Transcript What is an electric current?

```ELECTRICAL CIRCUITS
S.MORRIS 2006
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The CELL
The cell stores chemical energy and transfers it to
electrical energy when a circuit is connected.
When two or more cells are
connected together we call this a
Battery.
The cells chemical energy is used
up pushing a current round a
circuit.
What is an electric current?
An electric current is a flow of microscopic particles
called electrons flowing through wires and
components.
+
-
In which direction does the current flow?
from the Negative terminal to the Positive terminal of a cell.
simple circuits
Here is a simple electric circuit. It has a cell, a lamp
and a switch.
cell
wires
switch
lamp
To make the circuit, these components are connected
together with metal connecting wires.
simple circuits
When the switch is closed, the lamp lights up. This is
because there is a continuous path of metal for the
electric current to flow around.
If there were any breaks in the circuit, the current
could not flow.
circuit diagram
Scientists usually draw electric circuits using symbols;
cell
lamp
switch
wires
Circuit Symbols:
Battery
Resistor
Light-bulb
Switch
Wire
Three general types of circuits:
Closed Circuit - There is a
complete loop with wires
going from one side of the
battery through a resistor(s)
to the other side of the
battery.
Open Circuit - There is not a
complete loop.
Short Circuit - There is a
complete loop, but it does
not contain any resistors.
Only Working
Circuit
Chapter 22: Current and Resistance
•Most modern day appliances require electricity to operate.
•The electricity generates a current which runs through the
appliance.
•The current is made out of charged particles
•Current is defined as the rate at which charge flows through a
surface area perpendicular to the direction of flow.
•For example: if the charges are flowing through a wire, the
area is the cross-sectional area of the wire
CURRENT
Q
I
t
Where: I = current (amps, A)

Q = change in charge or how much charge
flows (Coulombs, C)
t = time (seconds)
Example: The amount of charge that passes through the
filament of a certain lightbulb in 2.00 sec is 1.67 C. Find (a)
the current in the bulb and (b) the number of electrons that
pass through the filament in 5.00 sec.
Q 1.67C
I

 0.835A
t 2.00s
(b) First find the total charge in 5.00 sec. Then find how many
electrons that is (remember, 1 electron = 1.6x10-19C)
Q
0.835A 
5.00s
1electron
19
Q  4.175C 
 2.6110 electrons
19
1.6 10 C
Electric Current Flow
In real circuits it is electrons (-charges) that are
flowing. We call this kind of current “electron current”.
However, for simplicity we assume that all charge
carriers are positive. We draw current arrows in the
direction that such charges would move….that is, positive to
negative.
Historically this is because Ben Franklin
carriers in a metal being positive.
For an electric current to flow between two
points, two conditions must be met.
1. There must be a conducting path between the
points along which the charges can move.
WHY? The conducting material allows the
charges to move.
2. There must be a difference of electric potential
(volts) between the two points.
WHY? The potential difference makes the charges
want to move.
What does potential have to do with it?
Water doesn’t flow in the pipe when both ends
are at the same height. That is to say, water will not
flow when both ends of the pipe are at the same
potential (height).
But tip the pipe, increase the
potential of one end so there is a
difference in potential across the
ends of the pipe, and water will
flow.
And so it is with
electric current flow.
You get current to flow through a conductor by applying
a voltage across it. This voltage creates an electric field within
the conductor. This electric field in turn creates a force on all
the charges within the conductor. If charges are free to move,
as in a conductor, they do. In solids these charges are always
electrons.
Current flow is the measure of electrons moving within a
conductor.
Amperes, or amps, measures the amount of current that flows.
Amps is named after a 19th-century French physicist, AndreMarie Ampere, and abbreviated A.
-
+
An ampere is a fairly large amount of current:
0.1 A flowing between your hands across your
heart will kill you. (Fortunately, your body has
fairly high resistance so it takes a substantial
voltage to drive that much current.)
Voltage
Voltage, or electric potential, is measured in volts,
named after a physicist named Volta, and abbreviated V.
Most small batteries (size AAA, AA, C, D) are 1.5 V;
there is the familiar box-like 'transistor' 9 V battery, and car
batteries are 12 V.
In contrast, high-voltage lines have many thousands
of volts between them.
electricity is the source of electrons.
