Electric Circuits
Thomas Edison with first commercial light bulbs
Types of Materials
• Conductor: A conductor allows electric current to pass
through. Example: Copper, iron, nickel. Conductors are also
known as metals. Wires and strips of metals conduct
electricity with almost no resistance.
Some metals (lead, steel) don’t conduct as well as others.
(Sometimes, if the surface of the metal is very oxidized, the
metal appears not to conduct as well.)
• Graphite is a poor conductor (a “semiconductor”) – it allows
current, but with more resistance.
• Insulator: An insulator does not allow electric current to pass
through. Example: Air, glass, rubber, plastic, dry wood, etc.
Current
• Current is formed by moving charges. (Current
is measured in amperes, or “amps” for short.)
• For current to flow, you need:
(i) A closed path joining the two ends of the
supply.
(ii) A supply, such as a battery, to “pump” the
current back up to the “top” of the path after the
current flows “down” through devices such as
bulbs. The strength of the pump is measured in
volts.
Unlike water flowing
in a river, current
needs a pump to get
it back to the “top” so
it can make a round
trip (“circuit”).
Current flow
No Current !
No Current !
Current Flow
Outside the battery, current flows from one terminal to the other,
and then gets “pumped” back to the first terminal within the
battery.
For the circuit shown on the right, the current is the same in the
battery as in the wire.
Resistance
If the conducting
path has very
small resistance
(like a wire),
current will be
very large and the
wire will get hot!!
Adding a
resistance,
like a bulb
decreases
the current
and therefore
the heating of
the wire.
Now the
filament in
the bulb gets
very hot from
the current,
and glows!
Structure of a light bulb
Partial Vacuum
(low pressure,
non-reactive
gas)
The greater the current
through the filament, the
hotter and brighter it will be.
Filament is the high
resistance part ⇒ light
and heat.
Note: In an “incandescent”
light bulb, most of the energy
goes into heating the filament,
and relatively little comes out
as light: it wastes a lot of
energy. Incandescent bulbs
are being replaced with
“fluorescent” lights and “lightemitting diodes”, which
convert a much larger fraction
of their input energy into light;
i.e. they are much more
energy efficient.
Symbols used in circuit diagram
Battery
Light bulb
Use lines
for connecting wires and dots
connections between wires (e.g. at tees).
Switch
for
Resistance and current through an
ideal battery
3 units
12 units
3
12
Total
resistance =
4R
Total
resistance =
R
3
12
12
3
The current through a battery is inversely proportional to the total resistance
connected to its two terminals: doubling the resistance results in halving the
current, etc.
Resistance and current through a
device
• For simple devices, the voltage across the device is
proportional to the current through it: V / I = R (the
resistance).
This is called “Ohm’s Law”.
• R is measured in ohms: 1Ω = 1V / 1A
• It is correct to say an insulator has a very, very large
resistance, and a conductor has small resistance.
• For incandescent light bulbs, the resistance increases
somewhat as the filament’s temperature increases, and
therefore R varies with current.
(That is why bulbs often “blow-out” when they are first turned on, before
the filament heats up.)
Series Circuit
Two light bulbs
connected in
series
Three light bulbs
connected in
series
Four light bulbs
connected in
series
If you walk along a series circuit, starting from the positive end of the battery,
you will not encounter any branches and you will end up at the negative end
of the battery by the same track.
The “voltage” provided by the battery is therefore divided between the bulbs,
but for each circuit, the same current flows through each bulb and the battery.
Properties of series circuits
0.3 A
0.3 A
0.3 A
0.3 A
0.3 A
0.3 A
0.3 A
0.3 A
The current at every point in a series
circuit (including the battery) is equal,
because the same current flows
through each element.
(But this current depends on the total
resistance, e.g. number of bulbs in
the circuit.)
If this is a 3 Volt battery, if each
bulb is identical, and if the
connecting wires are perfect, what
is the voltage across each bulb?
