Current and Resistance
February 12, 2014
Physics for Scientists & Engineers 2, Chapter 25
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February 12, 2014
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Resistance and resistivity
!  Resistance R of a resistor is given by
ΔV
R=
i
ΔV
i=
or ΔV = iR
R
!  Units are Ohms
1V
1Ω=
1A
!  Resistivity is a property of the material
E
ρ=
with current density J = i
J
A
!  Resistance depends on resistivity and length and cross
section area of the resistor R = ρ L
February 12, 2014
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Physics for Scientists & Engineers 2, Chapter 25
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Temperature Dependence of
Resistivity
!  The resistivity and resistance vary with temperature
!  For metals, this dependence on temperature is linear over a
broad range of temperatures
!  An empirical relationship for the temperature dependence
of the resistivity of metals is given by
ρ − ρ0 = ρ0α (T −T0 )
!  where
•  ρ is the resistivity at temperature T
•  ρ0 is the resistivity at temperature T0
•  α is the temperature coefficient of electric resistivity
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Temperature Dependence of
Resistance
!  In everyday applications we are interested in the
temperature dependence of the resistance of various devices
!  The resistance of a device depends on the length and the
cross sectional area
!  These quantities depend on temperature
!  However, the temperature dependence of linear expansion
is much smaller than the temperature dependence of
resistivity of a particular conductor
!  The temperature dependence of the resistance of a
conductor is, to a good approximation,
R − R0 = R0α (T −T0 )
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Temperature Dependence
!  Our equations for temperature dependence deal with
temperature differences so that one can use
°C as well as K (T-T0 in K is equal to T-T0 in °C)
!  Values of α for representative metals are shown below
(more can be found in Table 25.1)
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Example: Which Metal is it?
!  A metal wire of 2mm diameter and length 300m has a
resistance of 1.6424 Ω at 20oC and 2.415 Ω at 150oC. Find
the values of α, R(0oC), ρ(0oC) and identify the metal!
!  Solution:
R(150oC)=2.415Ω=R(0oC)(1+α(150-0))
R(20oC)=1.6424Ω=R(0oC)(1+α(20-0))
First equation: R(0oC)=2.451Ω/(1+α(150-0)), substitute in
second equation, solve for α: α=3.9 10-3 oC-1
With α known, use any of the above equations to get
R(0oC)=1.5236Ω. Then use R(0oC)=ρ(0oC)L/A
ρ(0°C) ⋅ 300m
−8
1.5236Ω =
⇒
ρ
(0°C)
=
1.596
⋅10
Ωm
−3
2
0.25π (2 ⋅10 m)
ρ(20°C) = ρ(0°C)(1 + α ⋅ 20) = 1.72 ⋅10 −8 Ωm
Table 25.1: Copper!
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Other Temperature Dependence
!  Most materials have a resistivity that varies linearly with the
temperature under ordinary circumstances
!  However, some materials do not follow this
!  At very low temperatures the resistivity of
some materials goes to exactly zero
!  These materials are called
superconductors
•  Many applications, including MRI
!  The resistance of some semiconducting materials actually
decreases with increasing temperature
!  These materials are often found in high-resolution detection
devices for optical measurements or particle detectors
February 12, 2014
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Electromotive Force and Ohm’s Law
!  To make current flow through a resistor one must
establish a potential difference across the resistor
!  This potential difference is termed an
electromotive force or emf
!  A device that maintains a potential difference is called an
emf device and does work on the charge carriers
!  The emf device not only produces a potential difference but
supplies current
!  The potential difference created by the emf device is termed
Vemf
!  We will assume that emf devices have terminals that we can
connect and the emf device is assumed to maintain Vemf
between these terminals
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Circuit
!  Examples of emf devices are
•  Batteries that produce emf through chemical reactions
•  Electric generators that create emf from electromagnetic induction
•  Solar cells that convert energy from the Sun to electric energy
!  We will often use DC (direct current) power supplies, which
supply emf just like a battery
!  A circuit is an arrangement of electrical components
connected together with ideal conducting wires (i.e., having
no resistance)
!  Electrical components can be sources of emf, capacitors,
resistors, or other electrical devices
!  We will begin with simple circuits that consist of resistors
and sources of emf
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Ohm’s Law
!  Here a source of emf provides a potential
difference Vemf across a resistor with
resistance R
!  The relationship between the potential
difference and the resistance is given by
Ohm’s Law
Vemf = iR
!  The current in the circuit flows through the resistor, the
source of emf, and the wires
