With a partner, get a battery, light bulb,
and paper clip.
 Find the two ways to light up the light
bulb using just these three items.
 Draw pictures of what the set up looked
like if you notes


Charge flows from an item with a higher
potential difference to something with a
lower potential difference until the two
have equal charges
› The reason there is static discharge

Electric Current: the flow of charged
particles
If moving a charge against the electric
field, work is done on the charge
 The work done changes the charges
potential energy to a higher value
 The work is equal to the change in the
potential energy
 There is a difference in electric potential
between the two locations
 This difference is represented by ΔV and
called electric potential difference


By definition, electric potential difference
is the difference in electric potential (V)
between the final and initial location of
the charge
› And we drop the Δ as a historical
convention (not a correct one) and just use
V

Standard unit for electric potential
difference is the volt (V), named in honor
of Alessandra Volta
An electric potential difference between
two locations of 12 volts means that one
coulomb of charge will gain 12 joules of
potential energy when moved between
those two locations
 Because it is expressed in volts it is
sometimes referred to as voltage


Two models
› Conventional current: assumes positive
charges flow out of a positive terminal and
travel to the negative terminal.
› Electron Flow: what actually happens, is that
electrons flow out of negative terminals and
travel to the positive terminal

Charge will always flow from one more
charged body to another, but what
makes it continue to flow?
› You need a generator of some sort
› Most common generator is several galvanic
cells (or dry cells) connected together – they
form a battery
 Chemical energy generator
 Other include hydro, steam, and wind
generators
 Photovoltaic cell: solar cell

Electric Circuit: a closed loop in which
current flows
› Includes a charge pump (battery) which
increases potential energy and a device to
reduce potential energy (light bulb)
› Think of a charge pump as the work done by
a roller coaster to get the cars to the top of
the hill, they go down the other side naturally
 Once a positive charge gets from inside the
battery to the positive terminal, it flows
naturally to the negative terminal

In a 9V battery, there is 9
volts of potential difference
between the two terminals.
That means that the battery
must do 9 joules of work
moving a positive charge
from the negative wire to the
positive wire. The positive
charge then gains 9 joules of
potential energy in which it
can deliver to the light bulb
and then come back with
no energy and do it all
again.
When a charge moves through a circuit,
the amount of potential energy it loses is
qV (charge times potential difference)
 So the generator/charge pump/battery
needs to increase the charges potential
and the energy required to do that is qV
 The change in electric energy, E, is equal
to qV

› E = qV

Power is the measure of the rate at
which energy is transferred (P = E/t)
› Transferring 1 joule per second is 1 watt

The rate of flow of electric charge, or
electric current, I, is measured in
coulombs per second
› 1 coulomb per 1 second is 1 ampere, A
› Ammeters measure amperes
Suppose current is flowing at 3C/s (3A)
and the potential difference is 120V,
which means that each charge supplies
the motor/light bulb/etc with 120J
 To find the power delivered we multiply
the current and the potential difference

› P = IV

The power delivered in this situation
would be 360 W

A 6.0 V battery delivers a 0.50 A current
to an electric motor that is connected
across its terminals.
› What power is consumed by the motor?
› If the motor runs for 5 minutes, how much
electric energy is delivered?

The current though a lightbulb
connected across the terminals of a 120V outlet is 0.50 A. At what rate does the
bulb convert electric energy to light?

A car battery causes a current of 2.0 A
through a lamp while 12 V is across it.
What is the power used by the lamp?

The current through the starter motor of
a car is 210 A. If the battery keeps 12 V
across the motor, what electric energy is
delivered to the starter in 10.0 s?

Resistance, R, is the ratio of the potential
difference, V, to the current, I
› R = V/I
› Measured in volts per ampere, which is ohms
 Named after Georg Simon Ohm
› 1 Ω (ohm) is equal to 1 A of flow when 1 volt
is applied

Ohm discovered that a devices
resistance stays the same no matter the
potential difference that is applied
(Ohm’s Law)

Most wires used in circuits have a very
small resistance (they don’t reduce the
potential difference much)
› Factors of resistance – think garden hose
 Small diameter v. large diameter
 Short v. long

Resistors are devices use to have a
specific resistance

A 30.0 V batter is connected to a 10.0 Ω
resistor. What is the current in the circuit?

A lamp draws a current of 0.50 A when it
is connected to a 120 V source.
› What is the resistance of the lamp?
› What is the power consumption of the lamp?
P515
 Redraw the circuits you made at the
beginning of the hour with proper circuit
diagrams


Parallel
› When two devices are connected so they
are parallel to each other
 Voltmeters need to be connected this way

Series
› When two devices are connected so that
the current that flows through one also flows
through the other
 Ammeters need to be connected this way

Substituting new equations learned
PI R
2

A heater has a resistance of 10.0 Ω. It
operates on 120.0 V.
› What is the current through the resistance?
› What thermal energy is supplied by the
heater in 10.0 s?

A 30.0 Ω resistor is connected across a 60
V battery.
› What is the current in the circuit?
› How much energy is used by the resistor in 5
min?
1 kWh is equal to 1000 watts delivered
continuously for 3600 seconds (1 hour)
 A television set draws 2.0 A when
operated on 120 V.

› How much power does the set use?
› If the set is operated for an average of 7.0
h/day, what energy in kWh does it consume
per month (30 days)?
› At $0.11 per kWh, what is the cost of
operating the set per month?

P527: 21, 23, 29, 32, 41, 44