Transcript Current, Resistance, DC Circuits, Kirchoff`s Rules

Electricity and Magnetism
Lecture 07 - Physics 121
Current, Resistance, DC Circuits: Y&F Chapter 25 Sect. 1-5
Kirchhoff’s Laws: Y&F Chapter 26 Sect. 1
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Currents and Charge
Electric Current i
Current Density J
Drift Speed, Charge Carrier Collisions
Resistance, Resistivity, Conductivity
Ohm’s Law
Power in Electric Circuits
Examples
Circuit Element Definitions
Kirchhoff’s Rules
EMF’s - “Pumping” Charges, Ideal and real EMFs
Work, Energy, and EMF
Simple Single Loop and Multi-Loop Circuits using
Kirchhoff Rules
Equivalent Series and Parallel Resistance formulas
using Kirchhoff Rules.
Copyright R. Janow - Fall 2015
1
Definition of Current : Net flow rate of charge through some area
dq
i
dt
or
dq  i dt
 q(t) 
t
0 i(t' )dt'
 i x t (if i
is constant)
Units: 1 Ampere = 1 Coulomb per second
Convention: flow is from + to – as if free charges are +
Charge / current is conserved - charge does not pile up or vanish
At any Node
(Junction)
 iin   iout
i1
i3
i2
i1+i2=i3
Current is the same across each cross-section of a wire
i
+
Kirchhoff’s
Rules:
(summary)
-
Current density J may vary
[J] = current/area
 Current (Node) Rule: S currents in = S currents out at any node
 Voltage (Loop) Rule: S DV’s = 0 for any closed path
Energy or  Voltage sources (EMFs, electro-motive forces) or current sources
can supply energy to a circuit
Power
 Resistances dissipate energy as heat
in a circuit:  Capacitances and Inductances store energy in E or B fields
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Junction Rule Example – Current Conservation
7-1: What is the value of the
current marked i?
A.
B.
C.
D.
E.
1 A.
2 A.
5 A.
7 A.
Cannot determine
from information given.
1A
3A
2A
2A
3A
1A
5A
8A
 iin   iout
i =7A
6A
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Current density J: Current / Unit Area (Vector)
i
A’
A
+
J=i/A
(large)
Same current crosses larger or smaller
Surfaces, current density J varies
J’ = i / A’
(small)
Small current density
in this region
i
High current density
in this region
For uniform density:
i  J A or J  i/A

For non-uniform density: di  J  n̂dA
units: Amperes/m2
i
area
conductivity resistivity
What makes current flow?
Microscopic level


J  dA
JE
J  E
  1/ 
Drift speed: Do electrons in a current keep accelerating?
Yes, for isolated charge in vacuum. No, in a conducting solid, liquid, gas
Copyright R. Janow - Fall 2015
COLLISIONS with ions, impurities, etc. cause resistance
Charges move at constant drift speed vD:
E field in
solid wire
drives
current.
•
•
•
+
+
-
APPLIED FIELD = 0
Charges in random motion
flow left = flow right
APPLIED FIELD NOT ZERO
Moving charges collide with fixed ions
and flow with drift velocity vd
3
6
Thermal motions (random motions) have speed vth  10 m/s ( 2 kBoltzT)
Drift speed is tiny compared with thermal motions.
Drift speed in copper is 10-8 – 10-4 m/s.
For E = 0 in conductor: no current,
vD=0, J = 0, i = 0
For E not = 0
(battery voltage not 0):
J  E
-
+
n  density of charge carriers Units : # /volume
nvD  # of charge carriers crossing unit area per unit time


J  qnvD 
vD 
net charge crossing area A per unit time

E
qn
i 
charge/unit time
 E
 qnvD A
Note: Electrons drift rightward but vd and J are still leftward
Copyright R. Janow - Fall 2015
-19
|q| = e = 1.6 x 10
C.
EXAMPLE: Calculate the current density Jions for ions in a gas
Assume:
• Doubly charged positive ions
• Density n = 2 x 108 ions/cm3
• Ion drift speed vd = 105 m/s
Find Jions – the current density for the ions only (forget Jelectrons)
J  qnvD  2  1.6 x 10-19  2  108  105
coul/ion
 J  6.4
ions/cm3
m/s