Myth: Electrons come from the power utility
company.
Some people think that the electrical outlets in
the walls of their homes are a source of
electrons. They think that electrons flow from
the power utility through the power lines and
into the wall outlets of their homes.
Basis for the Myth: In a hardware
store you can buy a water hose that is
empty of water. But you can't buy a
piece of wire, an "electron hose," that
is empty of electrons.
The truth: The source of electrons in a
circuit is the conducting circuit
material itself. The outlets in homes
are AC. Electrons make no net
migration through a wire in an AC
circuit.
Power utilities do not sell electrons. They sell energy.
You supply the electrons.
So when you are jolted by an
electric shock, the electrons making
up the current in your body originate
in your body. Electrons do not come
out of the wire and through your body
and into the ground; energy does.
The energy simply causes free
electrons in your body to vibrate in
unison. Small vibrations tingle; large
vibrations can be fatal.
A resistor is anything that electricity
cannot travel through easily.
When electricity is forced through a resistor, often
the energy is changed into another form of energy, such as
light or heat. The reason a light bulb glows is that the
electricity is forced through tungsten, which is a resistor.
The energy is released as light and heat.
Electrical Resistance
Resistance is sort of like the friction associated with moving charge
or electrical current.
•Good conductors, like most metals, allow current to flow without
much loss.
•Poor conductors, like most non-metals, impede the flow of current
to a great extent.
•Superconductors like very cold niobium-tin, are special
substances that allow current to flow with essentially zero loss;
•Semiconductors, like silicon, are either good or poor conductors
depending on certain conditions.
Resistance
The resistance of a conductor that obeys Ohm’s law
depends upon four factors:
1. The material of which it is composed; the ability to carry
an electric current varies more than almost any other
physical property of matter.
2. Its length (L): The longer the conductor, the greater its
resistance.
3. Its temperature . Resistance generally increases as
temperature increases.
4. Its cross-sectional area (A). The thicker the conductor the
less its resistance.
It’s much easier to travel on an
open highway than a rocky trail.
Also, the longer & hotter
the trip, the harder it is.
It’s much easier to travel on an
open highway than a rocky trail.
The filament in a 60 W light bulb has a
larger resistance than the filament in a
100 W light bulb. If both filaments are
made out of tungsten, how do their
length’s & cross-sectional areas compare?
60 W (large resistance) = long & thin filament
100 W (small resistance) = short & thick filament
You are wiring up a circuit and want the
lowest resistance possible. What type of
wire should you choose?
Rubber or Copper
Thick or Thin
Lots of extra wire or just enough to get the job done.
Run the circuit at a high temperature or a low temperature

Ohm’s Law
Resistance quantifies how much current
you get across something per volt applied.
R = Resistance ()
V
R
I
I = Current (A)
or
V  IR
V = Voltage (V)
Resistance has units of Volts/Amp, which
get another name, ohms, represented by the
 omega Ω .
Greek letter
How much voltage difference does it take for .25amps
to flow through a resistance of 8.0 Ω ?
1) 1.0 V
2) 2.0 V
3) 4.0 V
4) 8.0 V
V  IR
V  (0.25 A)(8.0)
5) 32 V
POWER
P = IV
But V = IR
P = I2R
so….
What is the current flow to a 1.0x104W
hairdryer plugged into a 120V socket?
1) 83A
2) 10.A
3) 12. A
4) 16A
5) 20.A
Ohm’s Law Equations:
V = IR
P = IV
P = (V/R)V = V2/R
P = V2/R
Lets say you want to protect your new
speakers by putting a fuse in line with the
speakers. Assume you have 100.W speakers
and that they are 8.0 ohms. What size (amps)
fuse should you use so that it just blows at 100.
P = I2 R
Watts?
100W = I2 8
I2 = 100/8 = 12.5
I = 3.5 amp fuse
Two light bulbs operate from 120V, but one has a
power rating of 25W, and the other has a power rating of
100W. Which bulb has the higher resistance?
