Chapter 27: Current & Resistance
Current: Dead or Alive
DEATH:
• NEUROLOGIC CRITERIA: An individual with
irreversible cessation of all brain function,
including the brain stem, is dead.
• CARDIOPULMONARY CRITERIA: An individual
with irreversible cessation of circulatory and
respiratory function is dead.
Positive Charges move from HI to LOW potential.
HI V
LOW V
Negative Charges move from LOW to HI potential.
HI V
LOW V
HOW FAST DO ELECTRONS MOVE IN A CURRENT
CARRYING CONDUCTING WIRE??
Electron Speed is called the DRIFT Velocity.
v = p/m=Ft/m = qEt/m
qE
vd 

m
Drift velocity ~ .001 m/s !!!
Electric Fields travel at the speed of light!
Current
dQ
I
dt
I = Coulomb/second = Ampere
• Current flows from a higher potential to a
lower potential (electrons flow the opposite
way). Current carrying wires are neutral!
• DC current flows in one direction
• AC current oscillates back and forth
• Electrons have a drift velocity of .001m/s !
• Electric Fields travel at speed of light
Engine Current Problem
The starter motor of a car engine draws a current of
150 A from the battery. The copper wire to the
motor is 5.0 mm in diameter and 1.2 m long. The
starter motor runs for 0.80 s until the car engine
starts.
a. How much charge passes through the starter
motor?
(a) Current is defined as I = Q/t, so the
charge delivered in time t is
Q = It = (150 A)(0.80 s) = 120 C
Current is charge in motion
• Charge, e.g. electrons, exists in conductors with a
number density, ne (ne approx 1029 m-3)
v = p/m=Ft/m = qEt/m
qE
vd 

m
• Current density, J, is given by J = qenevd = qnv
J
I
 qnv
A
– unit of J is C/m2sec or A/m2 (A ≡ Ampere) and 1A ≡ 1C/s
– current, I, is J times cross sectional area, I = J pr2
– for 10 Amp in 1mm x 1mm area, J = 10+7 A/m2, and ve is about 10-3
m/s (Yes, the average velocity is only 1mm/s!)
Engine Current Problem Again
The starter motor of a car engine draws a current of 150 A
from the battery. The copper wire to the motor is 5.0 mm
in diameter and 1.2 m long. The starter motor runs for
0.80 s until the car engine starts.
a. How much charge passes through the starter motor?
b. What is the drift speed of the electrons? Show how the
units work out!

electron number density: 8.5 10 m
vd 
28
3

J
I
 qnv
A
J
I A
I
150 A
4

 2 

5.617

10
m/s
2
28

3

19
nq
ne p r ne p  0.0025 m  8.5 10 m
1.60 10 C



Fuses
If the current drawn exceeds safe levels, the fuse melts and the
circuit ‘breaks” – most house have switches, not fuses.
Fuse Problem
You need to design a 1.0 A fuse that ‘blows’ if the current
exceeds 1.0 A. The fuse material in your stockroom melts
at a current density of 500 A/cm2. What diameter wire of
this material will do the job?
JI A
A
p D2
4

4 1.0 A 
I
4I
D

 0.050 cm  0.50 mm
2
J
pJ
p 500 A/cm


Atomic Model
• E-field in conductor provided by a battery
• Charges are put in motion, but scatter in a very short
time from things that get in the way
– it’s crowded inside that metal
– defects, lattice vibrations (phonons), etc
• Typical scattering time  = 10-14 sec
• Charges ballistically accelerated for this time and then
randomly scattered
• Average velocity attained in this time is v = Ft/m = qE/m
• Current density is J = qnv so current is proportional to E
which is proportional to Voltage
• OHM’s LAW J = s E, s = conductivity
qE
vd 