106
cm3/m3
A./m2
Copyright R. Janow - Fall 2015
Increasing the Current
7-2: When you increase the current in a wire, what
changes and what is constant?
A.
B.
C.
D.
E.
The density of charge carriers stays the same, and the drift speed
increases.
The drift speed stays the same, and the number of charge carriers
increases.
The charge carried by each charge carrier increases.
The current density decreases.
None of the above
JE
J  E
J  qnvD
Copyright R. Janow - Fall 2015
Definition of Resistance : Amount of current flowing
through a conductor per unit of applied potential difference.
E
i
DV
L
A
R
i
DV
i
units : 1 Ohm  1   1 volt/amper e
G  R -1  " conductance" Units are " Siemens"   -1
R depends on the material & geometry
Note: C= Q/DV – inverse to R
Apply voltage to a wire made of a good conductor.
- Very large current flows so R is small.
Apply voltage to a poorly conducting material like carbon
- Tiny current flows so R is very large.
V
Circuit
Diagram
R
Resistivity “” : Property of a material itself
(as is dielectric constant). Does not depend on dimensions
• The resistance of a device depends on resistivity  and also depends on shape
• For a given shape, different materials produce different currents for same DV
• Assume cylindrical resistors
R  resistance 
L
A
For insulators:   infinity
  resistivit y 
RA
for a resistor
L
resistivit y units : Ohm - meters  .m
Copyright R. Janow - Fall 2015
Example: calculating resistance or resistivity
resistivity
resistance
L
R
A
proportional to length
inversely proportional
to cross section area
EXAMPLE:
Find R for a 10 m long iron wire, 1 mm in diameter
L
9.7 x10 -8 .m x 10 m
R

 1.2 
3
2
2
A
 x (10 / 2) m
Find the potential difference across R if i = 10 A. (Amperes)
DV  iR  12 V
EXAMPLE:
Find resistivity of a wire with R = 50 m,
diameter d = 1 mm, length L = 2 m
2
RA
50x10-3  x 10 m
3


x
 10 / 2   1.96x10 - 8 .m


L
2m
Use a table to identify material. Not Cu or Al, possibly an alloy
Copyright R. Janow - Fall 2015
Resistivity depends on temperature:
•
Resistivity depends on temperature: Higher temperature  greater thermal
motion  more collisions  higher resistance.
 in .m @ 20o C.
SOME SAMPLE
RESISTIVITY
VALUES
1.72 x 10-8 copper
9.7 x 10-8 iron
2.30 x 10+3 pure silicon
Reference
Temperature
Simple model of resistivity: a = temperature coefficient
Change the temperature from
reference T0 to T
Coefficient a depends
on the material
   0 (1  a(T - T0 ))
a  temperature coefficient
Conductivity is the reciprocal of resistivity
i
DV
L
R
Definition:
A
E
L
A
i
 
1
 J  E

units : " mho"  (.m) -1
Across resistor : DV  iR  JAR  JL  EL
E/J    -1
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Resistivity Tables
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Current Through a Resistor
7-3: What is the current through the resistor
in the following circuit, if V = 20 V and R =
100 ?
A.
B.
C.
D.
E.
20 mA.
5 mA.
0.2 A.
200 A.
5 A.
V Circuit
Diagram
R
DV  i R
Copyright R. Janow - Fall 2015
Current Through a Resistor
7-4: If the current is doubled, which of the following
might also have changed?
A.
B.
C.
D.
E.
The
The
The
The
The
voltage across the resistor might have doubled.
resistance of the resistor might have doubled.
resistance and voltage might have both doubled.
voltage across the resistor may have dropped by a factor of 2.
resistance of the resistor may have dropped by a factor of 2.
DV  i R
Note: there might be
more than 1 right answer
V Circuit
Diagram
R
Copyright R. Janow - Fall 2015
Ohm’s Law and Ohmic materials (a special case)
R  V /i
(R could depend on applied V)
Definitions of
resistance:
  1 /   J / E ( could depend on E)
Definition: For OHMIC conductors and devices:
• Ratio of voltage drop to current is constant – NO dependance on applied voltage.
•
i.e., current through circuit element equals CONSTANT resistance times applied V
Resistivity does not depend on magnitude or direction of applied voltage
Ohmic Materials
e.g., metals, carbon,…
Non-Ohmic Materials
e.g., semiconductor diodes
band gap
varying slope = 1/R
constant slope = 1/R
OHMIC
CONDITION
1
di

R dV
is CONSTANT
Copyright R. Janow - Fall 2015
Resistive Loads in Circuits Dissipate Power
+
1
L
O
A
D
V
-
• Apply voltage drop V across basic circuit element (load)
• Current flows through load which dissipates energy
• An EMF (e.g., a battery) does work, holding potential
V and current i constant by expending potential energy
i
2
As charge dq flows from 1 to 2 it loses P.E. = dU
dU  V dq  Vi dt
- potential is PE / unit charge
- charge = current x time
i
dU dU dq
Power dissipated  P 