1) 25W bulb
2) 100W bulb
4) Depends on the bulb
3) Both the same
Thermal Energy
The sole purpose of some appliances is to produce
heat (electric stove, toaster, hair dryer). These
appliances have a resistor in them which turns
electrical energy into thermal energy (or heat)
E = P*t = I2R*t = (V2/R)*t
P = I2R
P = V2/R
E = Thermal Energy (J
P = Power (W)
t = time (sec)
I = Current (A)
R = Resistance ()
A fuse is used to prevent
an appliance from
1 battery
receiving too much
current. If the current
1 battery holder
goes through a fuse first
2 wires
and is too much for the
1 VERY small pinch of steel fuse to handle, the fuse
will burn up, protecting the
wool
appliance.
Each mini-group needs:
There are two ways to put
resistors into a circuit.
1. Resistors can be in series
OR
2. Resistors can be in parallel
Resistors in Series
Resistors are considered to be in series if the
current must go through all of the resistors in
order.
The current (amps) through all resistors in series
is the same.
The voltage across resistors in series may be
different
The rate of electron flow (or current) is
determined by which resistor?
The resistor with the largest amount of ohms.
R1
Series Resistors
Itotal = I1 = I2 = I3
R2
R3
Amps
Req = Rtotal = R1 + R2 + R3
Voltage is calculated with
Ohm’s Law
Q
Example 1: A circuit has three
8.0 , 5.0  and a 12  resistors
in series along with a 24 V
battery.
Draw the circuit.
Calculate the total resistance of the circuit.
Calculate the total current through the circuit.
 What is the current through each resistor?
Calculate the voltage across each resistor.
Resistors in Parallel
 Resistors are considered to be in parallel if the current is
shared between multiple resistors.
 The current (amps) through all resistors in parallel may be
different.
 The voltage across all parallel resistors is the same.
 Will a resistor with a large resistance have more or less
current through it then a resistor with a small resistance?
The resistor with a large resistance will have a
smaller current then the resistor with the smaller
resistance.
Parallel Resistors
Current is calculated with
Ohm’s Law
1
Rtotal
1
1
1
 

R1 R2 R3
Vtotal = V1 = V2 = V3
Example 2: A circuit has three
resistors: 8.0 , 5.0  and a 12 
resistors in parallel along with a
24 V battery.
Draw the circuit.
Calculate the total resistance of the circuit.
Calculate the total current through the circuit.
 What is the voltage across each resistor?
Calculate the current across each resistor.
Electrical Outlets
Electrical outlets provide electric potential
(or the voltage) for any appliance plugged
in to it.
In the United States ALL outlets provide
120 V (in Europe it is 240 V)
Light bulbs are made to be the
only appliance plugged into a
socket.
The power rating of a light bulb (25 W or
100 W…) is as if that bulb was the only
bulb plugged in to a 120 V power source.
The resistance of a light bulb is calculated
by knowing the power rating and the
voltage (120 V)
Current and actual voltage used by a light
bulb depends on the circuit.
Example 3: What will the power output be if
an American-made 45 W light bulb is
plugged in to a 310 V power source?
Using 120 V, calculate the resistance of the light
bulb.
V2
1202
P
 45W 
 R  320
R
R
Using the resistance and the voltage of the new
source, calculate the new power
V2
3102
P
P
 P  3.0 x102W
R
320
As more identical resistors R are added to the
parallel circuit shown, the total resistance
between points P and Q …
R
1. Increases
2. Remains the same
Q
…
3. Decreases
P
As more identical resistors R are added to the
parallel circuit shown, the total resistance between
points P and Q …
1. Increases
2. Remains the same
3. decreases
R
P
Q
…
Q
When one bulb is unscrewed, the other bulb will
remain lit in which circuit…
1. I
2. II
3. Both
4. Neither
Circuit II
Circuit I
When one bulb is unscrewed, the other bulb will remain
lit in which circuit…
1. I
2. II
3. both
4. neither
Circuit I
Circuit II
A 25W bulb and a 100W bulb are connected in
series. Which bulb will glow brighter?
25W
100W
120V
The Light Bulbs are really Resistors
A) Calculate the resistance for each resistor shown.
B) Calculate the total resistance of the circuit.
C) Calculate the current through each resistor.
D) Calculate the power used by each resistor.
E) Calculate the voltage across each resistor.
25W
100W
120V
Part A.