m
I
J   s E  qnv
A
q 2 n
s
m
Resistivity
J sE
Property of bulk matter related to
resistance of a sample is the
resistivity (r) defined as:
E
r
J
r
E
1
s
j
A
where E = electric field and
J = current density in conductor.
For uniform case:
J 
I
,
A
L
V  EL
 V  EL  r JL  r I L  I  ρL 
A
q 2 n
s
m
 A 
L
V  IR where R  r
A
So, in fact, we can compute the resistance if we know a bit about the
material, and YES, the property belongs to the material!
e.g., for a copper wire, r ~ 10-8 W-m, 1mm radius, 1 m long, then R  .01W;
for glass, r ~ 10+12 W-m; for semiconductors r ~ 1 W-m
Resistance: Resistivity
L
Rr
A
• The LONGER the wire the GREATER the R
• The THINNER the wire the GREATER the R
• The HOTTER the wire the GREATER the R
Resistance: Dependence on
Temperature
The HOTTER the wire the GREATER the R
R  R0 (1  T )
R0  original resistance
  temperature coefficient of resistivity
T  temperature change (<100 C)
When are light bulbs more likely
to blow?
When hot or cold?
The HOTTER the wire the GREATER the R!
R  R0 (1  T )
At lower Resistance, the bulb draws more
current and it blows the filament!
Resistivity Values
Engine Current Problem
The starter motor of a car engine draws a current of
150 A from the battery. The copper wire to the
motor is 5.0 mm in diameter and 1.2 m long. The
starter motor runs for 0.80 s until the car engine
starts.
c. What is the resistance in the copper wire?
l
Rr r
A
l
1.2m
8
3

4

1.7
x
10
W
m

1.04
x
10
W
2
2
D
p (.005m)
p
4
Problem
I
J   s E  qnv
A
E
qnv
s
 r qnv
q 2 n
s
m
If the magnitude of the drift velocity of free electrons in a
copper wire is 7.84 × 10^ –4 m/s, what is the electric field
in the conductor? The number density for copper is
8.49 × 10^28 electrons/m3. What is the collision time?
E  0.181 V m
A rod is made of two materials. The figure is not drawn to scale. Each
conductor has a square cross section 3.00 mm on a side. The first
material has a resistivity of 4.00 × 10–3 Ω · m and is 25.0 cm long, while
the second material has a resistivity of 6.00 × 10–3 Ω · m and is 40.0 cm
long. What is the resistance between the ends of the rod?
R
r1
1
A1

r2
2
A2
4.00  10

R
3

W m
r1 1  r2
2
d2
  0.250 m   6.00  10
 3.00  10 m 
3
3
2
W m
  0.400 m  
378 W
Radial Resistance of a Coaxial
Cable: Leakage!
• Assume the silicon
between the conductors to
be concentric elements of
thickness dr
• The resistance of the
hollow cylinder of silicon
is
L
Rr
A
ρ
dR 
dr
2πrL
• The total radial resistance
is
ρ
b
R   dR 
ln  
a
2πL  a 
b
This is fairly high, which is desirable since you want the current to
flow along the cable and not radially out of it
Resistors
Ohms Law: J  s E
V = IR
Resistance QUESTION
How much current will flow
through a lamp that has a
resistance of 60 Ohms when
12 Volts are impressed
across it?
USE OHMS LAW: V = IR
V 12V
12V
I


 .2 A
R
60W 60V / A
What makes
the wires
Glow?
Electrical Power
•
•
•
•
As a charge moves from a to b, the electric
potential energy of the system increases by QV
The chemical energy in the battery must decrease
by this same amount
As the charge moves through the resistor (c to d),
the system loses this electric potential energy
during collisions of the electrons with the atoms
of the resistor
This energy is transformed into internal energy in
the resistor as increased vibrational motion of the
atoms in the resistor
The resistor emits thermal radiation which can
make it glow.
U  qV
dU dqV
dq
P

