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Physics 212
Lecture 24
Today’s Concept:
AC Circuits
Maximum currents & voltages
Phasors: A Simple Tool
Electricity & Magnetism Lecture 24, Slide 1
Your Comments
“If you do better on the final than one of your midterms the final mark replaces the
midterm mark, does this still count if you happen to miss the midterm?”
“next Wednesday, instead of doing a pre-lecture which will not be on the midterm,
could we please have a full review class? :D”
“I don't see how the length of the phasor has to do with the reactance... don't they
represent voltage? I looked at the voltage equation in previous slides and dont see
reactance in the equation anywhere. Absolutely no idea what they were talking about
with phasor diagrams...”
“why when where the hell we need these stuff outta school, I am actually really totally
super EXCITED about this”
“How much of this material is going to be present on the midterm? Given that there is
already a boatload of other material we need to know, I'm beginning to feel
overwhelmed by the consistent stream of new material being taught before the
midterm.”
Electricity & Magnetism Lecture 24, Slide 2
AC Circuits
Up until now, we have been using DC voltage sources, that is, batteries that supply a
constant voltage to our circuit.
AC voltage sources are used in most real life applications due to being more natural
to generate and ease of use in transformers. This occurs because induction relies on
the current changing with time, and a varying voltage source gives rise to that.
A typical AC voltage source is one that varies as a sine function over time:
𝑉 𝑡 = 𝑉𝑚𝑎𝑥 sin(𝜔𝑡)
𝜔 is the angular frequency.
The angular frequency can be related to the regular frequency or the period:
𝜔 = 2𝜋𝑓 = 2𝜋/𝑇
Lets look at how our usual circuit elements act when supplied by an AC source.
Electricity & Magnetism Lecture 24, Slide 3
Resistors in AC
𝑉 𝑡 = 𝑉𝑅 = 𝐼𝑅
𝜀 = 𝑉𝑚𝑎𝑥 sin(𝜔𝑡)
R
𝑉𝑅 𝑉𝑚
𝐼=
=
sin(𝜔𝑡)
𝑅
𝑅
𝐼
𝐼𝑚𝑎𝑥 = Vm/R
Electricity & Magnetism Lecture 24, Slide 4
Capacitors in AC
𝑉 𝑡 = 𝑉𝐶 = 𝑄/𝐶
𝑄 = 𝐶𝑉𝑚 sin(𝜔𝑡)
𝜀 = 𝑉𝑚𝑎𝑥 sin(𝜔𝑡)
𝑑𝑄
𝐼=
𝑑𝑡
C
𝐼 = 𝐶𝜔𝑉𝑚 cos(𝜔𝑡)
𝐼
𝐼𝑚𝑎𝑥 = Vm/XC
90o
where XC = 1/wC
is like the “resistance”
of the capacitor
XC depends on w
𝐼=
𝑉𝑚
𝜋
sin 𝜔𝑡 +
𝑋𝐶
2
Electricity & Magnetism Lecture 24, Slide 5
Inductors in AC
𝑉 𝑡 = 𝑉𝐿 = 𝐿
𝜀 = 𝑉𝑚𝑎𝑥 sin(𝜔𝑡)
𝑑𝐼 𝑉𝑚
=
sin(𝜔𝑡)
𝑑𝑡
𝐿
L
𝐼
𝑑𝐼
𝑑𝑡
𝐼=−
𝑉𝑚
cos(𝜔𝑡)
𝜔𝐿
𝐼𝑚𝑎𝑥 = Vm/XL
90o
where XL = wL
is like the “resistance”
of the inductor
XL depends on w
𝐼=
𝑉𝑚
𝜋
sin 𝜔𝑡 −
𝑋𝐿
2
Electricity & Magnetism Lecture 24, Slide 6
What about all three together?
