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Current
Electricity
Prepared in Dec 1998
Second editing in March 2000
Learning objectives
At the end of this unit you should be able to :
1. state the resistance = p.d. / current and use the
equation R = V / I.
2. describe an experiment to determine resistance
using a voltmeter and an ammeter and make
necessary calculation.
3. sketch and interpret the V/I characteristic graphs
for ohmic (metallic) and non-ohmic conductors
appreciate the limitations of Ohm’s Law.
Ohm’s Law
Ohm
’s Law
Ohm’s
• Brief History:
In 1826, a German scientist, Georg Simon
Ohm, discovered the relationship between
the current flowing through a metal
conductor and the potential difference
across its ends of the conductor.
(continue on next slide)
Investigating
Investigating
Ohm
’s Law
Ohm’s
• Connect a single cell
in series with a
nichrome wire. Add a
suitable ammeter.
• Then add a voltmeter
across (in parallel) the
nichrome wire.
• Note the readings of
ammeter and
voltmeter.
(continue on next slide)
Investigating
Investigating
Ohm
’s Law
Ohm’s
• Connect another cell
in series to assist the
first cell. Then record
the readings.
• Repeat these steps
with three cells, four
cells and …
• Plot the graph V
against I.
V
x
x
x
x
x
O
I
Ohm
’s Law
Ohm’s
The current I, passing through a
conductor is directly proportional
to the potential difference, V, between
its ends, provided that physical
conditions and temperature remain
constant.
Ohm’s Law
Ohm
’s Law
Ohm’s
• By Ohm’s Law, we have
IαV
or V/I = constant
where I = current, V =p.d.
• Plot the graph of I against V
or V against I
I
O
V
O
V
gradient = V/I
I
Ohm’s law
• By Ohm’s law we have
V/I = constant
• The metal wire (nichrome wire) tends to resist
the movement of electrons in it because
electrons collide with the ions in the wire. We
say that the wire has a certain resistance to the
current.
• Therefore, we may rewrite the relationship as
V/I = R
where R is called the resistance of the wire.
Ohm
’s law
Ohm’s
• The formula of Ohm’s law can be written as:
V= IR
• where V=p.d. ; I = current;
• Note:
– SI unit of R is ohm, Ω
– and
1Ω = 1V/1A
R=resistance
Resistors
Resistors (fixed)
• Fixed resistors, sometimes made of a
length of nichrome wire, can be used to
reduce the current in a circuit.
Resistors
Some resistors are made of coiled nichrome wire.
Others are made of carbon.
Rheostat
(variable
(variable resistor)
resistor)
• A variable resistor
or rheostat is used
to vary the current
in a circuit.
• As the sliding
contact moves, it
varies the length of
wire in the circuit
and hence the
resistance will be
changed.
Measuring
Measuring Resistance
Resistance
• Use the circuit as
shown.
• Use the rheostat to
adjust the current to a
convenient value and
note the readings on
the ammeter and
voltmeter.
• Use V=IR to find the
value of R.
Ohmic Conductor and
Non-ohmic Conductor
• Conductors (pure metal) that are
obeyed Ohm’s law is called Ohmic
Conductors.
• Materials that are not obeyed Ohm’s
law are called non-ohmic materials.
V/I Graphs
Ohmic
Ohmic
Conductor
Conductor
ss
Pure metal,
carbon and copper
sulphate
V
O
The uniform gradient
shows uniform
resistance
I
(a) Pure metal
V
O
I
(b) Copper sulphate solution
Non
-Ohmic
Non-Ohmic
Conductors
Conductors
V
O
(d) solid state diode
V
I
V
O
I
(c) Vacuum diode
O
I
(e) dilute sulphuric acid
with platinum electrodes
Non
-Ohmic Conductors
Non-Ohmic
Conductors
At low
temperature,
the tungsten
wire obey
Ohm’s Law
but at higher
temperature
it is not
obeyed the
Law.
V
Constant
resistance
Higher
resistance
due to higher
temperature
I
O
(f) filament bulb
Worked Examples
Example
Example
• A lamp draws a current of 0.25 A when it
is connected to a 240V source. What is
the resistance of the lamp ?
since
V = IR
240 = 0.25 R
R = 960Ω
Ω
Example
Example
• Calculate the current flowing through a
5Ω
Ω resistor when a potential difference of
2 V is applied across it.
since
V = RI
2 = 5I
I = 0.4 A
GCE O-Level
Past Examination Paper
Science (Physics)
All rights go to University of Cambridge Examinations Syndicate and other sources
GCE O
Nov 1996
15. What is the current in a 5 Ω resistor when the
potential difference between the ends of the
resistor is 2.5 V ?
A
B
C
D
0.5 A
2.0 A
2.5 A
12.5 A
A
November 1989
16.