 V
 I V
dt
dt
dt
POWER P  I V
Energy J
  Watt
 P 
time
s
P  I V  I ( IR)  I R
2
V
 V 
P  I V  
V 
R
 R 
2
• You pay for ENERGY not for ELECTRONS!
• Kilowatt-hour is the energy consumed in one
hour: [kWh]=J NOT TIME! Power x Time
If V = 120V, What is I?
USE P = IV=> I = P/V
Appliance
_
Power
Hair Dryer
1600 Watts
Electric Iron
1200 Watts
TV
100 Watts
Computer
45 Watts
Current (A)
If V = 120V, What is I?
USE P = IV=> I = P/V
Appliance
_
Power
Current (A)
Hair Dryer
1600 Watts
13.3 A
Electric Iron
1200 Watts
10 A
TV
100 Watts
.83 A
Computer
45 Watts
.38 A
Electric Bill:
Cost to run for 1 hr @
$.05 per 1 kw-hr ?
Cost = Power x Time x Rate
Appliance
_
Power
Hair Dryer
1600 Watts
Electric Iron
1200 Watts
TV
100 Watts
Computer
45 Watts
Cost______
Electric Bill:
Cost to run for 1 hr @
$.05 per 1 kw-hr ?
Cost = Power x Time x Rate
Appliance
_
Power
Cost______
Hair Dryer
1600 Watts
$0.08
Electric Iron
1200 Watts
$0.06
TV
100 Watts
$0.005
Computer
45 Watts
$0.003
QUESTION
The voltage and power on a light bulb read
“120 V, 60 W” How much current will flow
through the bulb?
USE: P = I V
I = P/V = 60 W/120 V = 1/2 Amp
QUESTION
The power and voltage on a light bulb read
“120 V, 60 W” What is the resistance of the
filament? (I = .5 A)
Hint: USE OHMS LAW: V = IR
R = V/I = 120 V/ .5 A = 240 W
Parallel Circuits
• The voltage of each device is the
full voltage of the EMF source
(the battery)
• The total current is divided
between each path:
V V
1 1
V
I  I1  I 2  
V(  ) 
R1 R2
R1 R2
RP
Equivalent Resistance:
1
1 1
1
    ....
RP R1 R2 R3
Series Circuits
• The current is the same in each device.
• The equivalent resistance of the circuit is the sum
of the individual resistances. R=R1+R2
Vtotal  IRtotal
 I ( R1 +R2 )  IR1 +IR2  V1 +V2 ,
V  V1 +V2 ,
RS  R1  R2  R3  ...
Series Circuits
• The current is the same in each device.
• The equivalent resistance of the circuit is the sum
of the individual resistances. R=R1+R2
V  V1 +V2 ,
RS  R1  R2  R3  ...
To find the current, use the total voltage and equivalent resistance:
V
I
RS
Circuits Problem:Bulbs in Series vs Parallel
A circuit contains a 48-V battery and two 240W light bulbs.
In which circuit does each bulb burn brighter?
RULE: THE MORE POWER DISSIPATED IN A BULB, THE
BRIGHTER IT IS.
P  IV
Find the power in each bulb when in series and in parallel.
Circuits Problem:Bulbs in Series vs Parallel
In Series:
The Voltage is divided in series: each bulb gets half: V = 24V.
V
V 2 (24V )2
P  IV  V 

 2.4W
R
R
240W
In Parallel:
Voltage is the same in each bulb: 48V.
V 2 (48V )2
P

 9.6W
R
240W
Parallel Bulbs Burn Brighter!
Circuits Problem:3 Bulbs in Series
If one more bulb is added to each circuit
(3 bulbs total), how does the brightness
of the bulbs change? Or not?
In the series circuit, the bulbs DIM. WHY?
V  V1  V2  V3
In series, each of the three equal bulbs gets one third of the Voltage
(V/3) that a single bulb would get.
V / 3

