Download 91 accracy invetigations.p65

Physics Factsheet
www.curriculum-press.co.uk
Number 91
Accuracy in Electrical Investigations
This Factsheet is not meant to be a survey of practical investigations
concerning electricity, or to provide complete methods for these
investigations.
Real and Ideal Meters
An ammeter is placed in series with a resistor. An ideal ammeter has
zero resistance, and does not reduce the current flow.
It is instead intended to look at the causes of inaccuracy in electrical
work, and to suggest some techniques for minimising and measuring
inaccuracy.
I
R
In practise, the ammeter will have some resistance. But in modern
electronics, current flow is tiny (implying a large circuit resistance),
so the resistance of the ammeter is not too important.
Resistors and Combinations of Resistors
Standard resistors (e.g. carbon film) are manufactured to stated
tolerances (maximum percentage errors) often described by a
coloured band on the resistor. A gold band means 5% maximum
error.
However, voltmeters are placed in parallel to the component being
measured, leading to real problems at times.
R
I1
Example 1: What is the maximum possible error, in ohms, for
a 22kΩ resistor with a stated tolerance of 5%?
Answer:
I
A
A
I
I
5
× 22000 = 1100Ω.
100
V
I2
Often we must combine resistors to achieve required values. It is
worth seeing the effect on the percentage error of such
combinations.
The ideal voltmeter would have infinite resistance, drawing no
current away from the main circuit. But a moving coil meter requires
a current through it to produce a deflection. If the current flow
through R is very small, the current sidetracked to the parallel
voltmeter could be significant, and lead to a measurable error in the
voltage reading.
Example 2: A 1000Ω resistor and a 2200Ω resistor (each of
5% tolerance) are wired in series, then in parallel. Find the
maximum percentage error in each combination.
Example 3 (a): Look at this circuit:
1000Ω
1000Ω
12V
2200Ω
2200Ω
50kΩ
Answer:
Series:
Maximum value = (1000 × 1.05) + (2200 × 1.05) = 3360Ω .
Nominal value = 1000 + 2200 = 3200Ω.
160
Percentage error =
× 100 = 5%.
3200
100kΩ
Find the p.d. across the 100kΩ resistor.
(b) Compare this second circuit.
12V
100kΩ
Parallel: Maximum value:
1
1
+
= 1.385 × 10-3, R = 722Ω .
1050
2310
1
1
Nominal value:
+
= 1.455 × 10-3, R = 688Ω .
1000
2200
Percentage error = 5% again.
50kΩ
V
100kΩ
Now find the p.d. across the 100kΩ resistor.
Answer:
12
= 8.0 × 10-5 A.
150 000
V(across 100kΩ) = IR = 8.0 × 10-5 × 1.0 × 105 = 8V.
1
1
2
(b) Parallel:
=
=
100 000
100 000
100 000
R = 50kΩ (same as other resistor)
V(across parallel combination) = 6V.
(a) I =
Combining resistors with the same tolerance should
lead to an effective resistance with the same tolerance. It should
be noted that these are maximum percentage errors. In practice,
some of the errors might well cancel each other. The effective
resistance of the combination would probably be closer to the
nominal value. But there is no guarantee of this.
1
Physics Factsheet
91. Accuracy in Electrical Investigations
Introducing the moving coil voltmeter across the 100kΩ resistor
has changed the p.d. across it from 8V to 6V.
This can lead to Joule heating problems and is very wasteful of
power.
The solution to this voltmeter problem is to use a digital voltmeter
in electronic circuits. These meters can have effective resistances
up in the megohm region.
Calculations with potential dividers are only accurate
if the load is much larger than the resistances in the divider
itself. In practical situations, a voltmeter must be used to
provide an accurate value for the output p.d.
Example 4: This circuit uses a digital voltmeter with an input
resistance of 20MΩ.
12V
R1
E
100kΩ
R2
50kΩ
R
V
V
20MΩ
Find the p.d. across the 100kΩ resistor.
Resistance and Temperature
We tend to assume that resistors have fixed resistances. But their
resistance values increase with temperature. As an example, for
copper the temperature coefficient of resistance can be approximately
given as:
∆R
where R is the resistance at 0oC
α = (R ∆θ) ∆θ is theo temperature rise.
o
Answer:
1
1
+
= 1.005 ×10-5, R = 9.95×104 Ω.
2.0×107
1.0×105
12
I=
= 8.03×10-5 A.
( 5.0×104 + 9.95×104 )
V(across parallel combination) = IR = 8.03×10-5 × 9.95×104 = 8.0 V.
(the same as without the meter in the circuit)
Parallel:
Example 5: What is the percentage increase in the resistance
of a length of copper wire when its temperature goes up by
25oC? (α = 0.0043oC-1)
Exam hint: When doing practical work, always consider the
effects of the measuring instruments on the values being
obtained. High input impedance devices, like digital voltmeters
and cathode ray oscilloscopes, can significantly increase
accuracy.
∆R
= α∆θ = 0.0043 × 25 = 0.11
Ro
The percentage increase is 11%.
Answer:
Potential Dividers
A potential divider allows us to produce an adjustable output p.d.
which is less than the fixed input p.d. from the supply.
