Electricity
BHS Physical Science
K Warne
Grade 9 Revision
• Revision Presentation
Electrical Circuits
The Ammeter measures
the ……………flowing in
the circuit. (…….. A)
The Voltmeter Measures
potential difference or
…………….. in volts. (V)
The Resistance of the
Resistor is given in
…………… (Ω).
V1 = V2
V1
Voltmeter
A Ammeter
Resistor
V2
Voltage across …………..= voltage in ……………
Electric Current in a Conductor




Conventional current
- __________________
Direct current moves in
_______________.
Alternating current ___________________
continuously
Maintaining a current
Conductor - ___________
 _____________________
 _____________________ SOURCE

+
-
Conventional current
+
e-
-
ee- ee- e+ e+
- e- + e-+ + e- + e-+ + + +
e
e
+ + -+ + + e- + e-+
+ + + e+
e
+ e-+ +
+ - + e-+ + - + e-+ +
e
e
e
+ + e-+ + + e- + + + + +
<------- electrons
-
Conventional current is the movement “_____________” ------>
of from + to - in a conductor.
Electric Current & Voltage

+
The VOLTAGE is how
much ENERGY they
have.
-
Conventional current
+
Current is the number of soldiers moving past a point.
e-
-
Potential Difference
The DIFFERENCE in POTENTIAL (energy) per unit ………………. of the
current flowing between two points in the circuit.
Measured by a ………………...
W or E (J)
………….(J)
V=
1.50
V
Volts =
Q (C )
………….(C )
Voltmeter
_
+
Resistor
The Voltmeter:
• is connected in ……………. to another component in
the circuit (the current does NOT flow through it)
• has a …………. resistance
• Is connected positive to positive - negative to
V
negative.
R
POTENTIAL DIFFERENCE
Example 1:
Calculate the potential difference
between two points if 20 J of
work are required to move a
charge of 2 C.
A
?
2 coulombs of charge
Resistance R
20J Energy lost
Example 2:
Calculate the work done in moving a
charge of 5 C through a potential
difference of 2 V.
2V
A
5 coulombs of charge
Resistance R
?? Energy lost or work done??
EMF - Electro Motive Force
• Emf is the
……………
amount of
…………… that
the cell can
produce (per unit
charge).
• Measured when
the current in the
circuit is ………..
Emf
400V
VCell
Open Circuit!!
Vcell = EMF
A I = 0A
Vex
Vcrt.= ….. v
Resistance R
Emf = …..cell
External Potential Difference
…………
Circuit!!
50V
Vcell
(EMF = 400V)
350V
A I = IA
50V
Vex= Vcell
Vex
400V
350V
Resistance R
Energy …………….. by
battery is lost by
resistance in the
circuit.
The EMF of the cell is
equal to the sum of the
………………..cell voltage
and the ………… voltage.
This continues until the
cells have no more
energy.
Emf = V….. + V……
Measuring Current
An Ammeter measures the ……………… flowing through the
circuit.
_
1 A
Ammeter
+
The current is the number of charges passing a point in one
second.
I = Q/t 1A = …..C/……s
The Ammeter:
• is connected in ……….in the circuit (the current flows
through it)
• has a …………resistance
• Is connected positive to positive - negative to negative.
A
Calculating Current
Calculate the current flowing through the circuit.
_
0.53
A
Ammeter
+
If 160 C of charge flow through the ammeter in 3s what current
is flowing?
I = Q/t
=
Parallel Circuits
• Adding resistors in
parallel…decreases the total
resistance.
1 = 1 + 1
Rt
R1
R2
1 = 1 + 1
Rt
…
…
1
1
=
+
Rt
2
1
1
+
2
2
Rt = …./…. = ……. Ω
Total R = …. Ω
R1
R1
2Ω
2Ω
R2 2 Ω
Total R = … Ω
R1
2Ω
R2
2Ω
R2 2 Ω
Total R = …… Ω
Parallel Circuits
• The voltage is EQUAL over the
resistances .
VT
V1
VT = …………….
R1
2V
….. V
V2
R2
•
The current flowing is divided
between the resistances and would
increase as more resistances are
added- more routes for the current
to flow.
4A
A
AT= ………………..
R1
A1
…. A
A2
….. A
R2
….. V
Parallel Circuits
