 12, Slide 9
t
Capacitors in AC circuit
VC = Q/C
always true
• VC,max = ImaxXC
• Capacitive Reactance:
C
XC = 1/(2pfC)
• Voltage across capacitor “LAGS” current.
Reactance (XC)
– VC goes up and down
just after I does.
Frequency does
affect Reactance!
Frequency
Physics 102: Lecture 12, Slide 10
I
t
VC
t
Inductors in AC circuit
VL = +L(DI)/(Dt) always true
• VL,max = ImaxXL
• Inductive Reactance:
L
XL = 2pfL
• Voltage across inductor “LEADS” current.
Reactance (XL)
– VL goes up and down
just before I does.
I
t
VL
Frequency does
affect Reactance!
Frequency
Physics 102: Lecture 12, Slide 11
t
L
ACT/Preflight 12.4, 12.5
The capacitor can be ignored when…
(a) frequency is very large
(b) frequency is very small
R
C
XC
w
very large w gives very small XC
The inductor can be ignored when…
(a) frequency is very large
(b) frequency is very small
very small w gives very small XL
Physics 102: Lecture 12, Slide 12
XL
w
AC Circuit Voltages
An AC circuit with R= 2 W, C = 15 mF, and L = 30 mH
has a current I(t) = 0.5 sin(8p t) amps. Calculate the
maximum voltage across R, C, and L.
VR,max = Imax R = 0.5  2 = 1 Volt
VC,max = Imax XC = 0.5  1/(8p0.015) = 1.33 Volts
VL,max = Imax XL = 0.5  8p0.03 = 0.38 Volts
1
1
𝑋𝐶 =
=
2𝜋𝑓𝐶 𝜔𝐶
Physics 102: Lecture 12, Slide 13
𝑋𝐿 = 2𝜋𝑓𝐿 = 𝜔𝐿
L
R
C
ACT: AC Circuit Voltages
An AC circuit with R= 2 W, C = 15 mF, and L = 30 mH
has a current I(t) = 0.5 sin(8p t) amps. Calculate the
maximum voltage across R, C, and L.
L
R
Now the frequency is increased so I(t) = 0.5 sin(16p t).
Which element’s maximum voltage decreases?
1) VR,max
2) VC,max
3) VL,max
Physics 102: Lecture 12, Slide 14
C
Stays same: R doesn’t depend on f
Decreases: XC = 1/(2pfC)
Increases: XL = 2pf L
Summary so far…
L
• I = Imaxsin(2pft)
• VR = ImaxR sin(2pft)
• VR in phase with I
R
VR
I
• VC = ImaxXC sin(2pft–p/2)
1
1
• VC lags I 𝑋𝐶 =
=
2𝜋𝑓𝐶 𝜔𝐶
• VL = ImaxXL sin(2pft+p/2)
• VL leads I
Physics 102: Lecture 12, Slide 15
C
VC
𝑋𝐿 = 2𝜋𝑓𝐿 = 𝜔𝐿
t
VL
Kirchhoff: generator
voltage
Vgen
L
R
C
Write down Kirchhoff’s Loop Equation:
Vgen(t) = VL(t) + VR(t) + VC(t) at every instant of time
I
However …
Vgen,max  VL,max+VR,max+VC,max
Maximum reached at different
times for R, L, C
We solve this using phasors
Physics 102: Lecture 12, Slide 16
VR
t
VC
VL
A reminder about sines and cosines
y
Recall: y coordinates
of endpoints are
• asin(q + p/2)
• asin(q)
• asin(q – p/2)
a
q+p/2
a
q
x
q-p/2
a
Physics 102: Lecture 12, Slide 17
Graphical representation of voltages
I = Imaxsin(2pft) (q = 2pft)
VL = ImaxXL sin(2pft + p/2)
VR = ImaxR sin(2pft)
VC = ImaxXC sin(2pft – p/2)
ImaxXL
q+p/2
L
R
C
Physics 102: Lecture 12, Slide 18
ImaxR
q
ImaxXC
q-p/2
Phasor Diagrams: A Detailed Example
• I = Imaxsin(2pft)
• VR = VR,maxsin(2pft)
t = 1 f=1/12
2pft = p/6
p/6
VR,maxsin(p/6)
Length of vector = Vmax across that component
Vertical component = instantaneous value of V
Physics 102: Lecture 12, Slide 19
Phasor Diagrams
• I = Imaxsin(p/3)
• VR = VR,maxsin(p/3)
t=2
2pft = p/3
VR,maxsin(p/3)
p/3
Length of vector = Vmax across that component
Vertical component = instantaneous value of V
Physics 102: Lecture 12, Slide 20
Phasor Diagrams
• I = Imaxsin(p/2)
• VR = VR,maxsin(p/2)
VR,max
t=3
2pft = p/2
VR,maxsin(p/2)=V0
p/2
Length of vector = Vmax across that component
Vertical component = instantaneous value of V
Physics 102: Lecture 12, Slide 21
Phasor Diagrams
• I = Imaxsin(4p/6)
• VR = VR,maxsin(4p/6)
t=4
2pft = 4p/6
VR,maxsin(4p/6)
4p/6
Length of vector = Vmax across that component
Vertical component = instantaneous value of V
Physics 102: Lecture 12, Slide 22
Phasor Diagrams
• I = Imaxsin(p)
• VR = VR,maxsin(p)
VR,maxsin(p)=0
t=6
2pft = p
VR,max
p
Length of vector = Vmax across that component
Vertical component = instantaneous value of V
Physics 102: Lecture 12, Slide 23
Phasor Diagrams
• I = Imaxsin(8p/6)
• VR = VR,maxsin(8p/6)
t=8
2pft = 8p/6
8p/6
VR,maxsin(8p/6)
Length of vector = Vmax across that component
Vertical component = instantaneous value of V
Physics 102: Lecture 12, Slide 24
Phasor Diagrams
• I = Imaxsin(10p/6)
• VR = VR,maxsin(10p/6)
t = 10
2pft = 10p/6
10p/6
VR,maxsin(10p/6)
Length of vector = Vmax across that component
Vertical component = instantaneous value of V
Physics 102: Lecture 12, Slide 25
AC circuit summary
Kirchoff’s Loop Equation always holds true:
Vgen = VL + VR + VC
However, Vgen,max  VL,max+VR,max+VC,max
Maximum reached at different times for R, L, C
I
VR
t
VR in phase with I
L
R
VL VC lags I
VC
VL leads I
Phasors represent instantaneous voltages
Physics 102: Lecture 12, Slide 26
C
See you next lecture.
Physics 102: Lecture 12, Slide 27