In principle we can solve this using what we already know:
𝑄
𝑑2 𝑄
𝑑𝑄
𝜀 = +𝐿 2 +𝑅
𝐶
𝑑𝑡
𝑑𝑡
𝜀 = 𝑉𝑚𝑎𝑥 sin(𝜔𝑡)
We could solve this as a differential equation.
𝜀
C
L
R
But it is easier to solve it using some things
we already know about the solution and a
technique known as phasors.
Electricity & Magnetism Lecture 24, Slide 7
What about all three together?
We know there is only one
current, so getting everything in
terms of that would be good.
VCm = Im XC
We know that the voltage
across each individual piece
must vary with the same
frequency as the battery.
V 90o behind I
𝜀
C
L
R
VRm = Im R
V in phase with I
VLm = Im XL
V 90o ahead of I
The voltage across the inductor
will lead the current by 90o,
while the voltage across the
capacitor will lag behind the
current by 90o.
We also know that the sum of these
voltages (accounting for phase) must
add to the voltage from the battery.
We can imagine the current has the
form 𝐼 𝑡 = 𝐼𝑚 sin 𝜔𝑡 − 𝜙
Electricity & Magnetism Lecture 24, Slide 8
What about all three together?
𝐼 𝑡 = 𝐼𝑚𝑎𝑥 sin(𝜔𝑡 − 𝜙)
At some time, the current is at
its maximum, and the voltage
drops across the inductor and
capacitor are zero.
VCm = Im XC
V 90o behind I
𝜀
C
L
R
VRm = Im R
V in phase with I
VLm = Im XL
V 90o ahead of I
At some other time, the voltage
drop across the capacitor is
maximized, and the voltage
drop across the inductor is
maximally negative at that time.
We can visualize the voltage across
one component (the function
sin(𝜔𝑡 − 𝜙)) as a point moving up
and down along an axis.
Or instead visualize it as the y-projection
of a point moving in a circle.
Electricity & Magnetism Lecture 24, Slide 9
Phasors
Phasors make this
simple to see
𝐼 𝑡 = 𝐼𝑚𝑎𝑥 sin(𝜔𝑡 − 𝜙)
Im XL
Vm = Im XC
V 90o behind I
𝜀
Im R
C
L
R
Vm = Im XL
V 90o ahead of I
Im XC
Vm = Im R
V in phase with I
Each vector has a length corresponding
to that particular components
maximum amplitude
Electricity & Magnetism Lecture 24, Slide 10
Phasors
If we imagine the entire configuration rotating
counterclockwise about the middle, moving at
ω radians/sec, the y-projection is the actual real
value of the potential for that component. The
x-component is an imaginary piece we have
added on to make this visualization possible.
VL=Im XL
VR=Im R
VC=Im XC
This means that the battery 𝜀
on this picture is just the
vector sum of the other three.
The battery voltage should also be in this
picture somewhere, and it must be the case
that the y-components 𝑉𝐿𝑦 , 𝑉𝐶𝑦 , 𝑉𝑅𝑦 add to the
battery 𝜀.
Since it must also be true after rotating by 90
degrees that the y-components still add up, the xcomponents 𝑉𝐿𝑥 , 𝑉𝐶𝑥 , 𝑉𝑅𝑥 must also add to the
battery 𝜀.