!  The change in potential of the current in the circuit must
occur in the resistor
!  The change is called the potential drop across the resistor
February 12, 2014
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Ohm’s Law
!  We can represent the previous circuit in a different way,
making it clearer where the potential drop happens and
showing which parts of the circuit are at which potential
!  On the next slide, the top part of the figure shows the
previous circuit while the bottom part shows the same
circuit but with the vertical dimension representing the
value of the electric potential at different points around the
circuit
!  The potential difference is supplied by the source of emf,
and the entire potential drop occurs across the single
resistor
!  Ohm’s Law applies for the potential drop across the resistor
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February 12, 2014
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Resistors in Series
!  Circuits can contain more than one resistor and/or more
than one source of emf
!  The analysis of circuits with multiple resistors requires
different techniques
!  Resistors connected such that all the current in a
circuit must flow through each of the resistors are connected
in series
!  For example, two resistors R1 and R2
in series with one source of emf with
potential difference Vemf are shown
in this circuit
February 12, 2014
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Resistors in Series
!  The potential drop across resistor R1 is ΔV1
!  The potential drop across resistor R2 is ΔV2
!  The potential drops must sum to the potential difference
Vemf = ΔV1 + ΔV2
!  The current flowing through each resistor is the same
Vemf = iR1 + iR2 = i ( R1 + R2 ) = iReq
Req = R1 + R2
!  The two resistors in series can be replaced with an
equivalent resistance equal to the sum of the two resistances
!  Generalized for n resistors in series:
n
Req = ∑ Ri
i=1
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February 12, 2014
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Internal Resistance of a Battery
!  When a battery is not connected in a circuit, the voltage
across its terminals is Vt
!  When the battery is connected in series with a resistor with
resistance R, current i flows through the circuit
!  When current is flowing, the potential difference, Vemf,
across the terminals of the battery is less than Vt
!  This drop occurs because the
battery has an internal resistance
Ri that can be thought of as being
in series with the external resistor
Vt = iReq = i ( R + Ri )
!  The battery terminals are points
A and B
February 12, 2014
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Internal Resistance of a Battery
PROBLEM
!  Consider a battery that has Vt = 12.0 V when it is not
connected to a circuit
!  When a 10.0 Ω resistor is connected with the battery, the
potential difference across the battery’s terminals drops to
10.9 V
!  What is the internal resistance of the battery?
SOLUTION
!  The current flowing through the external resistor is
ΔV 10.9 V
i=
=
= 1.09 A
R 10.0 Ω
February 12, 2014
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Internal Resistance of a Battery
!  The current flowing through the complete circuit must be
the same as the current flowing through the external resistor
Vt = iReq = i ( R + Ri )
Vt
( R + Ri ) = i
Vt
12.0 V
Ri = − R =
− 10.0 Ω = 1.0 Ω
i
1.09 A
!  The internal resistance of the battery is 1.0 Ω
!  You cannot determine if a battery is still functional by
simply measuring the potential difference across the
terminals
!  You must connect a resistance across the terminals and then
measure the potential difference
February 12, 2014
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Resistors in Parallel
!  Instead of being connected in series so that all the current
must pass through both resistors, two resistors can be
connected in parallel, which divides the current between
them
!  The potential drop across
each resistor is equal to the
potential difference provided
by the source of emf
!  We can better visualize the potential drops, we can look at
the same circuit in 3D
February 12, 2014
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February 12, 2014
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Resistors in Parallel
!  The voltage drop across each resistor is equal to the
potential difference provided by the source of emf
!  Using Ohm’s Law we can write the current in each resistor
Vemf
i1 =
R1
Vemf
i2 =
R2
!  The total current in the circuit must equal the sum of these
currents
i = i1 + i2
!  Which we can rewrite as
⎛ 1 1⎞
Vemf Vemf
i = i1 + i2 =
+
= Vemf ⎜ + ⎟
R1
R2
⎝ R1 R2 ⎠
February 12, 2014
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Resistors in Parallel
!  We can then rewrite Ohm’s Law for the complete circuit as
i = Vemf
1
Req
!  .. where
1
1 1
= +
Req R1 R2
!  We can generalize this result for two parallel resistors to
n parallel resistors
n
1
1
=∑
Req i=1 Ri
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