A circuit is set up as shown in the diagram.
Assuming that the ammeter has negligible
resistance, what is the value of the resistor R?
A
B
C
D
0.5Ω
Ω.
1.5Ω
Ω
5Ω
6Ω
Ω
C
GCE O
Nov 1994
16. In the circuit the reading on the ammeter is 2 A.
Hint:
p.d. across
3Ω
Ω=2x3
= 6V
What is the value of the potential difference
across resistor X ?
A 1.5 V
C 3 V
B 2V
D 6V
C
O’ level Physics
Nov 1995
16. Some students set up the circuit shown to
investigate how a variable resistor affects an
electrical circuit
How will the readings
on the meters be
affected as the
resistance of the
variable resistor is increased?
A
B
C
D
ammeter reading
decrease
decrease
increase
increase
voltmeter reading
decrease
increase
decrease
increase
A
GCE O
Nov 1997
16. The diagram shows a circuit in which PQ is a
piece of resistance wire with a total resistance
of 12 Ω. R is a sliding contact joined to P with
normal contact wire.
What will be he seen
as R is moved along
the resistance wire
from Q towards P ?
Hint:
A The lamp filament will blow.
P = VI
V = IR
B The lamp will become brighter.
V is
C The lamp will become less bright.
constant
D the lamp will remain at constant brightness
C
Nov 1998
14. The voltage-current graphs for four electrical
devices are shown. Which graph shows the
resistance increasing as the current increases?
Hint:
R =gradient of the graph increases
A
Nov 1998
7.
An electrical light bulb draws a current of 0.5A
when connected to the 240V main supply.
(a) Calculate the power of the light bulb.
[2]
Since P = VI
= 240 x 0.5 = 120 W
(b) Calculate the resistance of the filament of the
[2]
bulb.
Since V =RI
R = V/I = 240 / 0.5 = 480 Ω
(continue on next slide)
(Cont. …) Q7
Nov 1998
(c) Explain why the filament reaches a constant
temperature, even though heat is produced
continually as the current flows through the
[2]
bulb.
When current flows through the filament,
the filament is heated until it is white hot.
Both heat and light are emitted. Also, the
amount of energy converted to heat is constant
since power is constant (P = VI). Therefore
temperature is constant at this stage.
Nov 1997
11. The VlI characteristic graphs for two resistors A
and B are shown in the diagram below.
I/A
(continue in next slide)
November 1992
7. The diagram shows a
circuit containing two
identical lamps and
a resistor. Each lamp
is marked 1.5 V 0.4
A. This refers to the
conditions when the
lamps are a normal brightness. The lamps can be
hint operated at the normal brightness by using a 6 V
supply and a resistor., R.
(a) What is the potential difference between Y and
1.5 V
Z? ______________________________
[1]
(continue on next slide)
(Cont. …) Q. 7
November 1992
7(b) What is the potential difference between X and
6.0 - 1.5 = 4.5 V
Y? _______________________
[1]
In series
(c) What is the value of the resistor of R ?
Since
therefore,
V = IR
R = V / I = 4.5 / 0.8 = 5.63 Ω
[2]
Lamps :
0.4A x2
(d) How much electrical energy is converted by
[3]
one lamp in one minute?
lamp:
E = VIt = 1.5 x 0.4 x 60 One
1.5V, 0.4A
= 36 J
GCE
Nov 1991
4 The graph shows the variation of current with
voltage for the filament of a light bulb.
(a) Over what
range of
voltage does
Ohm’s Law
apply?
Range of ammeter : 0 - 1 A
Range of voltmeter : 0 - 10 V
(continue on next slide)
(Cont. …)
Q.4
Nov 1991
4(b) Calculate the resistance of the filament when
[2]
a current of 0.25A flows through it.
When I = 0.25 A
we have V = 1 V.
By V = IR
R = V/ I
= 1 / 0.25
=4Ω
(continue on next slide)
Nov 1990
6. The circuit diagram below shows a 3.0 Ω resistor
and a 6.0 Ω resistor connected to a cell of e.m.f. 2.0 V.
Calculate
6.0 Ω
3.0 Ω
2.0 V
(a)
the current flowing in the 3.0 Ω resistor,
Since V = IR
I3 = 2 / 3 = 0.67 A
(continue in next slide)
[1]
(Cont. …)
Q. 6
Nov 1990
6(b) the current flowing in the 6.0 Ω resistor, [1]
(c) the current delivered by the cell,
[1]
[2]
(d) the power being supplied by the cell.
b).
Since V = IR
therefore I6 = 2 / 6 = 0.33 A
c).
As Rc = (6 x 3) / (6 + 3) = 2.0 Ω
therefore I = 2 / 2 = 1.0 A
d).
P = VI = 2 x 1 = 2.0 W