Joule heating causes significant changes in resistance
in metallic conductors.
There are various ways of minimising Joule heating effects:
(a) keep the current very small.
(b) ensure adequate ventilation or use a cooling fan
(c) use physically large components, or combinations to share the
heating effect.
R1
E
R2
V
10Ω
10Ω
R
The output p.d. is given by V = R + 2R × E.
1
2
10Ω
However the current drawn by the load can affect the size of the
output voltage.
E
10Ω
(d) use cooling e.g. immersing a resistance wire in water.
R1
R2
10Ω
R
The load, R, now forms a parallel branch with R2, lowering the output
voltage. (This theory is exactly the same as in the previous situation
where the voltmeter is in parallel with the 100kΩ resistor.)
To minimise this effect, R2 must be much smaller than the load, R.
However, if R2 is very small, then the current and power dissipation
through R1 and R2 will be large.
2
Physics Factsheet
91. Accuracy in Electrical Investigations
3. Assume the temperature coefficient of resistance of copper is
0.0043oC-1.
(a) Find the increase in resistance when a length of copper wire
of resistance 0.24Ω at 0oC is heated from 10oC to 45oC.
(b) Express this as a percentage change.
Metre bridges
Wheatstone bridges and metre bridges are very useful in making
accurate determinations of resistance and of the emf of cells. Their
accuracy relies on the fact that at the balance point, there is no
deflection on the meter.
X (Unknown)
4. If you decided to reduce Joule heating in a circuit by using very
thin resistance wire to reduce the current flow, suggest two
other factors you would have to consider (in terms of accuracy).
R
5. Here is another look at a metre bridge.
A
!1
!2
B
X (Unknown)
Null deflection means no errors due to current flowing through the
meter. In our diagram, X
l
= l1
R
2
However there can still be sources of inaccuracy. At A and B there
can be contact resistances. These add to the resistances of wire
lengths l1 and l2. These end errors are most significant if l1 or l2 is
very small.
A
!2
B
The wire has a total resistance of 6.0Ω. There is a contact resistance
of 0.010Ω at A and at B. The wire is 1.00m long.
(a) If the balance point is at the centre of the wire, find the
resistances of l1 and l2.
(b) Find the percentage errors in Rl1 and Rl2 due to the contact
resistances.
Rl
(c) Find the percentage error in Rl1 .
2
(d) If the unknown resistance is changed, and the balance point
is just 1.0cm from A, find the resistances of l1 and l2.
(e) Again find the percentage errors in Rl1 and Rl2.
Rl
(f) Now find the percentage error in Rl1
2
When using metre bridges, make R approximately equal
to X. This will give a balance point near the middle of the
wire, reducing the effect of end errors on l1 and l2.
Graphical work
Quite often, we can use graphical work to improve accuracy, making
use of best fit lines.
5. (a) l1 = 3.0Ω, l2 = 3.0Ω.
(b) % error l1 = % error (l2 = (0.010 / 3.0) × 100 = 0.33%
(c) % error l1 / l2 = 0.66%
(d) l1 = 1/100 × 6.0 = 0.060Ω
l2 = 99/100 × 6.0 = 5.94Ω
(e) % error l1 = (0.010 / 0.060) × 100 = 16.7%
% error l2 = 0.17%
(f) % error l1 / l2 = 16.9%.
This is, of course, applicable to all fields of physics and other
sciences. But it is always best to improve the accuracy of the
investigation itself. Don’t rely on graphical and other techniques to
sort out experimental weaknesses.
Questions
1. Use the method of maximum and minimum values to find the
maximum percentage error for this combination. The tolerance
of each resistor is 5%.
4. Examples might be (a) increased percentage errors from electrical
measuring instruments when dealing with very small currents, and (b)
increased percentage errors in measuring the diameter of the wire.
100Ω
!1
R
100Ω
3. (a) DR = αRoθ = 0.036Ω
(b) Percentage change = (0.036 / 0.24) × 100 = 15%.
2. (a) R = V/I = 120 000Ω.
(b) 1/R = 1 / 100 000 + 1 / 120 000. R = 54 500Ω.
Percentage change = 46%.
(c) 1/R = 1 / 100 + 1 / 120 000. R = 99.9Ω
Percentage change = 0.1%.
100Ω
100Ω
2. (a) A voltmeter draws a current of 0.1mA when a p.d. of 12V is
applied across it. Find its resistance.
(b) Assume the resistance it exerts is a constant. Find the change
in effective resistance when it is connected across a 100kΩ
resistor. Express this as a percentage change.
1. Nominal value: Series 200Ω in each branch, then Parallel gives R =
100Ω.
Maximum value: Increase each resistor to 105Ω. Find that effective R
= 105Ω.
Maximum percentage error is 5Ω for the combination, as well as for
each resistor.
100kΩ
Answers:
V
(c) Find the percentage change in effective resistance when the
voltmeter is connected across a 100Ω resistor.
3
Acknowledgements:
This Physics Factsheet was researched and written by Paul Freeman
The Curriculum Press,Bank House, 105 King Street,Wellington, Shropshire, TF1 1NU
Physics Factsheets may be copied free of charge by teaching staff or students, provided that their school is a registered
subscriber. ISSN 1351-5136