V
1 V 2 Psinglebulb
P



R
R
9 R
9
2
2
Note: P=VI but I is due to the equivalent Resistance: I = V/Rs =V/3R
So the Current through each is 1/3 the current through a single bulb and
P=VI=V/3 x I/3 = VI/9 = P/9. The bulbs burn 1/9 as bright!
Circuits Problem:3 Bulbs in Parallel
If one more bulb is added to each circuit
(3 bulbs total), how does the brightness
of the bulbs change? Or not?
In the parallel circuit, the bulbs DO NOT DIM.
WHY?
In parallel, each of the three equal
bulbs gets the full voltage of the battery source.
V2
P
 Psinglebulb
R
Is this getting something for nothing?
NO! Parallel circuits drain the battery faster!
Parallel Circuits
•As the number of branches is increased, the
overall resistance of the circuit is DECREASED.
•Overall resistance is lowered with each added
path between any two points of the circuit.
•This means the overall resistance of the circuit is
less than the resistance of any one of the
branches!!!! (Weird?)
•As overall resistance is lowered, more current is
drawn. This is how you blow fuses!
Circuits Problem:Bulbs in Series vs Parallel
If a bulb burns out - what happens to the other bulb in
each circuit? Does it go out? Is it brighter? Dimmer? Or?
In the series circuit, the burned out bulb will short the circuit and
the other bulb will go out.
In the parallel circuit the other bulb will have the same brightness.
QUESTION
The power rating for two light bulbs read
30W and 60W. Which bulb has the greatest
resistance at 120V?
P V / R  R V / P
2
2
R  (120V ) / 30W  480W
2
R  (120V ) / 60W  240W
2
Which burns brighter and why?
Quick Quiz
For the two lightbulbs shown in this figure,
rank the current values at the points, from
greatest to least.
Ia = I b > Ic = Id > Ie = If .
The 60 W bulb has the lowest
resistance and therefore draws the
most current and burns brightest!
Resistance Question
• Two cylindrical resistors, R1 and R2, are made of identical material.
R2 has twice the length of R1 but half the radius of R1.
– These resistors are then connected to a battery V as shown:
I1
I2
V
– What is the relation between I1, the current flowing in R1 ,
and I2 , the current flowing in R2?
(b) I1 = I2
(a) I1 < I2
(c) I1 > I2
• The resistivity of both resistors is the same (r).
• Therefore the resistances are related as:
R2  r
L2
2 L1
L
r
 8 r 1  8 R1
A2
( A1 / 4)
A1
• The resistors have the same voltage across them; therefore
I2 
V
V
1

 I1
R2 8 R1 8
Superconductivity
• 1911: H. K. Onnes, who had figured
out how to make liquid helium, used it
to cool mercury to 4.2 K and looked at
its resistance:
• At low temperatures the resistance of
some metals0, measured to be less
than 10-16•ρconductor (i.e., ρ<10-24 Ωm)!
–Current can flow, even if E=0.
–Current in superconducting rings can flow for years with no
decrease!
• 1957: Bardeen (UIUC!), Cooper, and Schrieffer (“BCS”)
publish theoretical explanation, for which they get the
Nobel prize in 1972.
– It was Bardeen’s second Nobel prize (1956 – transistor)
Superconductivity
• 1986: “High” temperature superconductors are
discovered (Tc=77K)
– Important because liquid nitrogen (77 K) is much cheaper than
liquid helium
– Highest critical temperature to date 138 K (-135˚ C = -211˚ F)
• Today: Superconducting loops are used to produce
“lossless” electromagnets (only need to cool them, not
fight dissipation of current) for particle physics.
[Fermilab accelerator, IL]
• The Future: Smaller motors, “lossless” power
transmission lines, magnetic levitation trains,
quantum computers?? ...
Superconductivity
• 1911: H. K. Onnes, who had figured
out how to make liquid helium, used it
to cool mercury to 4.2 K and looked at
its resistance:
• At low temperatures the resistance of
some metals0, measured to be less
than 10-16•ρconductor (i.e., ρ<10-24 Ωm)!
–Current can flow, even if E=0.
–Current in superconducting rings can flow for years with no
decrease!
• 1957: Bardeen (UIUC!), Cooper, and Schrieffer (“BCS”)
publish theoretical explanation, for which they get the
Nobel prize in 1972.
– It was Bardeen’s second Nobel prize (1956 – transistor)
Superconductivity
• 1986: “High” temperature superconductors are
discovered (Tc=77K)
– Important because liquid nitrogen (77 K) is much cheaper than
liquid helium
– Highest critical temperature to date 138 K (-135˚ C = -211˚ F)
• Today: Superconducting loops are used to produce
“lossless” electromagnets (only need to cool them, not
fight dissipation of current) for particle physics.
[Fermilab accelerator, IL]
• The Future: Smaller motors, “lossless” power
transmission lines, magnetic levitation trains,
quantum computers?? ...