• Adding resistors in
parallel…decreases the total
resistance.
1
1
1
=
+
Rt
R1
R2
• The voltage is EQUAL over the
resistances .
VT = V1 = V 2
•
The current flowing is divided
between the resistances and would
increase as more resistances are
added- more routes for the current
to flow.
AT= A1 + A2
VT
A
V1
A1
A2
R1
2Ω
V2
R2
2Ω
The current will divide in
such a way that the
potential lost by both all
branches of current will
be the same.
Series Circuits
12v
• Adding resistors in series…increases
the total resistance - because all the
current flows through all the resistors.
Rt = R1 + R2
• The total potential difference (voltage)
is the sum of the potential differences
of the resistors – the total potential
loss must equal the all the potential
lost along the way.
Vt = V 1 + V 2
The potential differences will be
proportional to the
resistances.
• The current flowing is the same all
over the circuit and would decrease as
more resistances are added -
A = A2 = A3
VT
V
VTT
A
A
vV22
V11
V
R1
1Ω
A3
A2
R2
3Ω
Worked Example
4.1
12v
A
6Ω
4Ω
12 Ω
8Ω
Calculate I
2Ω
Current & Resistance
CURRENT: An electrical current is a movement of …………….
through a conducting material from positive to negative. (?!)
_
+
RESISTANCE
• Electrical charge experiences ……………………as it moves
through a conductor.
• The resistance is due to ……………….. with particles in the
metal atoms and ions.
• The moving charges lose ……………………in the collisions which
…………….up the conductor.
Ohm’s Law - Practical
AIM:
– Investigate the relationship between the …………………….. across a
resistor and the ………………..flowing through it.
– Determine the …………………..of a resistor.
rheostat
METHOD:
1.
Set up the circuit as shown.
2.
Using the rheostat vary the
current in the circuit, obtain a
range of readings for the
potential difference across R for
different currents.
3.
A
V
RESULTS>>
Resistance R
Ohm’s Law - Results
Analysis - Graph
Results
I (A)
V (V)
Draw a graph of your results.
Ohm’s Law - Analysis
I (A)
0.80
0.81
0.85
0.90
0.95
V (V)
3.40
3.50
3.70
3.90
4.20
V/I
4.3
4.3
4.4
4.3
4.4
The ratio V/I produces a
constant value - for any resistor
This is the Resistance of the
resistor.
The Unit of measurement for
resistance is the Ohm - symbol
(Ω)
Current vs Voltage
Voltage (V)
4.40
The SLOPE of the graph gives
the RESISTANCE.
4.20
4.00
3.80
3.60
3.40
3.20
3.00
0.75
Slope
Rise
run
0.80
0.85
0.90
Current (A)
0.95
1.00
= DY/ DX
=
Current, voltage & resistance
The relationship between the
…………… through a resistor,
the …………... drop across
the resistor and the
resistance of the resistor is
expressed by the following
equation:
-
+
A
V
R
-
+
V
Rx ..
Calculate the voltage drop across
a 2  resistor when a current of
1.5 A is flowing..
We define the unit of resistance; one …….() is one volt
per ampere.
R = V/I
1 = …V/…A
Worked Example
4.1
12v
A
6Ω
4Ω
12 Ω
8Ω
Calculate I = V/R = 12/6 = 2 A
2Ω
Ohm’s Law
Factors affecting Resistance
-
+
1. Material
2. Length
3. Temperature
A
+
V
-
Effects of Current
Electric current generates heat in a conductor.
_
+
A small current (0.1A) would have only a few
charges flowing.
_
+
A large current (15A) would have a large number of charges
flowing and generate far more heat.
As a conductor heats up the RESISTANCE INCREASES.
Effects of electric current
An electric current that flows in a conductor has a number of
effects:
1.
HEATING The friction caused by the current causes the
conductor to heat up. The greater the current the more
heat is generated.
2.
MAGNETIC EFFECT - A magnetic field is generated
around any conductor when an electric current flows
through it.