V
voltmeter
If this is a 3 Volt battery, if each
bulb is identical, and if the
connecting wires are perfect,
what is the voltage across each
bulb?
V
voltmeter
If they are identical, each bulb will have the same voltage. Since there
are 4 bulbs, and the total voltage across them is the battery voltage,
3V, there will be a “voltage drop” of ¾ V across each bulb.
Series Circuit:
The total resistance is the sum of the resistances of each light bulb.
The current through the circuit is inversely proportional to this total resistance
R
R
0.3 A
0.4 A
1.2 A
0.3 A
0.4 A
1.2 A
1.2 A
0.4 A
R
0.3 A
R
R
0.4 A
1.2 A
1.2 A
1.2 A
0.4 A
0.3 A
R
0.3 A
R
0.3 A
0.4 A
0.4 A
0.3 A
Total resistance = R
Total resistance = 3R
R
Total resistance = 4R
Current = 1.2 A
Current = 0.4 A
Current = 0.3 A
[Here, we assumed that the resistance of a bulb is constant, and doesn’t
increase with current, so the results are only approximately correct.]
Parallel Circuit
Two light bulbs
connected in
parallel
Three light bulbs
connected in
parallel
Four light bulbs
connected in
parallel
Each terminal of a bulb in a parallel circuit is connected to a terminal of a
bulb in another branch of the circuit, and to the battery.
Therefore, the full voltage of the battery will be across each bulb!
Properties of Parallel Circuit
0.8 A
0.2 A
0.2 A
0.2 A
0.2 A
0.8 A
Current through the battery is the sum of the currents in each branch.
Note: These are different battery and bulbs than in previous pictures!
Properties of Parallel Circuit
0.8 A
1A
0.2 A
0.2 A
0.2 A
0.2 A
0.2 A
0.2 A
0.8 A
0.2 A
0.2 A
0.2 A
1A
Adding (or removing) one branch will not affect the current in the other
branches (for an ideal battery), but will affect the total current through the
battery.
(This is the way appliances are wired in buildings.)
Problem 1: Three Identical Bulbs –
rate their brightness
Problem 1: Three Identical Bulbs – rate
their brightness
IA
IB
IC
• All the current from the
battery flows through bulb
C.
• The current through C
splits, with half going
through A and half going
through B.
(IA = IB = IC/2)
• The more current that flows
through a bulb, the brighter
it is. Therefore, C is
brighter than A or B, which
are equally bright.
Note: Direction of arrows is arbitrary choice: you could also say IA and IB come
together to make IC.
Problem 2: Three Identical Bulbs –
rate their brightness
Problem 2: Three Identical Bulbs –
rate their brightness
V
IA
IBC
• The full voltage of the
battery appears across bulb
A and also across the series
combination of B and C.
• The same current (IBC) flows
through B and C
• The resistance of B and C in
series is greater than the
resistance of A, so less
current flows in the B-C
branch than in A: IBC < IA.
• The more current that flows
through a bulb, the brighter
it is. Therefore, A is brighter
than B or C, which are
equally bright.
If both batteries are identical (and
ideal) and all bulbs identical, which is
brightest?
#2
#1
#2
#1
• C is the brightest bulb in circuit #1 and A is the brightest bulb in
circuit #2, so we need to compare C1 with A2.
• A2’s leads are across the battery, so it gets all the voltage of the
battery: V(A2) = V(battery).
• In circuit #1, the battery’s voltage is divided between C and the AB
parallel combination, so V(C1) < V(battery).
• Therefore V(A2) > V(C1), and A2 is brighter than C1.
Batteries in Series
If two batteries are placed in series (i.e. negative terminal
on one connected to the positive of the second), the total
voltage provided is the sum of the voltages.
For example, if these are each 1.3 V batteries, the total
voltage is 2.6 V.
