Chapter 26
Current and Resistance
Current and Resistance
Electric Current
Resistance
Ohm’s Law
Power
Semiconductors and
Superconductors
“Ampere was the Newton of
electricity.”
- James C. Maxwell
David J. Starling
Penn State Hazleton
PHYS 212
Current and Resistance
Chapter 26
Current and Resistance
Electric Current
Resistance
Ohm’s Law
Objectives
(a) Determine the current and current density for a system
and relate these quantities to each other and to the
motion of the charges in the system.
(b) Define resistance, resistivity and conductivity, and
relate them to each other, and to the voltage/current in
a system using Ohm’s Law.
(c) Determine the power in a DC or AC electric circuit and
relate the power to observable features; be able to
compare the power in different situations.
Power
Semiconductors and
Superconductors
Chapter 26
Current and Resistance
Electric Current
Current — the net charge (in coulombs) that passes through
a fixed plane per unit of time (seconds).
Electric Current
Resistance
Ohm’s Law
dq
∆q
or i =
i=
∆t
dt
Another way to write this relationship is
Z
Z t
q = dq =
idt0 .
0
(1)
Power
Semiconductors and
Superconductors
(2)
Electric Current
I
The unit of current is the ampere (A)
I
1 A = 1 C/s = 1 coulomb per second
I
Current can be thought of as a stream of moving
Chapter 26
Current and Resistance
Electric Current
Resistance
Ohm’s Law
Power
positive charges
I
In actuality, the electrons move the opposite way
I
This is the bulk motion, and is actually very slow
I
The microscopic motion is much faster
I
Electrons bounce around randomly at ≈100 m/s
Semiconductors and
Superconductors
Chapter 26
Current and Resistance
Electric Current
Electric Current
Resistance
Charge is conserved, so current is conserved:
Ohm’s Law
Power
Semiconductors and
Superconductors
i0 = i1 + i2
I
The orientation of the conductors is irrelevant
(3)
Chapter 26
Current and Resistance
Electric Current
What if the size of the conductor changes?
Electric Current
Resistance
Ohm’s Law
Power
Semiconductors and
Superconductors
I
The current does not change (charge is conserved)
I
Therefore, the density of current changes
I
Current density: J = i/A (A/m2 )
I
Non-uniform current density:
Z
i = J dA
(4)
Electric Current
Question 1
When lightning strikes, the current flows from the ground
upward to the clouds above. What is the direction of the
electric field of the lightning?
(a) upward
(b) downward
(c) perpendicular to the current at every point
(d) parallel to the ground
Chapter 26
Current and Resistance
Electric Current
Resistance
Ohm’s Law
Power
Semiconductors and
Superconductors
Electric Current
Chapter 26
Current and Resistance
Electric Current
Example 1
Resistance
Ohm’s Law
Power
Find the total current through a wire of radius R = 2 mm
and current density J = 2 × 105 A/m2 . Then, find the
current through just the part of the wire with r > R/2.
Semiconductors and
Superconductors
Electric Current
Chapter 26
Current and Resistance
Electric Current
Example 2
Resistance
Ohm’s Law
Power
What is the drift speed vd of the electrons in a copper wire
of radius 0.9 mm with a current of 17 mA? Assume that
there is 1 charge carrier per copper atom, so that
n = 8.49 × 1028 electrons/m3 .
Semiconductors and
Superconductors
Resistance
Chapter 26
Current and Resistance
Electric Current
Resistance
Ohm’s Law
Power
Semiconductors and
Superconductors
Resistance slows down the motion of charge
(think: friction)
I
Definition: R = V/i
I
Apply a voltage V, get a current i
I
Unit of R: 1 ohm = 1 Ω = 1 V/A
I
Bigger R means we need bigger V for same i
I
Bigger R means we get smaller i for same V
Resistance
Chapter 26
Current and Resistance
Electric Current
Resistance
Can we predict the resistance of a particular object?
Ohm’s Law
Power
I
Let’s define resistivity as ρ = E/J.
I
We apply an elecric field, we get a current density
I
Units: Ωm.
I
Bigger ρ means smaller currents.
I
This is a property of the material
I
Resistance is a property of an object
I
How are they related?
Semiconductors and
Superconductors
Chapter 26
Current and Resistance
Resistance
Electric Current
Resistance
Ohm’s Law
Power
Semiconductors and
Superconductors
Can we predict the resistance of this object?
