BEE 2123
TUTORIAL 1
ERROR MEASUREMENT SOLUTION
1.
A batch of resistors that each have a
nominal resistance of 330 Ω are to be
tested and classified as ±5% and ±10%
components. Calculate the maximum
and minimum absolute resistance for
each case.
Answer:
330 ± 5 % = 330 ± 16.5Ω
330 ± 10% = 330 ± 33Ω
then, 330 + 16.5 = 346.5Ω
330 – 16.5 =313.5Ω
330 + 33 = 363Ω
330 – 33 = 297Ω
2.
The
resistors
in
Problem
1
are
specified at 25°C, and their temperature
coefficient is –300ppm/°C. Calculate the
1
maximum and minimum resistance for
these components at 100°C.
Answer:
T = 25° C, R = 330Ω ± 5%
R max = 346.5Ω
R min = 313.5Ω
∆R/ °C:
at R max, 346.5 x (–300/1,000,000 ) =
-0.10395 Ω /°C
at R min, 313.5 x (–300/1,000,000 ) =
-0.09405 Ω /°C
∆T = 100 –25 = 75 °C
∆R Total:
at R max, ∆ R = 0.10395 x 75 =
-7.79625ohm
at R min, ∆ R = 0.09405 x 75 =
-7.05375ohm
2
R max at 100 °C: R + ∆ R = 346.5Ω
-7.79625ohm= 338.7Ω
R min at 100 °C is R + ∆ R = 313.5Ω
-7.05375ohm = 306.4Ω
3.
Three of the resistors referred to in
Problem 1 are connected in series. One
has a ±5% tolerance, and the other two
are ±10%. Calculate the maximum and
minimum values of the total resistances.
Answer:
( R1 ± ∆ R1) + (R2 ± ∆ R2 ) + ( R3 ± ∆ R3 )
= ( R1 + R2 + R3 ) ± (∆ R1 + ∆ R2 +
∆ R3)
= ( 330 ± 5% ) Ω +( 330 ± 10% ) Ω +
( 330 ± 10% ) Ω
= (330 + 330 + 330) ± (16.5 + 33 + 33)
Ω
= 990 ± 82.5 Ω
RTmax= 1072.5 Ω
= 907.5 Ω
and
RTmin
3
4.
A
dc
power
supply
provides
currents to four electronic circuits. The
currents are 37 mA, 42 mA, 13 mA and
6.7 mA. The first two are measured with
an accuracy of ±3%, and the other two
are
measured
with
±1%
accuracy.
Determine the maximum and minimum
levels of the total supply current.
Answer:
I1 = 37 ± 3% mA
I2 = 42 ± 3% mA
I3 = 13 ± 1% mA
I4 = 6.7 ± 1% mA
I1 + I2 + I3 + I4 = ( I1 + I2 + I3 + I4 ) ±
(∆ I1 + ∆ I2 + ∆ I3 +∆ I4)
= 37 + 42 + 13 + 6.7 ± (1.11 + 1.26 +
0.13 + 0.067) mA
ITmax = 98.7 + 2.567 mA = 101.267 mA
4
ITmin = 98.7 - 2.567 mA = 96.13 mA
5.
Two currents from a different sources
flow in opposite directions through a
resistor. I1 is measured as 79 mA on a
100 mA analog instrument with an
accuracy
of
±3%
of
full
scale.
I2,
determined as 31 mA, is measured on a
digital
instrument
with
a
±100
µA
accuracy. Calculate the maximum and
minimum levels of the current in R1.
Answer:
3% FSD x 100mA = 3mA
then ,
I1 = 79 ± 3mA
I2 = 31 mA ± 100µ A
= 31 ± 0.1 mA
so,
5
I1 – I2 = (I1 ± ∆ I1) – ( I2 ± ∆ I2 ) = ( I1 –
I2 ) ± (∆ I1 + ∆ I2) = ( 79 – 31 ) ± ( 3
+ 0.1) mA
= 48 ± 3.1 mA
ITmax = 51.1 mA
6.
ITmax = 44.9 mA
The voltages at opposite ends of a 470
Ω ±5% resistor are measured as V1 = 12
V
and
V2
=
5
V.
The
measuring
accuracies are ±0.5 V for V1 and ±2%
for V2. Calculate the level of current in
the resistor, and specify its accuracy.
