DEE 2313
Instrumentation & Managements
DEE 2313: Instrumentation and Managements
Tutorial: Chapter1 (with solution)
1. The expected value of the voltage across a resistor is 50V, however, measurement yields
a value of 49V
Calculate:
(a) The absolute error.
(b) The percent error.
(c) The relative accuracy.
(d) The percent accuracy.
Solution:
(a) e
Yn
X n = 50 V – 49 V = 1 V
(b) % error
(c) A
1
=
Yn
1
X 100% = 2 %
50
Xn
Yn
1
50
49
50
=0.98
(d) a = 0.98 X 100% = 98%
2. If the average of set of voltage reading is 30.15V. Calculate the precision of one of the
reading that was equal to 29.9V.
Solution:
Y n = 30.15 V, X n = 29.9V
Precision
1
Xn
Xn
Xn
1
30 .15 29 .9
= 0.992
30 .51
3. The following set of measurement was recorded in the laboratory. Calculate the precision of
the 4th measurement.
Measurement no.
1
2
3
4
5
6
7
8
9
10
Measurement value (V)
98
102
101
97
100
103
98
106
107
99
DEE 2313
Instrumentation & Managements
Solution:
X n = 97 V
1011
101.1V
10
Xn Xn
Precision 1
Xn
Xn
97 101 .1
=1 – 0.04 = 0.96
101 .1
1
4. The following table of values represents a meter output in terms of the angular displacement
of the needle, expressed in degree, for a series of identical input current. Determine the
worst-case precision of the readings.
Iin
(µA)
10
10
10
10
10
10
10
10
Output Displacement
(degrees)
20.10
20.00
20.20
19.80
19.70
20.00
20.30
20.10
Solution:
The average output, determined by adding the output values and dividing by eight, is equal to
20.02 degrees. The fifth reading differs from the average by the greatest amount; therefore, the
worst-case precision is related to and found from this reading.
Precision
1
Xn
Xn
Xn
1
19 .70 20 .02
=1 – 0.016 = 0.984
20 .02
5. For the following data compute:
x1 = 50.1
x2 = 49.7
x3 = 49.6
x4 = 50.2
(a)
(b)
(c)
(d)
(e)
The arithmetic mean.
The deviation of each value.
The algebraic sum of the deviations.
Calculate the average deviation for the data.
Compute the standard deviation for the data.
DEE 2313
Instrumentation & Managements
Solution:
(a)
The arithmetic mean
x
(b)
(c)
x1
x2
x3
x4
4
50.1 49.7 49.6 50.2
4
199.6
4
49.9
The deviation of each value from the mean is computed as
d1
d2
d3
50 .1 49 .9 0.2
49 .7 49 .9
0.2
49 .6 49 .9
0.3
d4
50 .2 49 .9
0.3
The algebraic sum of the deviation is
d tot
0.2 0.2 0.3 0.3
0
(d)
D
0.2
0 .2
0.3
0.3
4
1.0
4
0.25
(e)
S
(0.2) 2
(0.2) 2 (0.3) 2
4 1
(0.3) 2
0.26
3
0.294
6. A set of independent current measurement were taken by six observer and the results were
records as:
12.8A, 12.2 A, 12.5A, 13.1A, 12.9A and 12.4A
Calculate the:
(a)
Arithmetic mean
(b)
Deviation from mean
(c)
Average deviation
(d)
Standard deviation
Solution:
(a)
x
(b)
d1
12.8 12.2 12.5 13.1 12.9 12.4
6
12 .8 12 .65
0.15 A
DEE 2313
Instrumentation & Managements
d2
d3
12 .2 12 .65
12 .5 12 .65
0.45 A
0.15 A
d4
d5
13 .1 12 .65
12 .9 12 .65
0.45 A
0.25 A
d6
12 .4 12 .65
0.25 A
(c)
Average deviation, D
(d)
S=
0.15
2
d1
0.45
d2
d3
d4
d5
6
2
0.15
2
0.45
2
d6
0.283
0.25
2
0.25
2
6 1
= 0.339 A
7. A 300V voltmeter is specified to be accurate within ± 2% at full scale. Calculate the Limiting
error when the instrument is used to measure a 120V source.
Solution:
The magnitude of limiting error at full scale
=±
2
X 300 V = ± 6V
100
Therefore, the limiting error at 120V
=±
6
X 100% = ± 5%
120
8. An ammeter read 70mA on its 100mA range. Determine the limiting error if the ammeter is
guaranteed to be accurate within ± 1% of its full-scale reading.
Solution:
The magnitude of limiting error at full scale
=±
1
X 100mA = ± 1mA
100
Therefore, the limiting error at 70mA
=±
1
X 100% = ± 1.43%
70
9. A voltmeter and an ammeter are to be used to determine the power dissipated in a resistor.
Both instruments are guaranteed to be accurate within ±1% at full-scale deflection. If the
voltmeter reads 80 V on its 150-V range and the ammeter reads 70 mA on its 100-mA range,
determine the limiting error for the power calculation.
Solution:
DEE 2313
Instrumentation & Managements
The magnitude of the limiting error for the voltmeter, 0.01 x 150 V = 1.5 V
The limiting error at 80 V,
1.5
x 100% = 1.86%
80
The magnitude of the limiting error for the ammeter, 0.01 x 100 mA = 1 mA
The limiting error at 70 mA,
1.5
x 100% = 1.43%
70
The limiting error for the power calculation is the sum of the individual limiting errors
involved; therefore
Limiting error = 1.86% + 1.43% = 3.29%