I
The resitance: R = V/i
I
The voltage: V = EL
I
The current density: J = i/A
I
The resistivity:
ρ=
E
V/L
VA
A
=
=
=R
J
i/A
i L
L
(5)
Chapter 26
Current and Resistance
Resistance
Electric Current
Resistance
Ohm’s Law
Or, solving for R,
Power
L
R=ρ
A
I
We can find the resistance of a wire if we know its
(a) composition (i.e., ρ)
(b) length
(c) cross-sectional area
I
Resistivity ρ is more general than resistance R
(6)
Semiconductors and
Superconductors
Resistance
Chapter 26
Current and Resistance
Electric Current
Resistance
Ohm’s Law
Power
Semiconductors and
Superconductors
Resistance
Chapter 26
Current and Resistance
Electric Current
Resistance
Ohm’s Law
Example 3
Power
Semiconductors and
Superconductors
A rectangular block of iron has dimensions 12 x 12 x 150
mm3 . A potential difference of V is applied to opposite ends
of the block. What is the resistance of the block if
(a) V is applied to the square ends, and
(b) V is applied to the larger ends.
Resistance
Question 2
A copper wire is fabricated that has a gradually increasing
diameter along its length as shown. If an electric current is
moving through the wire, which quantities change along the
length of the wire?
(a) current (i)
(b) current (i) and current density (J)
(c) current density (J)
(d) resistivity (ρ) and current (i)
(e) current (i), resistivity (ρ) and current density (J)
Chapter 26
Current and Resistance
Electric Current
Resistance
Ohm’s Law
Power
Semiconductors and
Superconductors
Resistance
Question 3
When a potential difference is applied to a copper wire, a
current of 1.5 A passes through the wire. If the wire was
removed from the circuit and replaced with an identical
copper wire but with twice the diameter, what current would
flow through the new wire?
(a) 0.38 A
(b) 0.75 A
(c) 1.5 A
(d) 3.0 A
(e) 6.0 A
Chapter 26
Current and Resistance
Electric Current
Resistance
Ohm’s Law
Power
Semiconductors and
Superconductors
Ohm’s Law
Ohm’s Law is an assertion that the current through a device
is always directly proportional to the potential difference
applied to the device.
Chapter 26
Current and Resistance
Electric Current
Resistance
Ohm’s Law
Power
Semiconductors and
Superconductors
I
Many devices behave this way.
I
We often write this as V = iR
Chapter 26
Current and Resistance
Ohm’s Law
Electric Current
Resistance
Ohm’s Law
Power
V = iR
(7)
Key Ideas:
I
The resistance R tells us the current response of the
system.
I
Viewed another way, the resistance R tells us how
much the potential drops as charge move through the
system.
Semiconductors and
Superconductors
Chapter 26
Current and Resistance
Power
Power is the rate of energy transfer
Electric Current
Resistance
Ohm’s Law
Power
Semiconductors and
Superconductors
I
When charge ∆q moves through the circuit, its
potential changes by V
I
But ∆q = i∆t
I
So ∆U = (∆q)V = (i∆t)V
I
Or,
P=
∆U
= iV
∆t
(8)
Chapter 26
Current and Resistance
Power
Electric Current
Resistance
What is the power through a resistor?
Ohm’s Law
Power
I
P = iV in general
I
Voltage drop across a resistor: V = iR
I
Power dissipated by a resistor:
P = i2 R = V 2 /R
I
Each version has R.
I
The units of power are watts, W.
Semiconductors and
Superconductors
(9)
Resistance
Question 4
Chapter 26
Current and Resistance
Electric Current
Resistance
The insulated wiring in a house can safely carry a maximum
current of 18 A. The electrical outlets in the house provide
an alternating voltage of 120 V. A space heater when
plugged into the outlet operates at an average power of 1500
W. How many space heaters can safely be plugged into a
single electrical outlet and turned on for an extended period
of time?
(a) zero
(b) one
(c) two
(d) three
(e) four
Ohm’s Law
Power
Semiconductors and
Superconductors
Power
Chapter 26
Current and Resistance
Electric Current
Resistance
Example 4
Ohm’s Law
Power
Semiconductors and
Superconductors
You are given a uniform wire made of Nichrome with a
resistance of 72 Ω. Find how much energy is dissipated in
the wire if
(a) V = 120 V is applied across the wire;
(b) the wire is cut in half and V = 120 V is applied across
each piece.
Semiconductors and Superconductors
Not all materials conduct charge in the same way
I
I
Semiconductors have a very high resistivity
Chapter 26
Current and Resistance
Electric Current
Resistance
Ohm’s Law
I
made of an insulator with small impurities
Power
I
the process is called “doping”
Semiconductors and
Superconductors
Superconductors have no resistivity
I
usually only work at low temps (about -450◦ F)
I
this is a quantum effect