25W Bulb
100W Bulb
2
V
P
R
V2
R
P
120
R
25
2
R  576
120 2
R
100
R 144
B) Calculate the total circuit resistance Rtotal
Rtotal = R1 + R2
Series Resistors
= 576 + 144
= 720 
144 
576 
120V
C) Calculate the total circuit current (I)
V 120volts
I 
 .167 amps
720
R
720 
 .167 amps
 .167 amps
120V
D) Calculate the Power used by each resistor.
25 W Bulb
100 W Bulb
P = I2R
P = I2R
= .1672 x 576
= 16 watts
= .1672 x 144
= 4 watts
144 
576 
 .167 amps
120V
E) Calculate the Voltage across each resistor.
25W Bulb
100W Bulb
V = IR
V = IR
= .167 x 576
= 96 volts
= .167 x 144
= 24 volts
120 volts
96 volts
24 volts
144 
576 
 .167 amps
120V
E) Consider the Percent Power Needed to
Light Each Bulb
100 W Bulb
4 watts
 100  4 percent
100 watts
25 W Bulb
16 watts
 100  64 percent
25 watts
Q
The circuit below consists of two identical light bulbs
burning with equal brightness and a single 12V battery.
When the switch is closed, the brightness of bulb A…
A
1. Increases
2. Decreases
3. Remains unchanged
The circuit below consists of two identical light bulbs
burning with equal brightness and a single 12V battery.
When the switch is closed, the brightness of bulb A…
1. Increases
2. decreases 3. remains unchanged
When the switch is closed,
bulb B goes out because all of
the current goes through the
wire parallel to the bulb. Thus,
the total resistance of the circuit
decreases, the current through
bulb increases, and it burns
brighter.
A
Q
Which bird is in trouble when the switch is closed?
1) Bird 1
2) Bird 2
3) Neither
4) Both
1
2
Which bird is in trouble when the switch is closed?
1) Bird 1 2) bird 2 3) neither
1
2
4) both
Charge flows through a light bulb. Suppose a wire
is connected across the bulb as shown. When the wire
is connected…
1. All the charge continues to
flow through the bulb, and
the bulb stays lit.
2. Half the charge flows
through the wire, the other
half continues through the
bulb.
3. Essentially all the charge
flows through the wire and
the bulb goes out.
4. None of these.
Q
Analyze the circuit:
A) Calculate Rtotal
B) Calculate the current through each resistor.
C) Calculate the voltage through each resistor.
16
Parallel:
120V
1
R1 23
R123 

1 1
1
4
 

16 32 32 32
32
 8
4
16
32
32
120V
Series:
R123-4=8+16
R1234=24
16
8
Make
These are in parallel so their
voltage is the same along with
chart:
the total voltage
R
16
120V
16
32
32
All these numbers
will be the same.
1
16
2
16
3
32
4
32
234
8
1234
24
I
V
120
Make
These are in series so their
current is the same along with
chart:
the total current
R
16
120V
8
All these numbers
will be the same.
1
16
2
16
3
32
4
32
234
8
1234
24
I
V
120
Fill out the chart with V=IR
V = IR
V = IR
120 = I (24)
V = (5) (16)
I=5A
V = 80 V
V = IR
V = IR
V = (5) (8)
40 = I (16)
V = 40 V
I = 2.5 A
V = IR
40 = I (32)
I = 1.25 A
1
R
16
I
5
V
80
2
16
2.5
40
3
32
1.25
40
4
32
1.25
40
234
8
5
40
1234
24
5
120
Another way to do the problem (without
the chart)
I=V/R
120V
I=120v/24
I=5 amps
24
V=IR
V=(5)(16)
V=80volts
120V
16
80volts
5amps
V=IR
V=(5)(8)
V=40volts
8
40volts
120volts
I=V/R
=40volts/16 
=2.5 amps
120V
I=V/R
=40volts/32 
=1.25 amps
5 amps
5 amps
16
16
32
80volts
32
40
volts
When the series circuit shown is connected, Bulb A is
brighter than Bulb B. If the positions of the bulbs were
reversed…
1. Bulb A would again be brighter
2. Bulb B would be brighter
3. They would be equal brightness
When the series circuit shown is connected, Bulb A is
brighter than Bulb B. If the positions of the bulbs were
reversed…
1. Bulb A would again be brighter
2. Bulb B would be brighter
3. They would be the same
The bulbs are connected in series, so the same
current passes through both of them. Different
brightnesses indicate different filament resistances.