 Vi
dt dq dt
V  iR 
P  i2R  V 2 /R
Active sign convention:
Voltage rise:
+
V
-
[Watts] for any load
resistors only
Pabsorbed = dot product of Vrise with i
element absorbs power
for V opposed to i
+
1
-
2
element supplies power
for V parallel to i
+
V
V
1
i
P = - Vi
i
P = + Vi
2
Note: Some EE circuits texts use passive convention:
model voltage drops Vdrop as positive.Copyright R. Janow -
Fall 2015
EXAMPLE:
EXAMPLE:
EXAMPLE:
Copyright R. Janow - Fall 2015
Basic Circuit Elements and Symbols
Basic circuit elements have two terminals
Generic symbol:
2
1
Ideal basic circuit elements:
Passive:
resistance,
capacitance,
R
C
or
inductance
C
L
Active: voltage sources (EMF’s), current sources
Ideal independent voltage source:
• zero internal resistance
• constant vs for any load
Ideal independent current source:
• constant is regardless of load
DC
vs
+
-
+
DC
battery E
AC
vs
is
Dependent voltage or current sources: depend on a control voltage or current
generated elsewhere within a circuit
Copyright R. Janow - Fall 2015
Circuit analysis with resistances and EMFs
CIRCUIT ELEMENTS ARE CONNECTED TOGETHER AT NODES
CIRCUITS CONSIST OF JUNCTIONS
(ESSENTIAL NODES)
... and …
ESSENTIAL BRANCHES (elements
connected in series,one current/branch)
i1
i
i2
i
ANALYSIS METHOD: Kirchhoff’S LAWS or RULES
Current (Junction) Rule:
Loop (Voltage) Rule:
Charge conservation
 DV  0 (closed loop) Energy conservation
DV
L
R
  resistivit y
i
A
dW
dU
P
 iV
P  i2R (resistor)
dt
dt
RESISTANCE:
R
POWER:
OHM’s LAW:
iin  iout
i
slope
= 1/R
R is independent of DV or i
Copyright R. Janow - Fall 2015
EMFs “pump” charges to higher energy
•
EMFs (electromotive force) such as batteries supply energy:
– move charges from low to high potential (potential energy).
– maintain constant potential at terminals
– do work dW = Edq on charges (source of the energy in batteries is chemical)
– EMFs are “charge pumps”
Variable name: script E.
•
Unit: volts (V).
•
•
Types of EMFs: batteries, electric generators, solar cells, fuel cells, etc.
DC versus AC
E
i
work done
dW

unit charge
dq
Power supplied by EMF:
+
E
-
R
i
CONVENTION:
Current flows CW through circuit
from + to – outside of EMF
from – to + inside EMF
P  power 
dW
dt
dW  E dq  E i dt  P dt
 
Pemf   E i  E  i
Power is dissipated by resistor:
PR  - i V  - i2 R  - V2 / R
Copyright R. Janow - Fall 2015
Ideal EMF
device
• Zero internal battery resistance
• Open switch: EMF = E
no current, zero power
• Closed switch: EMF E is also
applied across load circuit
• Current & power not zero
Real EMF
device
• Open switch: EMF still = E
• r = internal EMF resistance in
series, usually small ~ 1 
• Closed switch:
• V = E – ir across load, Pckt= iV
• Power dissipated in EMF
Pemf = i(E-V) = i2r
Multiple
EMFs
Assume EB > EA (ideal EMF’s)
Which way does current i flow?
• Apply kirchhoff Laws to find current
• Answer: From EB to EA
• EB does work, loses energy
• EA is charged up
• R converts PE to heat
• Load (motor, other) produces
motion and/or heat
R
Copyright R. Janow - Fall 2015
The Kirchhoff Loop Rule Generates Circuit Equations
• The sum of voltage changes = zero around all closed loops in a multi-loop circuit)
• Traversing one closed loop generates one equation. Multiple loops may be needed.
• A branch is a series combination of circuit elements between essential nodes.
• Assume either current direction in each branch. Minus signs may appear in the result.
• Traverse each branch of a loop with or against assumed current direction.
• Across resistances, write voltage drop DV = - iR if following assumed current direction.
Otherwise, write voltage change = +iR.
• When crossing EMFs from – to +, write DV = +E. Otherwise write DV= -E
• Dot product i.E determines whether power is actually supplied or dissipated
EXAMPLE:
Single loop (circuit with battery (internal resistance r)
Follow circuit from a to b to a, same direction as i
E
E - ir - iR  0
EE
i
r R
Potential around the circuit
Power in External Ckt
P = iV = i(E – ir)
P = iE – i2r
circuit
battery battery
dissipation drain dissipation
Copyright R. Janow - Fall 2015
Example: Equivalent resistance for resistors in series
Junction Rule: The current through all of the resistances in series (a single
essential branch) is identical. No information from Junction/Current/Node Rule
i  i1  i2  i3
Loop Rule: The sum of the potential differences around a
closed loop equals zero. Only one loop path exists:
 - iR 1 - iR 2 - iR 3  0
i

