current, resistance and power. battery +- negative electrode positive electrode electrolyte
TRANSCRIPT
Current, Resistance and Power
Battery
+-Negative Electrode
Positive Electrode
electrolyte
Battery
+-Negative Electrode
Positive Electrode
electrolyte
Current
Simple flow of charge
-
I
dt
dQI
(Note Convention)
MKS unit - Ampere
Current
See Active Figure 27.09
Current
+
++
++
+
n - concentration of charges per unit volume
vd – drift velocity
A – cross sectional area of conducting wire
q – charges carried by each particle
A
Charge carriers
+
++
++
+
tvl d
A
After some time t, the particles will pass beyond a particular point on the wire
The volume contains the passing charges can be found with vd, A and t
tAvAlVol d
+
++
++
+
tvl d
A
Volume of passing charges
qVolnQ
NqQ
)(
+
++
++
+
qnAvt
QI
tqnAvQ
d
d
Current Density and Ohm’s Law
A
IJ
Current Density
Current per unit Area
Current Density and Ohm’s Law
EJ
Ohm’s Law
- conductivity
AdJI
A more familiar form
A
lIV
l
VE
A
I
EJ
1
Resistivity
IRV
A
lR
where
Resistance
Volt/amp= MKS Unit
Ohm’s Law: Macroscopic form
Note Dependencies
• If you double the area (ie. Adding an addition wire) the effective resistance halves
• If you add the wire to the length the effective resistance doubles
• The resistivity is an intrinsic property of the material the resistor is made of. If you change material keeping physical geometry the same, the resistance changes
Ohmic (or linear) device
V
V
Non-Ohmic (or nonlinear) device
Slope = 1/R
Microscopic View of Conductor
smmCmmatoms
Amps
nAq
Iv
qnAvt
QI
d
d
/37.0106.110/1048.8
51926328
e
d
m
en
Em
qEnqnqvJ
EJ
2
Independent of Electric Field: OhmicBut can depend on conditions which effect , such as temperature
copper Cross sectional area Electron Charge
See Active Figure 27.09
acceleration
Time between collisions
Resistivity vs. Temperature
(T) – characteristic of material
T
T
metallic semi-conductors
insulators
T
))(1( 00 TT
T0
0 – resistivity at T0
– thermal coefficient of resistivity
Superconductors
• Many materials will below a specific characteristic temperature, Tc, have a pronounced decrease in resistivity.
Power
Battery – “works” to push current through circuit
Powersource = VI
V – Potential Source
I – Current sent from source through circuit
V
I
Thermal energy dissipated through resistors
RI
Rate of Thermal Energy dissipation through Resistor
R
VRIIVP
dt
Vdq
dt
dU
dt
dWP
22
Voltage drop across resistor
Example Problem: Suppose we wanted to design a small heater for your to work before your car warmed up. We want 500Watts using the 12V of your car battery. How much Nichrome wire with a crossectional area of 0.1 cm2 do we need?
29.0
12500
2
22
RR
Vwatts
R
VRIIVP
cm
cm
cmA
R
290
1.0
10029.0
2
r
Battery (Source)
- actual potential difference between electrodes of battery (EMF)
r – internal resistance of battery
R
By attaching the battery to a circuit including a load resistor R, the current drawn through the battery will
effect the actual potential difference in the battery
r
Battery (Source)
I
Kirchoff’s Voltage Loop Theorem
• The algebraic sum of the changes in electric potential encountered in a complete traversal of the circuit must be zero.
• A circuit is closed path through which current (electrons) may be forced to move through circuit elements (resistors).
V = - Ir
r
I
R
I
Jumping from the negative to the positive end of the battery, the potential increases by , but after going
through the resistor, the potential drops by IR
Battery voltage terminal to terminal
r
I
R
I
)(
0
RrI
IRIr
To find the current…
Kirchoff’s Voltage Loop
Resistors in Series and Parallel:Equivalent Resistance
Resistors in Series
R2
R1
R3
I
I
0)(
0
321
321
RRRI
IRIRIR
Resistors in Series
Rseries
I
I
)(
0
321 RRRR
IR
series
series
Kirchoff’s Junction Theorem
• At any junction (point where current can split) the algebraic sum of the currents into and out of the wires of the junction must add to zero.
• By convention the current into a junction is positive and the current out of a junction is negative.
Resistors in Parallel
I
I
R3R2R1
Resistors in Parallel
I
I
R3R2R1
I2I1 I3
321 IIII
Resistors in Parallel
I
I
R3R2R1
I2I1 I3
1111 0R
IRI
Resistors in Parallel
I
I
R3R2R1
I2I1 I3
2222 0R
IRI
Resistors in Parallel
I
I
R3R2R1
I2I1 I3
3333 0R
IRI
Resistors in Parallel
I
I
Rparallel
I
parallelparallel R
IIR 0
Resistors in Parallel
321
321
321
1111
RRRR
RRRR
IIII
parallel
parallel
42 V
2
12
4
2 1
Solve for the currents going through each of the resistors by circuit reduction (equivalent resistance)
Break circuit down into series and parallel resistors
42 V
2
12
4
2 1
I1
I2I
Currents in the various branches
42 V
2
12
4
2 1
I1
I2I
Find equivalent resistance for the Series Resistors
42 V
12
6 3
I1
I2I
Find Equivalent Parallel Resistance
42 V
12
2
I
Find Equivalent Series Resistance
42 V14
I
Find Equivalent Series Resistance
42 V14
I
AI
I
3
01442
42 V
2
12
4
2 1
I1
I2I
AI
VII
III
2
612423
0211242
1
1
11
42 V
2
12
4
2 1
I1
I2I
AIII
III
112
21
0
0)2()6(
0)4()6(
321
311
2211
III
II
II
Kirchoff’s Analysis
Solve simultaneously for the unknown currents