Electric Current

Current and Resistance

Sunday, July 24, 2011

Current

Convention : Current
depicts flow of
positive (+) charges


Current

depicts flow of

Area

+


Current

depicts flow of

Area

+

Ammeter
(measures current)


Current

depicts flow of

Area

+ +
+
Ammeter
(measures current)


Current

A measure of how much charge passes through an amount of time

+
+
+ Ammeter
(measures current)


Current

Count how many charges flow through

+ +
+


Current

Expand surface to a volume

+ +
+


Current


+ +
+
Area = A


Current


+ +
+
Area = A
length = !x


Current


+ + Total volume
V = (A)(!x)
+
Area = A
length = !x


Current


+ + Total volume
V = (A)(!x)
+
Area = A
length = !x

Number of charges = (charge density or charge per volume)*(volume)
Number of charges = (n) * (A!x)


Current


+ + Total volume
V = (A)(!x)
+
Area = A
length = !x

Number of charges = (charge density or charge per volume)*(volume)
Number of charges = (n) * (A!x)

Total amount of charge = (number of charges)*(charge)
!Q = (n A !x)*(q)

Current

!Q = (n A !x)*(q)

+ + Total volume
V = (A)(!x)
+
Area = A
length = !x


Current

!Q = (n A !x)*(q)
but charges have drift velocity vd = !x/!t

+ + Total volume
V = (A)(!x)
+
Area = A
length = !x = vd !t


Current

!Q = (n A !x)*(q)

+ + Total volume
V = (A)(!x)
+
Area = A
length = !x = vd !t

!Q = (n A vd !t)*(q)


Current

!Q = (n A !x)*(q)

+ + Total volume
V = (A)(!x)
+
Area = A
length = !x = vd !t

!Q = (n A vd !t)*(q)
!Q/!t = (n A vd)*(q)
I = n q vd A


Current

This is the reason why large wires are
needed to support large currents


Resistance

Current density (J)
current per area


Resistance

Current density (J)
current per area

Direction of current (flow of positive charges)
is same with direction of electric field


Resistance

Current density (J)
current per area


conductivity


Resistance

Current density (J)
current per area


conductivity (material property)

resistivity (material property)


Resistance

Current density (J)
current per area


conductivity

resistivity

Current is proportional to conductivity but
inversely proportional to resistivity!


Resistance



Resistance

Current is proportional to the electric potential
(specifically potential difference)


Resistance


Ohm’s Law Potential difference
Resistance
current


Resistance


Resistance
current

a much better form
than ΔV = I R


Resistance


Resistance
current

a much better form Increasing !V increases I
than ΔV = I R Increasing R decreases I


Resistance


Resistance
current

a much better form Increasing !V increases I
than ΔV = I R Increasing R decreases I

!V = I R Increasing R does not increase !V
Current (I) is increased because !V is increased

Resistance


Resistance

Important points:

same with capacitance, resistance does not
depend on !V and I

Resistance depends on material property
resistivity ", length of wire l and cross
sectional area A

conventional current is flowing positive (+) charges though
in reality electrons flow

direction of the current I is same as direction of electric field


Recent Equations
→
→ →E
J = σE =
ρ
→ →
J = nq v d A
→
→ I
J =
A
∆V
I=
R
ρl
R=
A


Exercise
Rank from lowest to highest amount of current

Derive the equation R = "L/A
from V = IR, J = E/" = I/A, V = EL


Resistance and Temperature
ρl
R=
A
ρ = ρ0 (1 + α∆T )
∆T = T − T0

T0 is usually taken to be 25 °C

T ↑ ρ↑


Power

∆U
P =
∆t
∆(q∆V )
P =
∆t
(∆q)(∆V )
P =
∆t
∆q
P = ∆V
∆t
P = I∆V


Power

P = I∆V
∆V
I=
R

V2
P = P = I 2R
R


Exercises
The electron beam emerging from a certain high-energy electron accelerator
has a circular cross section of radius 1.00 mm. (a) The beam current is 8.00 µA.
Find the current density in the beam, assuming that it is uniform throughout. (b)
The speed of the electrons is so close to the speed of light that their speed can
be taken as c = 3.00 x 108 m/s with negligible error. Find the electron density in
the beam. (c) How long does it take for Avogadroʼs number of electrons to
emerge from the accelerator?

An aluminum wire having a cross-sectional area of 4.00 x 10-6 m2 carries a current
of 5.00 A. Find the drift speed of the electrons in the wire. The density of aluminum
is 2700 kg/m3. Assume that one conduction electron is supplied by each atom.
Molar mass of Al is 27 g/mol.

Four wires A, B, C and D are made of the same material but of different lengths
and radii. Wire A has length L but has radius R. Wire B has length 2L but with
radius ½R. Wire C has length ½L but with radius 2R. Wire D has length ½L but
with radius ½R.

Rank with increasing resistance

A 0.900-V potential difference is maintained across a 1.50-m length of tungsten
wire that has a cross-sectional area of 0.600 mm2. What is the current in the
wire?
resistivity of tungsten is 5.6 x 10-8 Ω-m


Exercises

An electric heater is constructed by applying a potential difference of 120 V to a
Nichrome wire that has a total resistance of 8.00 Ω. Find the current carried by
the wire and the power rating of the heater.

