Effect of length and thickness on resistance

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Experiment 2c
Class:
Name:
(
) Date:
2c Effect of length and thickness on
resistance
Objective
To find out how the resistance of a metal wire is affected by its
length and thickness (cross-sectional area).
Background information
1
The resistance of a conductor is defined as:
voltage across conductor
Resistance =
current through conductor
or
2
3
R=
V
I
By measuring the current I through a conductor when a known
voltage V is applied across it, the resistance of the conductor
V
can be determined from the formula R = .
I
The graph in Figure 2c-1 is a straight line
passing through the origin, which means x
and y are directly proportional, i.e.
y
y0
slope = x = k
0
y0
y∝x
or y = kx (k is a constant)
origin
x0
x
Fig 2c-1
Apparatus
❏ 2 eureka wires of different thickness (0.5 m, mounted on a wooden strip)
❏ 1 voltmeter
❏ 1 ammeter
❏ 1 battery box
❏ 1 switch
❏ several connecting leads (one with a crocodile clip at one end)
24
New Physics at Work (Second Edition)
© Oxford University Press 2007
Class:
Name:
(
Procedure
✐ Before the experiment,
discuss with Ss the
need for control of
variable—keep thickness
of the wire constant when
investigating the effect
of length on resistance
and keep length constant
when investigating the
effect of thickness of
wire.
Experiment 2c
) Date:
Effect of length
1
Set up the apparatus as shown in Figure 2c-2:
(a) Connect one of the eureka wires in series with an ammeter, a
battery box and a switch.
(b) Connect a voltmeter across the wire using a crocodile clip as a
sliding contact at one end.
Be careful! Do not touch the hot
wire! Disconnect the circuit once
the readings are taken. Otherwise,
the eureka wire may overheat.
battery box
switch
ammeter
voltmeter
crocodile clip
two 0.5 m-eureka wires of different
thickness mounted on wooden strip
+
eureka wire
A
length of
wire
+
V
Fig 2c-2
2
(a) Vary the length of the wire connected by sliding the crocodile
clip along the wire.
(b) Take the ammeter and voltmeter readings. Record the results in
Table 2c-1 on p.26.
New Physics at Work (Second Edition)
© Oxford University Press 2007
25
Experiment 2c
Class:
Name:
(
) Date:
3
Repeat several times with other lengths. Record the results in
Table 2c-1 and calculate the resistance of different lengths of the
wire.
✎
Results:
Length of wire l / m
0.10
0.20
0.30
0.40
0.48
Voltage across wire V / V
0.48
0.90
1.30
1.70
2.10
Current through wire I / A
0.50
0.50
0.50
0.50
0.50
Resistance of wire R / Ω
0.96
1.80
2.60
3.40
4.20
Table 2c-1
4
Plot a graph of the resistance R of wire against the length l of wire in
Figure 2c-3.
resistance of wire R / 8
4
3
2
1
0
0.1
0.2
0.3
0.4
0.5
length of wire l / m
Fig 2c-3
26
New Physics at Work (Second Edition)
© Oxford University Press 2007
Class:
Name:
(
Experiment 2c
) Date:
Effect of thickness (cross-sectional area)
5
Measure the current and voltage of equal length of the two wires of
different thickness. Record the results in Table 2c-2 and calculate the
resistance of the two wires.
✎
Results:
Thin wire
Thick wire
Voltage across wire V / V
1.00
1.00
Current through wire I / A
0.24
1.00
Resistance of wire R / Ω
4.20
1.00
Table 2c-2
Discussion
✎
How is the resistance of the wire related to its length?
The resistance of the wire is directly proportional to its length.
✎
How does the resistance of the wire change with its cross-sectional
area?
The resistance of the wire decreases as its cross-sectional area increases.
1For metal wires of the same thickness and material, the
directly proportional
resistance is ___________________________________
to the length of
the wires.
2For metal wires of the same length and material, the larger the
smaller
cross-sectional area of the wires, the ________________________
the resistance.
New Physics at Work (Second Edition)
© Oxford University Press 2007
27
Experiment 2c
Class:
Name:
(
) Date:
Further thinking
✎
Design an experiment to show whether the resistance of a metal wire
is inversely proportional to its cross-sectional area. Write down the
necessary procedures.
Use several eureka wires of different diameters. Measure the current and voltage of
equal length of the wires to calculate their corresponding resistances. Plot a graph
•
•
1
4
=
. If the
of the resistance of the wire against
cross-sectional area
π × (diameter)2
graph is a straight line passing through the origin, the resistance of a metal wire is
inversely proportional to its cross-sectional area.
28
New Physics at Work (Second Edition)
© Oxford University Press 2007
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