Batteries in Series
Therefore, a way to increase the current through a device (e.g. bulb, LED,
motor, … -- shown as box “X” in the figures below) is to add batteries in
series. When the current through a bulb or LED increases, the intensity of
the light increases. When the current through a motor increases, its
speed increases. In some cases (e.g. the LED) the intensity doesn’t
simply double when you double the number of batteries – these are called
non-linear devices. In the case of the LED, it emits (almost) no light with
one battery, but glows intensely with two.
“X” attached to
one battery.
“X” attached to two
batteries.
“X” attached to
three batteries
I
A
Insert ammeter in
series with device, to
measure current
through it
V
Insert voltmeter in parallel
with device to measure
voltage “across” it
Non-Ideal Batteries
2A
1A
1A
Almost I(max)
Real batteries are “non-ideal”: they have a maximum current they can put
out. Therefore, if you put too many bulbs in parallel, the current through
each will drop. Similarly, if you put a short circuit in parallel with the bulbs,
so much current will flow through the short circuit that very little current is
left to flow through the bulbs.
I(max)
switch open:
no current
thru switch
switch
closed
almost no current
thru bulbs
When a battery “dies”, the first thing that happens is that its maximum current drops.
Current has Direction
•
•
•
Electric current corresponds to the flow of
charged particles (usually electrons), and
therefore has a “direction”.
However, the heat generated in an
electrical wire and the light generated by
an incandescent bulb only depend on the
amount of current and not on the direction
it is flowing.
Two devices for which the direction of the
current does matter are the electric motor
and light emitting diode (or LED). In
particular, the direction the motor is turning
switches when the voltage and current
direction are switched. Current can only
flow one way through the LED, lighting it
up – i.e. it will only allow current and light
up if hooked up in the correct direction to
the battery.
Symbol for a motor:
Symbol for an LED
The energy provided by the battery is dissipated in the circuit,
mostly in the light bulb. Similarly, water flowing in a river may
lose its energy in turning the turbine of a generator or water
wheel at a mill.
Energy and Power
• Power is the rate of change of energy:
Power = Energy / time
• The power delivered by an electrical energy source (e.g.
battery) is its voltage times the current flowing through it:
P = V·I
• The power dissipated in any device (e.g. motor, light bulb)
is the voltage across it times the current flowing through it:
P = V·I
• The unit of power is the Watt: 1 W = 1 V · 1 A
[Your utility company charges you for the energy you use:
Energy = power x time. The units they use are kWh = kiloWatt·hours:
1 kWh = 1000 (V· A · hr)]
The voltage in our homes is not supplied by a battery but by a
generator at the power station and a series of “transformers”
which change the voltage to a convenient level, typically 110120 V. In the problem below, we will assume the voltage is 120
V (to keep the math easy). When you buy a “60 W” bulb, it
assumes that it will be connected to 120 V, as shown here.
Sample Problem: What is the
resistance of a “60 W” bulb?
120 V
Hint: first find the current.
Sample Problem: What is the
resistance of a “60 W” bulb?
120 V
Hint: first find the current.
Power = V · I → I = P/V
I =60 W / 120 V = ½ A
Now find the resistance using Ohm’s Law.
Sample Problem: What is the
resistance of a “60 W” bulb?
120 V
Hint: first find the current.
Power = V · I → I = P/V
I =60 W / 120 V = ½ A
Now find the resistance using
Ohm’s Law.
R=V/I
R = 120 V / 0.5 A
R = 240 Ω
Exam
• The exam will include a circuit problem (like homework
problem) and a general question about components of a
circuit.
• 15 Multiple Choice Questions including some on Ohm’s
Law: resistance and power (with easy arithmetic, so no
calculators needed).
MC example: Several light bulbs are connected in
parallel to an ideal battery. What happens if one of them
burns out?
a) The brightness of the others won’t change.
b) The others will become brighter.
c) The others will stay lit, but become dimmer.
d) The others will go out.