A 500-W heating coil designed to operate from 110 V is made of Nichrome wire
0.500 mm in diameter. (a) Assuming that the resistivity of the Nichrome remains
constant at its 20.0°C value, ﬁnd the length of wire used. (b) What If? Now
consider the variation of resistivity with temperature. What power will the coil of
part (a) actually deliver when it is heated to 1200°C?
ρ = 1.50 x 10-6 Ω-m


A 500-W heating coil designed to operate from 110 V is made of Nichrome wire
0.500 mm in diameter. (a) Assuming that the resistivity of the Nichrome remains
constant at its 20.0°C value, ﬁnd the length of wire used. (b) What If? Now
consider the variation of resistivity with temperature. What power will the coil of
part (a) actually deliver when it is heated to 1200°C?
ρ = 1.50 x 10-6 Ω-cm


More exercises

A certain lightbulb has a tungsten filament with a resistance of 19.0 Ω when cold
and 140 Ω when hot. Assume that the resistivity of tungsten varies linearly with
temperature even over the large temperature range involved here, and find the
temperature of the hot filament. Assume the initial temperature is 20.0°C.
4.5 x 10-3 C-1

The cost of electricity varies widely through the United States; $0.120/kWh is
one typical value. At this unit price, calculate the cost of (a) leaving a 40.0-W
porch light on for two weeks while you are on vacation, (b) making a piece of
dark toast in 3.00 min with a 970-W toaster, and (c) drying a load of clothes in
40.0 min in a 5 200-W dryer.


A certain lightbulb has a tungsten filament with a resistance of 19.0 Ω when cold
and 140 Ω when hot. Assume that the resistivity of tungsten varies linearly with
temperature even over the large temperature range involved here, and find the
temperature of the hot filament. Assume the initial temperature is 20.0°C.
4.5 x 10-3 C-1


The cost of electricity varies widely through the United States; $0.120/kWh is
one typical value. At this unit price, calculate the cost of (a) leaving a 40.0-W
porch light on for two weeks while you are on vacation, (b) making a piece of
dark toast in 3.00 min with a 970-W toaster, and (c) drying a load of clothes in
40.0 min in a 5 200-W dryer.

$0.120 $0.120 1kW 1hour $3.33 × 10−8
= =
1kWh 1kWh 1000W 3600secs 1Joule
∆U ∆U 1week 1day 1hour ∆U
(a) P =
∆t
=
2weeks 7days 24hours 3600secs
=
1209600secs
∆U
40.0W =
1209600s
$3.33 × 10−8
∆U = 48384kJ 4.84 × 107 J = $1.61
1Joule

(b) $5.82 × 10− 3
(c) $0.416

Electromotive Force

The electromotive force is denoted as “ε”
A force that moves charges

The emf ε is the maximum possible voltage
that the battery can provide.

ε = ∆V in batteries

Direct current - current that is constant in direction and magnitude


Resistors in Series

∆V
Recall: I=
R

use the equation to
calculate the equivalent
resistance Req


Resistors in Series

Convert
to simple
equivalent
circuit


Resistors in Series

I1 I2

∆V1 ∆V2

Conservation of matter = Current is conserved
I = I1 = I2


Resistors in Series

I1 I2

∆V1 ∆V2

I = I1 = I2
Conservation of energy
∆V = ∆V1 + ∆V2


Resistors in Series

I1 I2

∆V1 ∆V2

Conservation of matter = Current is conserved Ohms Law
I = I1 = I2
∆V
Conservation of energy I=
R
∆V = ∆V1 + ∆V2


Resistors in Series

I1 I2
∆V = I1 R1 + I2 R2

∆V1 ∆V2 ∆V = IR1 + IR2
∆V = I(R1 + R2 )
∆V = IReq
Req = R1 + R2

I = I1 = I2
∆V
R
∆V = ∆V1 + ∆V2


Resistors in Parallel
1. Imagine positive charges pass ﬁrst I1 I2
through R1 and then through%R2.
Compared to the current in R1, the
current in R2 is ∆V1 ∆V2
(a) smaller
(b) larger
(c) the same.
2. With the switch in the circuit of closed (left),
there is no current in R2, because the current
has an alternate zero-resistance path through
the switch. There is current in R1 and this
current is measured with the ammeter (a
device for measuring current) at the right side
of the circuit. If the switch is opened (right),
there is current in R2. What happens to the
reading on the ammeter when the switch is
opened?
(a) the reading goes up
(b) the reading goes down
(c) the reading does not change.



∆V
Recall: I=
R

use the equation to
calculate the equivalent
resistance Req



Convert
to simple
equivalent
circuit


I1
∆V1

I2
∆V2

I = I1 + I2


I1
∆V1

I2
∆V2

I = I1 + I2
Conservation of energy
∆V = ∆V1 = ∆V2


I1
∆V1

I2
∆V2

I = I1 + I2
∆V
R
∆V = ∆V1 = ∆V2


I1
∆V1 I = I1 + I2
∆V ∆V1 ∆V2
I2 = +
∆V2 R R1 R2
∆V ∆V ∆V
= +
R R1 R2
1 1 1
= +
R R1 R2

I = I1 + I2
∆V
R
∆V = ∆V1 = ∆V2


Recall:

Ohms Law Capacitance
∆V
I= Q = C∆V
R
Series

Parallel


Exercise

Find the current passing through each resistor
Find the voltage drop (potential difference) through each resistor


Kirchhoff’s Rules

Junction Rule
“conservation of matter”

Loop Rule
“conservation of energy”

Σ ∆V = 0
closed loop

Exercise

In solving complicated circuit problems
apply Junction rule first (conservation of current)

You may assign any direction of current as long as it
is reasonable (does not violate common sense!) A

Then apply the loop rule
B

Write down the equations for loop rules concerning
loop A, B, C and the outer loop of the circuit following
C
clockwise direction. (there must be four equations!)


Electric Current

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