Bulb A is NOT brighter because it is “first in line”
for the current of the battery! After all, electrons
deliver the energy, and they flow from negative to
positive --- in the opposite direction!
Example: Find the voltage and
current for each resistor.
6
3
3
6
4
12
18 volts
2
6
3
3
6
4
12
18 volts
2
3
3
3
4
12
18 volts
2
3
3
3
4
12
18 volts
2
6
3
4
12
18 volts
2
6
3
1
Rtotal
1
Rtotal
1
1
 
R1 R2
1 1
 
4 12
Rtotal 3
4
12
18 volts
2
6
3
3
1
Rtotal
1
Rtotal
1
1
 
R1 R2
1 1
 
4 12
Rtotal 3
18 volts
2
6
3
3
18 volts
2
6
3
5
18 volts
6
1
Rtotal
3
1
1
 
6 5
5
Rtotal  2.73
18 volts
3
2.73
18 volts
3
2.73
18 volts
5.73
18 volts
Now, find the total current flowing
5.73
V
I
R
18volts
I
5.73
I  3.14 amps
18 volts
9.42volts
3
3
V=IR
V=(3.14)(3)
4
V=9.42
12
18 volts
6
6
2
18-9.42
8.57volts
9.42volts
3
3.14 amps
6
3
6
4
12
18 volts
2
18-9.42
8.57volts
9.42volts
3
3.14 amps
6
4
12
18 volts
2
18-9.42
8.57volts
9.42volts
3
3.14 amps
6
4
12
18 volts
2
18-9.42
8.57volts
9.42volts
6
3
3.14 amps
5
18 volts
18-9.42
8.57volts
9.42volts
6
3
V
I
R
5
8.57volts
I
6
I  1.43amps
18 volts
18-9.42
8.57volts
9.42volts
3
3.14 amps
6
1.43 amps
5
1.71 amps
18 volts
18-9.42
8.57volts
9.42volts
6
1.43 amps
3
4
3.14 amps
1.71
amps
V=IR
V=(1.71)(2)
V=3.42volts
12
18 volts
3.42Volts
2
1.71
amps
18-9.42
8.57volts
9.42volts
6
5.15 volts
3
4
3.14 amps
1.71
amps
12
18 volts
1.43 amps
3.42Volts
2
1.71
amps
18-9.42
I=V/R
I=5.15volts/12
9.42volts
I= 0.43 amps
8.57volts
6
5.15 volts
3
4
3.14 amps
1.71
amps
12
0.43 amps
18 volts
1.43 amps
3.42Volts
2
1.71
amps
18-9.42
I=V/R
I=5.15volts/4
9.42volts
I= 1.28 amps
Or…
8.57volts
1.71 amps – 0.43 =
6 1.28 amps
5.15 volts
3
4
3.14 amps
1.71
amps
12
0.43 amps
18 volts
3.42Volts
2
1.71
amps
6
3
3
6
4
12
18 volts
2
6
3
3
6
4
12
18 volts
2
6
3
3
6
4
12
2
18 volts
Q
Given: R1=1; R2=2 ; R3=3 . Rank the bulbs
according to their relative brightness
1.
R1 > R 2 > R 3
2.
R 1 > R 2 = R3
3.
R 1 = R 2 > R3
4.
R1 < R 2 < R 3
5.
R 1 = R 2 = R3
R1
R2 R3
15
Given: R1=1; R2=2 ; R3=3 . Rank the bulbs
according to their relative brightness
1.
R1 > R 2 > R 3
2.
R 1 > R 2 = R3
3.
R 1 = R 2 > R3
4.
R1 < R 2 < R 3
5.
R 1 = R 2 = R3
R1
R2 R3
2
V
P  IV  I R 
R
2
Q
If the four light
bulbs in the figure below
are identical, which
circuit puts out more
total light?
1. I
2. II
3. Same
Circuit II
If the four light
bulbs in the figure below
are identical, which
circuit puts out more
total light?
1. I
2. II
Circuit II
3. Same
The resistance of two light bulbs in parallel in
smaller than that of two bulbs in series. Thus the
current through the battery is greater for circuit I
than for circuit II.
Since the power dissipated is the product of
current and voltage, it follows that more is
dissipated in circuit I.
```