Electricity & Magnetism Lecture 24, Slide 11
The Voltages still Add Up
Im XC
So now we are essentially adding vectors:
C
em
L
Im XL
R
Im XL
Im XL
Im R
em
Im R
Im XC
Im R
Im R
Im XC
Im XC
Im XL
em
Electricity & Magnetism Lecture 24, Slide 12
Make this Simpler
Im XC
C
em
L
Im XL
R
Im XL
Im XL
Im R
em
Im R
Im XC
Im R
Im XC
Electricity & Magnetism Lecture 24, Slide 13
Make this Simpler
Im XC
C
em
L
Im XL
R
Im XL
Im R
em = Im Z
Im R
Im(XL - XC)
Im R
Z = impedance
Im XC
Electricity & Magnetism Lecture 24, Slide 14
Make this Simpler
Im XC
C
em = Im Z
f
em
L
Im(XL - XC)
Im XL
R
Im R
Im R
Z = R 2  ( X L - X C )2
R
Impedance Triangle
X L - XC
tan (f ) =
R
Electricity & Magnetism Lecture 24, Slide 15
Summary
VCm = Im XC
VLm = Im XL
𝑋𝐿 = 𝜔𝐿
VRm = Im R
em
Im XC
1
𝑋𝐶 =
𝜔𝐶
C
em
L
Im XL
R
= Im Z
Im R
Im = em / Z
Z = R  X L - X C 
2
X L - XC
tan (f ) =
R
2
Z = R2  ( X L - X C )2
f
R
Electricity & Magnetism Lecture 24, Slide 16
Summary
Im XC
𝜀 𝑡 = 𝑉𝑚 sin(𝜔𝑡)
𝐼 𝑡 = 𝐼𝑚 sin(𝜔𝑡 − 𝜙)
X L - XC
tan (f ) =
R
Z = R2  ( X L - X C )2
f
R
Since 𝑋𝐶 =
1
and 𝑋𝐿 = 𝜔𝐿
𝜔𝐶
Which one is “stronger” can depend on frequency.
C
em
L
Im XL
R
Im R
If 𝑋𝐿 > 𝑋𝐶 then the inductor is
“stronger” and the current lags
behind the voltage (𝜙 > 0)
If 𝑋𝐶 > 𝑋𝐿 then the capacitor is
“stronger” and the current
leads the voltage (𝜙 < 0)
Electricity & Magnetism Lecture 24, Slide 17
Resonance
Im XC
There is an optimal frequency for
the system to oscillate at.
C
Given 𝜀𝑚 , R, L, and C fixed, what
frequency should we choose if we want
to maximize our output current?
Im = em / Z
1
𝑋𝐶 =
𝜔𝐶
Z = R  X L - X C 
2
em
2
𝑋𝐿 = 𝜔𝐿
We should choose the frequency that sets 𝑋𝐿 = 𝑋𝐶
L
Im XL
R
Im R
𝑋𝐿 = 𝑋𝐶
1
= 𝜔𝐿
𝜔𝐶
𝜔0 =
1
𝐿𝐶
Electricity & Magnetism Lecture 24, Slide 18
Resonance
Resonance frequency
𝜔0 =
Im XC
1
𝐿𝐶
C
em
L
Im XL
R
At this frequency 𝑍 = 𝑅 and the circuit behaves as
though the capacitor and inductor are transparent.
Im R
The response of the system is at its maximum at the resonance frequency.
This is the same frequency we saw with oscillating LC circuits, it is the natural
frequency of the system in the absence of a battery.
Electricity & Magnetism Lecture 24, Slide 19
Calculation
Consider the harmonically driven series LCR circuit shown.
Vmax = 100 V
Imax = 2 mA
VCmax = 113 V
The current leads generator voltage by 45o
L and R are unknown.
C
V ~
L
R
What is XL, the reactance of the inductor, at this frequency?
Electricity & Magnetism Lecture 24, Slide 20
Calculation
Consider the harmonically driven series LCR circuit shown.
Vmax = 100 V
Imax = 2 mA
VCmax = 113 V
The current leads generator voltage by 45o
L and R are unknown.
C
V ~
L
R
What is XL, the reactance of the inductor, at this frequency?
Compare XL and XC at this frequency:
A) XL < XC
B) XL = XC
C) XL > XC
VL
D) Not enough information
This information is determined from the phase
Current leads voltage
VL = ImaxXL
VC = ImaxXC
IR
45o
VR (phase of current)
VC
V
V
leads
Electricity & Magnetism Lecture 24, Slide 21
Calculation
Consider the harmonically driven series LCR circuit shown.