R1  R 2  R 3
The equivalent circuit replaces the series resistors with a
single equivalent resistance: same E, same i as above
 - iR eq  0
i

R eq
CONCLUSION: The equivalent resistance for a series
combination is the sum of the individual resistances and is
always greater than any one of them.
R eq  R 1  R 2  R 3
R eq 
n
 Ri
i1
inverse of series capacitance rule
Copyright R. Janow - Fall 2015
Example: Equivalent resistance for resistors in parallel
Loop Rule: The potential differences across each of the 4 parallel branches
are the same. Four unknown currents. Apply loop rule to 3 paths.
E - i1R1  0
i1 
E - i2R 2  0
E
E
E
, i2 
, i3 
R1
R2
R3
E - i3R 3  0
i not in
these
equations
Junction Rule: The sum of the currents flowing in equals the
sum of the currents flowing out. Combine equations for all the
upper junctions at “a” (same at “b”).
 1
1
1 

i  i1  i2  i3  E 


R
R
R
2
3
 1
The equivalent circuit replaces the series resistors with a
single equivalent resistance: same E, same i as above.
i
 - iR eq  0

R eq
CONCLUSION: The reciprocal of the equivalent resistance for a
parallel combination is the sum of the individual reciprocal
resistances and is always smaller than any one of them.
1
1
1
1



R eq
R1 R 2 R 3
1

R eq
n
1
R
i1 i
inverse of parallel capacitance rule
R eq 
R1R 2
R1  R 2
Copyright R. Janow - Fall 2015
Resistors in series and parallel
7-7: Four identical resistors are connected as shown in
the figure. Find the equivalent resistance between
points a and c.
A. 4 R.
B. 3 R.
C. 2.5 R.
D. 0.4 R.
E.Cannot determine
from information given.
c
R

R
R
a
1

R eq
n
1
R
i1 i
R eq 
R
n
 Ri
i1
Copyright R. Janow - Fall 2015
Capacitors in series and parallel
7-8: Four identical capacitors are connected as shown
in figure. Find the equivalent capacitance between
points a and c.
A. 4 C.
B. 3 C.
C. 2.5 C.
D. 0.4 C.
E. Cannot determine
from information given.
c
C

C
C
C
a
n
1
1

Ceq i1 Ci
Ceq 
n
 Ci
i 1
Copyright R. Janow - Fall 2015
EXAMPLE:
Find i, V1, V2, V3, P1, P2, P3
R1= 10 
+
-
i
E =7V
R2= 7 
R3= 8 
EXAMPLE:
Find currents and voltage drops
i
i1
+
-
E =9V
i2
R1
R2
i
R eq 
R1R 2
R1
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 R2
EXAMPLE: MULTIPLE BATTERIES
SINGLE LOOP
i
+
-
R1= 10 
E1 = 8 V
+
-
i
E2 = 3 V
R2= 15 
A battery (EMF) absorbs power (charges up) when I is opposite to E
Pemf
 
  Ei  E  i
Copyright R. Janow - Fall 2015
EXAMPLE: Find the average current density J in a copper
wire whose diameter is 1 mm carrying current of i = 1 ma.
i
10-3 amps
J 

 1273
3
2
A
 x (.5 x 10 m)
amps/m
2
Suppose diameter is 2 mm instead. Find J’:
J' 
i
J

 318 amps/m2
A' 4
Current i is unchanged
Calculate the drift velocity for the 1 mm wire as above?
J  enCu vd
where
nCu 
3
# conduction electrons/m
 8.49 x 1028
J
1273
-8


9
.
37
x
10
m / s About 3 m/year !!
19
28
enCu 1.6x10 x 8.49x10
So why do electrical signals on wires seem
to travel at the speed of light (300,000 km/s)?
vd 
Calculating n for copper: One conduction electron per atom
6.023 x 1023 atoms/mole
3
6
3 3
nCu  1 electron / atom x
x8.96 gm/cm x 10 cm /m
63.5 gm/mole
 8.49 x 1028
electrons/m
3
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