Announcements: RL  RV  RLC circuits Homework 06: due next Wednesday … Maxwell’s equations / AC current. Physics 1502: Lecture 23 Today’s Agenda. X X X X X X X X X. Induction. SelfInductance, RL Circuits. long solenoid. Energy and energy density. In series (like resistors). a.
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Announcements:
RL  RV  RLC circuits
Homework 06: due next Wednesday …
Maxwell’s equations / AC current
X X X X X X X X X
Induction
SelfInductance, RL Circuits
long solenoid
Energy and energy density
In series (like resistors)
a
a
a
a
L1
L2
Leq
Leq
L1
L2
b
b
b
b
And finally:
And:
J. C. Maxwell (~1860) summarized all of the work on electric and magnetic fields into four equations, all of which you now know.
However, he realized that the equations of electricity & magnetism as then known (and now known by you) have an inconsistency related to the conservation of charge!
Gauss’ Law
Faraday’s Law
Gauss’ Law
For Magnetism
Ampere’s Law
I don’t expect you to see that these equations are inconsistent with conservation of charge, but you should see a lack of symmetry here!
Gauss’ Law:
!
Remember:
Iin
Iout
changing electric
flux
FE
Can we understand why this “displacement current” has the form it does?
circuit
Big Idea: The Electric field between the plates changes in time.
“displacement current” ID = e0 (dfE/dt)= the real current I in the wire.
R
C
L
e
~
w
AC Current
S
S
S
N
N
N
End View
Side View
Water
Add resistance to circuit:
R
C
L
e
Answer:Yes, if we can supply energy at the rate the resistor dissipates it! How? A sinusoidally varying emf (AC generator) will sustain sinusoidal current oscillations!
R
+
+
C
L


We could solve this equation in the same manner we did for the LCR damped circuit. Rather than slog through the algebra, we will take a different approach which uses phasors.
R
C
L
e
~
Before introducing phasors, per se, begin by considering simple circuits with one element (R,C, or L) in addition to the driving emf.
R
i
R
e
~
m
m / R
Note: this is always, always, true… always.
0
0
m
m / R
t
0
t
0
x
Þ
Voltage across R in phase with current through R
Consider a simple AC circuit with a purely resistive load. For this circuit the voltage source is e = 10V sin (2p50(Hz)t) and R = 5 W. What is the average current in the circuit?
R
e
~
A) 10 A
B) 5 A
C) 2 A
D) √2 A
E) 0 A
Consider a simple AC circuit with a purely resistive load. For this circuit the voltage source is e = 10V sin (2p50(Hz)t) and R = 5 W. What is the average power in the circuit?
R
e
~
A) 0 W
B) 20 W
C) 10 W
D) 10 √2 W
Average values for I,V are not that helpful (they are zero).
Thus we introduce the idea of the Root of the Mean Squared.
In general,
So Average Power is,
Voltage across C lags current through C by onequarter cycle (90°).
C
e
~
Cm
m
Is this always true?
YES
0
0
m
Cm
0
t
t
0
x
Þ
Þ
A circuit consisting of capacitor C and voltage source e is constructed as shown. The graph shows the voltage presented to the capacitor as a function of time.
Which of the following graphs best represents the time dependence of the current i in the circuit?
e
t
i
i
i
(c)
(a)
(b)
t
t
t
Voltage across L leadscurrent through L by onequarter cycle (90°).
L
e
~
m
m L
0
0
Yes, yes, but how to remember?
m
m L
0
t
t
0
x
x
Þ
Þ
A phasor is a vector whose magnitude is the maximum value of a quantity (eg V or I) and which rotates counterclockwise in a 2d plane with angular velocity w. Recall uniform circular motion:
y
y
Þ
Þ
Þ
The projections of r (on the vertical y axis) execute sinusoidal oscillation.
w
x
Suppose:
i
w
i
0
wt
t
w
i
i
0
wt
w
i
i
wt
0
ß
A series LCR circuit driven by emf e = e0sinwt produces a current i=imsin(wtf). The phasor diagram for the current at t=0 is shown to the right.
At which of the following times is VC, the magnitude of the voltage across the capacitor, a maximum?
t=0
f
i
i
t=0
t=tc
t=tb
i
i
(c)
(a)
(b)
Here all unknowns, (im,f) , must be found from the loop eqn; the initial conditions have been taken care of by taking the emf to be: e = em sinwt.
R
C
L
e
~
From these equations, we can draw the phasor diagram to the right.
R
C
L
e
~
Þ
w
Þ
w
R
C
L
e
~
The unknowns (im,f) can now be solved for graphically since the vector sum of the voltages VL + VC + VR must sum to the driving emf e.
Ohms >
Þ
ß
y
x
imXL
em
f
imR
Z
 XLXC

imXC
R
“Full Phasor Diagram”
“ Impedance Triangle”
From this diagram, we can also create a triangle which allows us to calculate the impedance Z: