Physics 169
Kitt Peak National Observatory
Tuesday, March 8, 16
Luis anchordoqui
1
5.1 Ohm’s Law and Resistance
ELECTRIC CURRENT ☛
is defined as flow of electric charge through a cross-sectional area
dQ
i =
dt
Unit ☛ Ampere
[A = C/s]
Convention
① Direction of current is direction of flow of positive charge
② Current is NOT a vector ☛ but the current density is a vector
~j = charge flow per unit time per unit area
Z
~
~j · dA
i =
Tuesday, March 8, 16
2
Drift Velocity
i=
ˆ
j · dA
Consider a current i flowing through a cross-sectional area A
ty :
r a current i flowing through
sectional area A:
time
)t, total
charges passing
through
segment:passing through segment
In time
charges
t ☛ total
Q = q A(VdQt)=n qA(v
⇤ ⇥
⌅
Volume of charge
passing through
d
t)n
q ☛ charge of current carrier
q is charge n
of the
carrier,
n is density
of charge
carrier
☛current
density
of charge
carrier
per unit
volume
!
t volume
Q
= nqAvd
) Current Q i =
Current: i =
= nqAvd t
t
vd
Current Density ~j = nq~
Current Density:
Tuesday, March 8, 16
j = nqvd
3
Note
For metals ☛ charge carriers are free electrons inside
)
~j =
ne~vd for metals
) Inside metals ~j and ~vd are in opposite direction
We define a general property of materials ☛ conductivity (
Note
In general
~j =
)
~
E
is NOT a constant number
but rather a function of position and applied
~ -field
E
1
⇢
Resistivity ( ) is more commonly used property defined as ⇢ =
Unit of ⇢ : Ohm-meter (⌦m) where Ohm ( ⌦ ) = Volt/Ampere
OHM’S LAW
Ohmic materials have resistivity
that are independent of applied electric field
e.g. metals (in not too high
Tuesday, March 8, 16
~ -field)
E
4
Example
:
Example
Consider
a resistor (ohmic material)
Consider a resistor (ohmic material) of
L and
oflength
length
cross-sectional
area A
L and
cross-sectional
area A.
) Electric field inside conductor
Z b
Z
Electric field inside ~
conductor:
V b Va =
E · d~s = E
ˆ
a
V =
i
Current density j =
A
i
Current density:
j=
L
0
⇤ · d⇤s = E · L
E
V
dx = EL ) E =
L
V
E=
)
A
E
=
j
=
V
1
·
L i/A
L
E
⇢ =
j
V
1
⇢ =
·
L
i/A
V
L
= R = ⇢
i
A
R ☛ resistance of conductor
V
L
= R = of Ohm’s Law
Note V = iR is NOT a statement
i
A
but it’s just a definition for resistance
Tuesday,
MarchR
8, 16
5
where
is the resistance of the conductor.
ENERGY
ININ
CURRENT:
ENERGY
CURRENT
Assuming a charge Q
Assuming a ENERGY
chargeIN CURRENT:
Q enters
with Vpotential
V1 with
withenters
potential
leaves
Assuming
a charge
Q enters
1 and
potential
V1 and leaves
V2 with
and leaves
potential
otential
V2 : withwith
potential V2 :
) Potential energy lost in wire
U lost
=in theQV
Potential energy
wire:2
Potential energy lost in the Uwire:
=
U =
QV1
QV
QV
Q(V
V
)
2
U =
Q(V
V )1
2
1
U unit
=timetime
Q V2
Q V1
) Rate of energy
lost
Rate of energy
lostper
per unit
Q 2 V1 )
UU =
U
QQ(V
=
(V (V
V)
V1 )
2
t= t
t
t
Rate of energy lostJoule’s
per heating
unit time
Power dissipated
Power dissipated in conductor
Joule’s heating ☛ P = Pi =· i · VV= in=conductor
2
2
1
1
2
U V2 Q
V
2
=
(V2 V1 )
For a resistor
For a resistor
R R,R,P = iPR ==
i
R
=
R
t
t
R
2
Tuesday, March 8, 16
5.9
DC Circuits
6
2
a resistor R,
9
P =i R=
5.2 DC Circuits
V2
R
DC
Circuits
A battery is a device that supplies electrical energy
attery is a device that supplies electrical energy
maintain
a currentinina acircuit
to to
maintain
a current
uit.
moving from point 1 to 2, elecpotential energy increase by
= Q(V2 V1 ) = Work done by E
ne
E = Work done/charge = V2
V1
In moving from point 1 to 2 electric potential energy increase by
U =
Define
Q(V2
V1 ) = Work done by E
E = Work done/charge = V2
Tuesday, March 8, 16
V1
7
Example :
Example
o
Va = Vc
Va =
V
Vb =c Vd
Vb = Vd
By Definition
assuming(1) perfect conducting wires.
assuming perfect conducting wires
By Definition: Vc
Va
Vc (2)
Also, we have assumed
E
R
Vdresistance
= iRinside battery.
zero
E = iR
Resistance in combination :
Va
)
Vd = iR
Vb = E
⇥
Vb = E
E = iR
)
i=
E
i =
R
Potential
(P.D.)
Also ☛ we
havedifferece
assumed
zero resistance inside battery
Tuesday, March 8, 16
Va
Vb = (Va
Vc ) + (Vc
Vb )
8
E
Resistance in combination :
R
Resistance in combination
Also, we have assumed(2) zero resistance inside battery.
E = iR
⇥
i=
Resistance in combination :
Potential difference
Potential differece (P.D.)
Va
Vb = (Va
Va
=
Potential differece (P.D.)
Vc ) + (Vc
Vb = (Va Vc ) + (Vc
= iR1 + iR2
Vb )
Vb )
iR1 + iR2
Equivalent Resistance
) Equivalent Resistance
Va Vb
= (Va
Vc ) + (Vc
Vb )
R = R1 + R2
for resistors in series
1
1 = 1iR1 + iR2
=
+
for resistors in parallel
R
R
R
R
=
R
+
R
for resistors
2
Equivalent Resistance 1 1 2
R = R1 + R2
11
11 1
==+
RR R1 RR2
1
Tuesday, March 8, 16
in series
for resistors in series
1
parallel
+for resistors infor
resistors in parallel
R2
9
I 2R # (3.93 A)2 (3.00
# 46.3
W
Example
For !)
real
battery
mple :!R #
e power delivered to the internal resistance is
☛
Notice that 40% of the power from the battery is
to the internal resistance. In part (B), this per
there
is an internal resistance
1.6%. Consequently, even though the emf rema
weresistance
cannotsignificantly
ignore re
the increasing that
internal
ability of the battery to deliver energy.
) battery,
E there
= i(R
ForWreal
is an
!r # I 2r # (3.93 A)2 (0.05 !) # 0.772
+ r)
internal resistance that
we cannot ignore.
E
i =
R you
+ can
r vary the load resistance and internal r
At the Interactive Worked Example link at http://www.pse6.com,
observing the power delivered to each.
heatingthe
inLoad
resistor R
ampleJoule’
28.2 s Matching
P = i · (P.D. across resistor R )
= i2 R
E = i(R + r)
ow that the maximum power delivered
E to the load resisi
=
ce R in Figure 28.2a occurs whenR +
the
r load resistance
P =
tches the internal resistance—that is, when R # r.
Joule’s heating in resistor R :
to zero,
and solving for R. The details are left as
2
E
R
for you to solve (Problem 57).
(R + r)2
Question ☛ What is value of R to obtain maximum Joule’s heating?
lution The powerP delivered
toacross
the load
resistance
is
= i · (P.D.
resistor
R)
2
ual to I Answer
R, where I is☛
given
Equation
2by want
We
to28.3:
find R that
Question:
= iR
E 2 R 2R2
P =2
!dP
# I R (R
# + r)2E 2
=
%
maximizes P
E 2 2R
(R & r) 2
3
dR
(R
+
r)
(R
+
r)
What is the value of R to obtain maximum Joule’s heating?
hen ! is plotted versus R as in Figure 28.3, we find that
2/4r atP.
Answer:a We
want dP
to find
R to
ER2# r. When R is
reaches
maximum
value
of maximize
0 )Eso
[(R +
Setting
2 that the
2 3power
ge, there
is very little =
current,
I 2Rr)
dP
E
2R
dR
(R + r) R3 is small,
= small.
ivered to the load resistor
dR is
(R + r)2 When
(R + r)
)
r
R loss
= of0 power
e current is large and there is significant
as energy is delivered to the)
internal
When
RE 2=resistance.
r
dP
# r, these effects
balance=to0 give
Setting
⇥ a maximum
[(R +transfer
r) 2R]of
=0
3
Tuesday, March 8, 16 dR
(R
+
r)
wer.
%
!
P
!max
Pmax
2R] = 0
r
2r
3r
R
Figure 28.3 (Example 28.2) Graph of the power
10! d
by a battery to a load resistor of resistance R as a func
KIRCHOFF’S LAWS:
(1) First Law (Junction Rule):
Total
current
entering
a junction
equal to the total
current leaving the
ANALYSIS
OF COMPLEX
CIRCUITS
KIRCHOFF’S
LAWS:
(1)
First Law
(Junction
Rule):
junction.
Total current entering a junction equal to the total current leaving the
KIRCHOFF’S
(1)
First LAWS
Law (Junction Rule):
junction.
Total
current entering
a junction
① First Law
(Junction
Rule):
equal to the total current leaving the
junction.
Total current entering a junction equal to total current leaving junction
②
(2) Second Law (Loop Rule):
(2) The
Second
(Loop Rule):
sumLaw
of potential
differences around a complete circuit loop is zero.
The sum of potential differences around a complete circuit loop is zero.
Convention
:
Second
Law (Loop
Rule):
Convention :
Law (Loop
Rule):
The sum(2)
ofSecond
potential
differences
around a complete circuit loop is zero
(i)
The sum of potential differences around a complete circuit loop is zero.
Convention(i)
Convention :
(i)
(i) V
Va >
b
(ii)
(ii)
(ii)
)
Va > Vb Potential
⇥ Potential
difference = iR
difference
Va > Vb ⇥ Potential difference = iR
i.e.
Potential drops
drops across
resistors resistors
i.e. i.e.
Potential
across
Potential drops across resistors
=
Va > Vb
Vb (ii)
> Va
>)
V
⇥
iR
Potential difference =
i.e. Potential drops across resistors
Potential
⇥ Potentialdifference
difference = +E=
+E
Vb
a
Vb > Va ⇥ Potential difference = +E
i.e.
rises
across
plate
i.e.Potential
Potential rises
across the
negativenegative
plate of the battery.
i.e. Potential rises across the negative plate of the battery.
Tuesday, March 8, 16
Example :
iR
of battery
11
5.9. DC CIRCUITS
68
Example
By junction rule:
By loop rule:
(6.1)
i1 = i2 + i3
By loop rule:
Loop A
)
Loop B
)
Loop C
i = i2 + i3
By junction 1rule:
2E0
i1 R
(5.1)
i2 R + E 0
Loop A ⇥ 2E0 i1 R i2 R + E0 i1 R = 0
Loop B ⇥
i3 R E0 i3 R E0 + i2 R = 0
3Loop C ⇥ 02E0 i1 R3 i3 R E0 0i3 R i1 R2= 0
i R
E
i R
2E
i R
i R
BUT ☛ (6.4) = (6.2) +
(5.2)
(5.3)
(5.4)
E +i R = 0
BUT:
(5.4) = (5.2) + (5.3)
General rule:
0 Need only
1 3 equations
3 for 3 current
0
)
i1 R = 0
E
i1 = i2 + i3
3E0 2i1 R i2 R = 0
(6.3)2E0 + i2R 2i3R = 0
i3 R
(6.2)
(6.3)
i1 R = 0 (6.4)
(5.1)
(5.2)
(5.3)
Substitute (5.1) into (5.2) :
Tuesday, March 8, 16
3E0 2(i2 + i3 )R i2 R = 0
⇥ 3E0 3i2 R 2i3 R = 0
(5.4)
12
General rule
Need only 3 equations for 3 current
i1 = i2 + i3
3E0
2i1 R
2E0 + i2 R
(6.1)
i2 R = 0
(6.2)
2i3 R = 0
(6.3)
Substitute (6.1) into (6.2)
3E0
)
2(i2 + i3 )R
i2 R = 0
3E0
2i3 R = 0
3i2 R
(6.4)
Subtract (6.3) from (6.4), i.e. (6.4)-(6.3)
3E0
( 2E0 )
)
Tuesday, March 8, 16
3i2 R
i2 R = 0
5 E0
i2 =
·
4 R
13
Substitute i2 into (6.3)
2E0 +
)
Substitute
⇣5
E0 ⌘
R
·
4 R
i3 =
2i3 R = 0
3 E0
·
8 R
i2 , i3 into (6.1)
i1 =
⇣5
4
3 ⌘ E0
7 E0
=
·
8 R
8 R
Note
A negative current means that
it is flowing in opposite direction from one assumed
Tuesday, March 8, 16
14
one assumed.
5.3 RC5.10
Circuits
RC Circuits
(A) Charging
a capacitor
with
battery
(A) Charging
a capacitor with
battery:
Using loopUsing
rulethe loop rule:
+E0
+E0
P.D
iR
|{z}
iR
⇧⌅⇤⌃
Q
C
across R⇧⌅⇤⌃
P.D.
across R
Q
= 0
= 0C
|{z}
P.D. P.D across
C
Note ☛
across C
of i so
is chosen
that the current
represents the rate at
Direction ofNote:
chosen
thatso current
represents
i isDirection
which the charge on the capacitor is increasing.
rate at which charge on capacitor is increasing
i
z}|{
dQ Q
1st order
E =R
+
Q
differentialdQ
eqn.
dt C
)
+
dQ E dt= R
⇥
=
dt
C
EC Q
RC
Integrate both sides and use the initial condition:
dQ
dt
t = 0, Q on capacitor = 0)
=
ˆ
ˆ
EC
Q
RC
dQ
dt
=
i
⇤⌃⇧⌅
Q
Tuesday, March 8, 16
0
1st order
differential eq.
t
EC
Q
0
RC
15
Integrate both sides and use initial condition
Z
Q
0
t = 0,
Z t
dQ
dt
=
EC
Q
0 RC
ln(EC
)
)
)
)
)
Tuesday, March 8, 16
Q)
Q
0
t
=
RC
Q on capacitor = 0
t
0
t
ln(EC
Q) + ln(EC) =
RC
⇣
⌘
1
t
ln
=
Q
RC
1
EC
1
1
Q
EC
= et/RC
Q
= 1
e
EC
Q(t) = EC (1
t/RC
e
t/RC
)
16
⇤
Note ☛
Q(t) = EC(1
e
t/RC
)
= At0,t Q(t
==
0)0)=
EC (11) = 01)
①Note:
At t (1)
= 0 , Q(t
= EC(1
= 0
② As (2)
t !
1, Q(t ! 1) = EC (1
0)
As t ⇥ ⌅ , Q(t ⇥ ⌅) = EC(1 0) = EC
③ Current
= EC
= Final charge on capacitor (Q0 )
dQ
i =
dt i⇣ = dQ
1 dt⌘⇥ t/RC
= EC = EC e 1 ⇤e t/RC
RC RC
E
E t/RC
i(t) = i(t)
e t/RC
=
e
R
R
⌅
n ⌃ i(t = 0) E= E = Initial current = i
i(t = 0)
=
Initial current 0= i0
R
⇧
i(t ⇥ ⌅) R
= 0
i(t ! 1) = 0
(3) Current:
Tuesday, March 8, 16
(4) At time = 0, the capacitor acts like short circuit when there is
17
From the definition of the natural logarithm, we can write this expression as
%
(28.14)
q(t ) ! C (1 " e "t/RC ) ! Q(1 " e "t/RC )
Charge as a function
a capacitor being ch
④ At time = 0 ☛ capacitor acts like short circuit
where e is the base of the natural logarithm and we have made the substitution from
when
there
zero ischarge on capacitor
t time
= 0,
the capacitor acts like short
circuit
whenis there
Equation
28.13.
We can
expression for the charging current by differentiating Equation
ro charge
onfind
theancapacitor.
28.14
with
respect!
time.
I ! dq/dt, weisfind
that charged and current = 0
1Using
⑤ As
time
☛ capacitor
fully
s time
⇥
⌅,
thetocapacitor
is
fully charged
and current = 0, it
ts like a open circuit.
it acts like a open
circuit
%
Current as a functio
I(t ) !
R
(28.15)
e "t/RC
a capacitor being ch
Plots of capacitor charge and circuit current versus time are shown in Figure 28.20.
Note that the charge is zero at t ! 0 and approaches the maximum value C as t : #.
The current has its maximum value I 0 ! /R at t ! 0 and decays exponentially to zero
as t : #. The quantity RC, which appears in the exponents of Equations 28.14 and
of theconstant
circuit. It represents the time interval
28.15,
called $time
⑥ is ccalled the timeisconstant
during which the current decreases to 1/e of its initial value; that is, in a time interval
$, I ! e"1I 0 ! 0.368I 0. In a time interval 2$, I ! e" 2I 0 ! 0.135I 0, and so forth.
It is in
time
takes$, the
forcharge
charge
to from
reach
0 .
Likewise,
a timeitinterval
increases
zero to C [1 " e"1] ! 0.632C
%
%
⌧ = RC
q
I
Cε
τ =RC
τ
(a)
1⌘
Q '
% 0.63 Q0
e
I0 = ε
R
I0
0.632C ε
Tuesday, March 8, 16
⇣
%1
0.368I0
t
τ
(b)
t
Figure 28.20 (a) Plot of capacitor charge versus time for the circuit shown in Figure
18
(6)
= RC is called the time constant. It’s the time it takes for
the charge to reach (1 1e ) Q0 ⇥ 0.63Q0
c
(B) Discharging
a charged capacitor
(B) Discharging a charged capacitor:
Note ☛ Note: Direction of i is chosen so that the current represents the rate at
Direction of which
chosen
sothethat
charge on
capacitor is decreasing.
i is the
current represents rate at whichdQcharge on capacitor is decreasing
i=
)
Loop Rule:
Loop Rule
Vc
dt
dQ
i V= iR = 0
Q dQ dt
c
⇤
+
R=0
C
dt
dQ
1
⇤
=
Q
dt
RC
iR = 0
Integrate both sides and use the initial condition:
t = 0, Q on capacitor = Q0
ˆ Q
ˆ t
dQ
1
=
dt
RC 0
Q0 Q
t
⇤ ln Q ln Q0 =
RC
Q⇥
t
⇤ ln
=
Q0
RC
Tuesday, March 8, 16
Q
t/RC
)
)
Q
dQ
+
R = 0
C
dt
dQ
1
=
Q
dt
RC
19
Integrate both sides and use initial condition
Z
Q
Q0
dQ
=
Q
1
RC
Z
0
)
)
i =
(
At
Tuesday, March 8, 16
dQ ⌘
dt
Q on capacitor = Q0
dt
)
⇣
t0 ,
t
)
)
ln Q
ln Q0 =
⇣Q⌘
t
ln
=
Q0
RC
Q
= e t/RC
Q0
Q(t) = Q0 e
Q0
i(t) =
e
RC
t
RC
t/RC
t/RC
1
Q0
t = 0) ) i(t = 0) =
·
R |{z}
C
V0
Initial P.D. across capacitor
i0 =
R
20
At
t = RC = ⌧
Tuesday, March 8, 16
1
Q(t = RC) = Q0 ' 0.37 Q0
e
21
5.4 Semiconductors
A semiconductor is a substance that can conduct electricity
under some conditions but not others
making it a good medium for control of electrical current
Specific properties of a semiconductor depend on impurities added to it (or dopants)
N-type semiconductor carries current in form of negatively-charged
in a manner similar to conduction of current in a wire
P-type semiconductor carries current predominantly as
e
e deficiencies
called holes
A hole has a positive electric charge equal and opposite to charge on
e
In a semiconductor material flow of holes occurs
in a direction opposite to flow of electrons
Tuesday, March 8, 16
22
Acronym LED stands for Light Emitting Diode
A diode is a combination of two semiconductors
one of which is doped n-type and other p-type
(region between n-type and p-type material)
When electrons and holes meet in junction
they can recombine to liberate energy
Tuesday, March 8, 16
23
5.5 Intuition
Each of the 12 edges of a cube contain a 1 Ω resistor
Calculate the equivalent resistance between two opposing corners
Tuesday, March 8, 16
24
Here is where the intuition comes into play
Color coding is used to help keep track of the resistors and associated nodes
Due to symmetry P.D. at the three nodes labeled "α" are equal
nocomes
current
flowsColor
between
a potential
difference
of associated
zero V, nodes
e theSince
intuition
into play.
coding isnodes
used towith
help keep
track of the
resistors and
e to symmetry,
at the
three nodes
labeled "the
α" arecircuit's
equal. Since
no current flows between
they canthe
bepotential
shorted(voltage)
together
without
affecting
integrity
potential difference of 0 V, they can be shorted together without affecting the circuit's integrity. The same can
The same can be done for the nodes labeled "β"
the nodes labeled "β."
A
pe
E
S
☛
ort those
obtain
the equivalent
below.
you can see,
there are two
sets of three
Oncenodes,
you you
short
those
nodes circuit
youshown
obtain
the As
following
equivalent
circuit
parallel, in series with one set of six resistors in parallel. So, you have 1/3 Ω in series with 1/6 Ω in series with
h equals
5/6March
Ω. 8, 16
Tuesday,
25
There are two sets of three resistors in parallel
u short those nodes, you obtain the equivalent circuit shown below. As you can see, there are two sets of three
in series with one set of six resistors in parallel
in parallel, in series with one set of six resistors in parallel. So, you have 1/3 Ω in series with 1/6 Ω in series with
which equals 5/6 Ω.
So ☛ you have 1/3 Ω in series with 1/6 Ω in series with 1/3 Ω
RF8, 16
Cafe
Tuesday, March
which equals 5/6 Ω
Method of Solving the Resistor Cube Problem
26