Vmax = 100 V
Imax = 2 mA
VCmax = 113 V
The current leads generator voltage by 45o
L and R are unknown.
C
V ~
L
R
What is XL, the reactance of the inductor, at this frequency?
What is Z, the total impedance of the circuit?
A) 70.7 kW
B)
50 kW
C) 35.4 kW
D) 21.1 kW
Vmax 100V
Z=
=
= 50k W
I max 2mA
Electricity & Magnetism Lecture 24, Slide 22
Calculation
Consider the harmonically driven series LCR circuit shown.
Vmax = 100 V
Imax = 2 mA
VCmax = 113 V
The current leads generator voltage by 45o
L and R are unknown.
C
V ~
R
What is XL, the reactance of the inductor, at this frequency?
Z = 50kW
sin(45) =.707
What is R?
A)
70.7 kW
L
B) 50 kW
C)
35.4 kW
D)
21.1 kW
cos(45) =.707
Determined from impedance triangle
R
45o
(XC - XL)
R
cos(45) =
Z
R = Z cos(45o)
= 50 kW x 0.707
= 35.4 kW
Electricity & Magnetism Lecture 24, Slide 23
Calculation
Consider the harmonically driven series LCR circuit shown.
Vmax = 100 V
Imax = 2 mA
VCmax = 113 V
The current leads generator voltage by 45o
L and R are unknown.
C
V ~
R
Z = 50kW
What is XL, the reactance of the inductor, at this frequency?
A) 70.7 kW
We start with the
impedance triangle:
R
45o
Z
B) 50 kW
C) 35.4 kW
XC - X L
= tan 45 = 1
R
L
D)
21.1 kW
R = 35.4kW
XL = XC - R
What is XC?
VCmax = ImaxXC
(XC - XL)
XL = 56.5 kW - 35.4 kW
113
XC =
= 56.5k W
2
Electricity & Magnetism Lecture 24, Slide 24
Review: CheckPoint 1.1
Draw Voltage Phasors
Imax XL
emax
Imax R
A
B
C
Electricity & Magnetism Lecture 24, Slide 25
Review: CheckPoint 1.2
Draw Voltage Phasors
Imax XL
emax
Imax R
A
B
C
Electricity & Magnetism Lecture 24, Slide 26
Review: CheckPoint 1.3
The CURRENT is THE CURRENT
Imax XL
f
emax
Imax R
f is the phase between the
generator and the current
A
B
C
D
Electricity & Magnetism Lecture 24, Slide 27
Review: CheckPoint 2.1
50
40
30
20
10
A
0B
C
IXL
e
IR
e
IR
IXL
What does the voltage phasor diagram look like
when the current is a maximum?
IXc
IXc
Electricity & Magnetism Lecture 24, Slide 28
IXL
IXc
Review: CheckPoint 2.2
e
IR
IR
50
40
e
IXc
IXL
30
20
10
A
B
C0
What does the voltage phasor diagram look like when the
capacitor is fully charged?
Electricity & Magnetism Lecture 24, Slide 29
IXL
IXc
CheckPoint 2.3
e
IR
IR
50
40
e
IXc
IXL
30
20
10 A
0
B
C
What does the voltage phasor diagram look like when the voltage
across the capacitor is at its positive maximum?
Electricity & Magnetism Lecture 24, Slide 30
Midterm Information
1) Midterm 2: Friday, July 17th in class (9:30-10:20am)

MATERIAL: Units 9 – 20 (i.e. up to and including today)
• Unit 18-20 material: multiple choice type questions only.

Format similar to the first midterm: multiple choice and written
questions.

Sample midterm will be posted later today.

As before, bring a “regulation” calculator and know what tutorial
section you belong to.
Electricity & Magnetism Lecture 24, Slide 31
Final Thoughts
 Written assignment due today.
 No prelecture for next class.
 Next class will be a review of midterm material (I have moved
the schedule around a bit).
Electricity & Magnetism Lecture 24, Slide 32