Chapter 18
Electric Circuits
Electromotive Force and Current
In an electric circuit, an energy source and an energy consuming device
are connected by conducting wires through which electric charges move.
Electromotive Force and Current
Within a battery, a chemical reaction occurs that transfers electrons from
one terminal to another terminal.
The maximum potential difference across the terminals is called the
electromotive force (emf).
The emf has units of Volts, e.g. a 12 V battery has an emf of 12 V
Zn – Cu cell in dilute acid (H2SO4) electrolyte
-
+
2H + + 2e− → H 2
Zn → Zn 2+ + 2e−
ZnSO4
precipitate
H2 gas
H2SO4
+ H 2O
Electromotive Force and Current
The electric current is the amount of charge per unit time that passes
through a surface that is perpendicular to the motion of the charges.
Δq
I=
Δt
One coulomb per second equals one ampere (A).
Electromotive Force and Current
If the charges move around the circuit in the same direction at all times,
the current is said to be direct current (dc) --> e.g. simple circuits with
batteries are normally dc.
If the charges move first one way and then the opposite way, the current is
said to be alternating current (ac) --> e.g. the current coming out of your
electric outlet is ac.
Electromotive Force and Current
Example: A Pocket Calculator
The current in a 3.0 V battery of a pocket calculator is 0.17 mA. In one hour
of operation, (a) how much charge flows in the circuit and (b) how much energy
does the battery deliver to the calculator circuit?
(a)
(
)
Δq = I (Δt ) = 0.17 ×10 −3 A (3600 s ) = 0.61 C
ΔV = -W/q
(b)
Energy = Charge ×
Energy
= (0.61 C )(3.0 V ) = 1.8 J
Charge
Electromotive Force and Current
Conventional current is the hypothetical flow of positive charges that would
have the same effect in the circuit as the movement of negative charges that
actually does occur.
Ohm’s Law
It is found experimentally that the
current flowing through a circuit is
proportional to the voltage across the
circuit, i.e.,
I is proportional to V
The resistance (R) is defined as the
ratio of the voltage V applied across
a piece of material to the current I through
the material.
Ohm’s Law
OHM’S LAW
The ratio V/I is a constant, where V is the
voltage applied across a piece of material
and I is the current through the material:
V
= R = constant
I
or
Resistance
SI Unit of Resistance:
volt/ampere (V/A) = ohm (Ω)
V = IR
Ohm’s Law
Resistance, R, represents to what extent
the current can flow freely in the circuit, i.e.
the larger R, the more the electrons scatter
with atoms in the material.
These scatterings slow down electrons and
transfer energy as heat to the material.
To the extent that a wire or an electrical
device offers resistance to electrical flow,
it is called a resistor.
Ohm’s Law
Example: A Flashlight
The filament in a light bulb is a resistor in the form
of a thin piece of wire. The wire becomes hot enough
to emit light because of the current in it. The flashlight
uses two 1.5-V batteries to provide a current of
0.40 A in the filament. Determine the resistance of
the glowing filament.
V
3.0 V
R= =
= 7.5 Ω
I 0.40 A
Resistance and Resistivity
For a wide range of materials, the resistance R of a piece of
material of length L and cross-sectional area A is
L
L
R=ρ
A
ρ
A
Resistivity, in units of ohm·meter (Ω m)
Intuitively, this equation makes sense:
For larger L --> R increases
For larger A --> R decreases
Resistance and Resistivity
L
R=ρ
A
Resistance and Resistivity
Example: Longer Extension Cords
The instructions for an electric lawn mower suggest that a 20-gauge extension
cord can be used for distances up to 35 m, but a thicker 16-gauge cord should
be used for longer distances. The cross sectional area of a 20-gauge wire is
5.2x10-7 m2, while that of a 16-gauge wire is 13x10-7 m2. Determine the
resistance of (a) 35 m of 20-gauge copper wire and (b) 75 m of 16-gauge
copper wire. The resistivity of Cu is 1.72 x 10-8 Ωm.
(a)
(b)
(
)
L 1.72 ×10 −8 Ω ⋅ m (35 m )
R=ρ =
= 1.2 Ω
-7
2
A
5.2 ×10 m
(
)
L 1.72 ×10 −8 Ω ⋅ m (75 m )
R=ρ =
= 0.99 Ω
-7
2
A
13 ×10 m
Resistance and Resistivity
The resistivity of a material depends on its temperature as,
ρ = ρ o [1 + α (T − To )]
Resistivity at
temperature T0
temperature coefficient
of resistivity
ρoL/A
or, using
L
R=ρ
A
R = Ro [1 + α (T − To )]
As your book discusses, if the temperature of some materials
is lowered below some critical temperature, TC, they become
Superconductors and ρ = 0, i.e. they have no resistance to
current flow and so currents can persist in them indefinitely,
e.g. for lead, TC = 7.20 K
Example. The heating element of a kitchen stove has a resistivity of
6.8 x 10-5 Ωm at T0=320oC and temperature coefficient of resistivity
of 2.0 x 10-3 (oC)-1. Find the resistance of the heater wire at 420oC.
First find ρ at T = 420oC:
ρ = ρ0[1 + α(T - T0)]
= (6.8 x 10-5)[1 + (2.0 x 10-3)(420 - 320)]
= 8.2 x 10-5 Ωm
Now find R at 420oC using this and the
dimensions of the heater wire shown:
ρL
R=
=
A
−5
8.2
×10
(
) (1.1)
3.1×10
−6
= 29 Ω
If the voltage across the element is 120 V
the current drawn at 420oC will be
I=
V 120
=
= 4.1 A
R 29