Transcript Current, Resistance and Power

Current, Resistance and
Power
Battery
Negative Electrode
-
electrolyte
+
Positive Electrode
Battery
Negative Electrode
-
electrolyte
+
Positive Electrode
Current
Simple flow of charge
See Active Figure 27.09
I
(Note Convention)
-
dQ
I
dt
Current
MKS unit - Ampere
Current
Charge carriers
+
+
+
+
+
A
+
n - concentration of charges per unit volume
vd – drift velocity
A – cross sectional area of conducting wire
q – charges carried by each particle
After some time t, the particles will pass
beyond a particular point on the wire
+
A
+
+
+
+
+
l  vd t
The volume contains the passing charges
can be found with vd, A and t
Vol  Al  Avd t
Volume of passing charges
+
A
+
+
+
+
+
l  vd t
Q  Nq
Q  (nVol )q
+
+
+
+
Q  nAvd tq
Q
I
 nAvd q
t
+
+
Current Density and Ohm’s Law
I
J
A
Current Density
Current per unit Area
Current Density and Ohm’s Law


J  E
Ohm’s Law
 - conductivity
 
I   J  dA
A more familiar form
J  E
I
V
 E  
A
l
 l 
V  I 
 A 

1

Resistivity
V  IR
Ohm’s Law:
Macroscopic form
where
R
l
A
Resistance
Volt/amp= W MKS Unit
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
I
Ohmic (or linear) device
V
Slope = 1/R
I
Non-Ohmic (or nonlinear) device
V
Microscopic View of Conductor
Q
I
 nAvd q
t
I
5 Amps
vd 

 0.37mm / s
28
3
-6
2
-19
nAq 8.48 10 atoms / m 10 m 1.6 10 C


copper

Cross sectional area
J  E
Electron Charge
acceleration
 qE 
J  nqvd  nq
  E
 m  Time between collisions
 e2 
Independent of Electric Field: Ohmic
  n 
But can depend on conditions which
 me 
effect , such as temperature
See Active Figure 27.09
Resistivity vs. Temperature
(T) – characteristic of material


T
metallic
T
semi-conductors
insulators

0
T0
T
  0 (1   (T - 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
I
Powersource = VI
V
V – Potential Source
I – Current sent from source through circuit
Thermal energy dissipated through
resistors
R
I
dW dU dq V
P


dt
dt
dt
2
V
P  IV  I 2 R 
R
Voltage drop across resistor
Rate of Thermal Energy dissipation
through 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?
2
V
P  IV  I R 
R
2

12V 
500 watts 
R
R  0.29W
2
R

A

100W  cm 
0.29 W 
0.1cm 2
  290 cm
Battery (Source)
e
r
e- actual potential difference between
electrodes of battery (EMF)
r – internal resistance of battery
Battery (Source)
e
r
I
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
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 = e - Ir
Battery voltage terminal to terminal
e
r
I
I
R
Jumping from the negative to the positive end of the
battery, the potential increases by e, but after going
through the resistor, the potential drops by IR
To find the current…
e
r
I
I
R
e - Ir - IR  0
e
I
(r  R)
Kirchoff’s Voltage Loop
Resistors in Series and Parallel:
Equivalent Resistance
Resistors in Series
R1
I
e
R2
I
R3
e - IR1 - IR2 - IR3  0
e - I ( R1  R2  R3 )  0
Resistors in Series
I
e
Rseries
I
e - IRseries  0
Rseries  ( R1  R2  R3 )
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
e
R1
I
R2
R3
Resistors in Parallel
I
e
I1
I2
R1
I3
R2
I
I  I1  I 2  I 3
R3
Resistors in Parallel
I
e
I1
I2
R1
I3
R2
I
e - I1R1  0  I1 
e
R1
R3
Resistors in Parallel
I
e
I1
I2
R1
I3
R2
I
e - I 2 R2  0  I 2 
e
R2
R3
Resistors in Parallel
I
e
I1
I2
R1
I3
R2
I
e - I 3 R3  0  I 3 
e
R3
R3
Resistors in Parallel
I
e
I
Rparallel
I
e - IR parallel  0  I 
e
Rparallel
Resistors in Parallel
I  I1  I 2  I 3
e
R parallel
1
R parallel

e
R1

e
R2

e
R3
1
1
1



R1 R2 R3
Solve for the currents going through each of the resistors by
circuit reduction (equivalent resistance)
12 W
42 V
1W
2W
2W
4W
Break circuit down into series and parallel resistors
Currents in the various branches
12 W
I
I2
42 V
I1
1W
2W
2W
4W
12 W
I
I2
42 V
I1
1W
2W
2W
4W
Find equivalent resistance for the Series Resistors
12 W
I
I2
42 V
3W
I1
Find Equivalent Parallel Resistance
6W
12 W
I
42 V
2W
Find Equivalent Series Resistance
I
42 V
14 W
Find Equivalent Series Resistance
I
42 V
14 W
42 - 14 I  0
I  3A
12 W
I
I2
42 V
I1
1W
2W
2W
4W
42 - 12 I - 1I1 - 2 I1  0
3I1  42 - 12 I  6V
I1  2 A
12 W
I
I2
42 V
I1
1W
2W
2W
4W
I  I1  I 2
I 2  I - I1  1 A
Kirchoff’s Analysis
e 1 - I1 (6W)  e 2  I 2 (4W)  0
e 1 - I1 (6W) - I 3 (2W)  0
I1  I 2 - I 3  0
Solve simultaneously for the unknown
currents