Download Lecture 22: RC circuit, EM waves intro

Phy2005
Applied Physics II
Spring 2016
Announcements:
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Test 2 Wednesday, March 23
2 practice tests posted on course Tests page
Review session in class March 21 in class + March 21 6pm NPB 2205
PH travel next week APS Spring meeting; Prof. Lee subs
Science news page
Link to NYT article
American Meteor Society
Last time
Use rms values of v and i for AC to evaluate average
power.
<P> = vrms2/R
= irms2R
= irmsvrms
Vind = - n ℓ dF/dt
= - n ℓ A(mon)di/dt
= - n2moA ℓ (di/dt)
L: self inductance: Vind=-L di/dt
geometrical quantity
[L] = Henry
Calculate L for a solenoid:
r = 1 cm
ℓ = 2.5 cm
n = 10 turns/cm
= 0.01 m
= 0.025 m
= 1000 turns/m
L = n2moA ℓ
= (1000)2(4p x 10-7)(p x 0.012)0.025
= 9.9 x 10-6 H
= 9.9 mH
Ex 22-4 A steady current of 5.0 A is flowing through a coil with 2
H self inductance. The current is suddenly stopped in 0.01 s.
How large an EMF would be induced in the coil?
Vind = -L (di/dt)
= -(2)(-5/0.01)
= 1000 V
Connect and disconnect circuits with inductive
elements especially carefully!!!!
Level
Time
RC Circuit
-
Q=CV
+
+
+
+
+
+
V
I
Constant I-source
Q
V
Vc
Vc
V
V=0
t = 0:
Vc = 0
I0 = (V – Vc)/R = V/R
t = t1:
Vc = V1 (>0)
I1 = (V –V1)/R (< I0)
t = t2:
Vc = V2 (> V1 >0)
t
I2 = (V – V2)/R (< I1 < I2)
Vc
Vc = V (1 – e-t/RC)
V
e = 2.71828183
0.63V
RC: time constant
t
tc = RC
Vc = V (1 – e-1)
= 0.63 V
When a capacitor with a capacitance C is connected to a constant voltage
source Vo through a series resistor R,
(i) The potential difference across the capacitor increases not linearly but
exponentially.
(ii) The potential difference reaches 63% of the maximum value (0.63Vo)
at time tc = RC.
V = Vo e-t/RC
0.37
t
RC
R
C
++Q
-Q
Vc = Q/C
When a capacitor (C ) initially charged with Q is
discharged through a series resistor R,
(i) The amount of charge on the capacitor (voltage)
decreases exponentially.
(ii) 63% of the total charge (0.63Q) will be
discharged at time tc = RC.
Ex 23-1 A 100 mF capacitor is fully charged with 5 V DC source.
This capacitor is discharged through 10 KW resistor for 1 s.
How much of charge is left in the capacitor in C?
Total amount of charge: Q = C V = (100 x 10-6 F)(5 V)
= 5 x 10-4 C
Time constant: t = RC = (100 x 10-6 F)(10 x 103 Ohm)
=1s
Since Q is proportional to V, after one time constant
(100 – 63)% of initial charge is left: Q = 0.37 (5 x 10-4) C
ACADEMIC HONESTY
Each student is expected to hold himself/herself to a high standard
of academic honesty. Under the UF academic honesty policy.
Violations of this policy will be dealt with severely. There will be
no warnings or exceptions.
Q1 (24.11) An uncharged 2-mF capacitor is connected in series
with a switch, a 5x106W resistor, and a 12V battery. Find the
current in the circuit just after switch is closed, and the time constant.
1.
2.
3.
4.
(I)
2.0 x 10-6 A
6.0 x 106 A
2.0 x 10-6 A
4.0 A
(t)
10 s
3.2 s
2.7 s
1.41 s
Electromagnetic WAVES
y = sin(kx)
x
Wavelength: l = 2p/k
Traveling sin(kx) with a velocity v:
y = sin{k(x-vt)}
t
period: T
Frequency: f = 1/T
v = f l
v = f l
Any traveling wave satisfies this equation.
Sound Wave
Sound is a longitudinal mechanical wave that travels through an elastic medium.
The speed of sound in air at 27C is 347 m/s. The middle A note (A4) has
440 Hz frequency. We can use the above equation to calculate the wavelength
of the middle A note sound wave.
l = v/f = (347/440) m
From quantities present in his equations Maxwell was able to
calculate the speed with which electromagnetism moves
Turned out to be c, the speed of light (known from Romer
astronomical expts. in 17th century.)
Light was known to be a wave (Young experiment 1801)
“We can scarcely avoid the inference that light consists of the
transverse undulations of the same medium which is the cause of
magnetic and electric phenomena” -- Maxwell
Fiat Lux!
Velocity of the wave solutions to Maxwell equations
v=
1
 o mo
=
1
(8.85 10 12 C 2 / N  m2)( 4p 10 7 N / A2 )
Note 0,m0 come from laboratory measurements of electricity
and magnetism
c=2.9986 x 108 m/s