Answer:
V1 = 12 ± 0.5 V
5 ± 2% V
V2 =
R1 = 470 ± 5%
then,
V
0.5V) – (5V ± 2%)
(12V ±
6
I = ---=
---------------------------------R
± 5% Ω
470
7± ( 0.5 + 0.1) V
= ------------------470 ± 5% Ω
7± 8.57% V
= ------------------470 ± 5% Ω
X = A/B  % error X = ± (% error A
+ % error B)
I = 0.01489 ± ( 8.57% + 5%) A
= 14.89 ± 13.57% mA
7.
A resistor R1 has a potential difference
of 25 V across its terminals, and a
current
of
63
mA.
The
voltage
is
measured on a 30 V analog instrument
with an accuracy of ±5% of full scale.
The current is measured on a digital
instrument with a ±1 mA accuracy.
7
Calculate
the
resistance
of
R1
and
specify its tolerance.
Answer:
Voltage error:
5% x 30V = 1.5V
Potential difference across the resistor:
25± 1.5 V = 25 ± 6%
63 ± 1 mA
R1
25 ± 6%
25V ± 6%
V
V
R = ---I
1.59% mA
=
--------------63 ±
8
= 396.83 ±
7.59% Ω
8.
A 470 Ω ±10% resistor has a potential
difference of 12 V across its terminals. If
the
voltage
is
measured
with
an
accuracy of ±6%, determine the power
dissipation in the resistor, and specify
the accuracy of the result.
Answer:
12 ± 6% V
470± 10% Ω
(12V ± 6%
V2
V)2
P = ---R
10% Ω
=
--------------470 ±
9
V2
P = ----
=
144
---------- =
0.306 W
R
470
X=A/B, % error X= ± (% error A + %
error B)
X=AB , % error X = ± B(% error A)
22% W
% ∆ P = ± (2(6) +10) % = ±
P = 0.306 ± 22% W
9.
The output voltage from a precision 12
V power supply, monitored at intervals
over a period of time, produced the
following readings: V1 = 12.001 V, V2 =
11.999 V, V3 = 11.998 V, V4 = 12.003 V,
V5 = 12.002 V, V6 = 11.997 V, V7 =
12.002 V, V8 = 12.003 V, V9 = 11.998 V
and V10 = 11.997 V. Calculate the
average
voltage
level,
the
mean
10
deviation, the standard deviation, and
the probable error in the measured
voltage at any time.
Vi
d= VVi 
1 12.001 0.001
2 11.999
0.001
3 11.998
0.002
4 12.003
0.003
5 12.002
0.002
6 11.997
0.003
7 12.002
0.002
8 12.003
0.003
9 11.998
0.002
1 11.997
0
∑ 120.00
0
0.003
0.022
D2
0.0000
01
0.0000
01
0.0000
04
0.0000
09
0.0000
04
0.0000
09
0.0000
04
0.0000
09
0.0000
04
0.0000
09
0.0000
54
11
V = ∑ Vi /n = 120.00/10 = 12.000 V
D = ∑ V- Vi/n = 0.022/10 = 2.2 mV
σ = √ (∑ d2 /n) = √ 0.000054/10 =
2.32 mV
Voltage error probability
0.6745 = 1.57
10.
Successive
= 2.32 x
measurements
of
the
temperature of a liquid over a period of
time produced the following data: T1 =
25.05°C, T2 = 25.02°C, T3 = 25.03°C, T4 =
25.07°C, T5 = 25.55°C, T6 = 25.06°C, T7 =
25.04°C, T8 = 25.05°C, T9 = 25.07°C, T10 =
25.03°C, T11 = 25.02°C, T12 = 25.04°C, T13
= 25.02°C, T14 = 25.03°C and T15 =
25.05°C.
Determine
the
average
temperature, the mean deviation from
average, the standard deviation, and the
probable measurement error.
12
ti
1
2
3
4
5
6
7
8
9
1
0
1
1
1
2
1
3
1
d= tit
25.0
0.02
5
25.0
0.05
2
25.0
0.04
3
25.0
0.00
7
25.5
0.48
5
25.0
0.02
5
25.0
0.03
4
25.0
0.02
5
25.0
0.00
7
25.0
0.04
3
25.0
0.05
2
25.0
0.03
4
25.0
0.05
2
25.0
0.04
d2
0.00
04
0.00
25
0.00
16
0.00
00
0.23
04
0.00
04
0.00
09
0.00
04
0.00
00
0.00
16
0.00
25
0.00
09
0.00
25
0.00
13
4
1
5
∑
3
25.0
5
376.
12
t = ∑ ti / n
0.02
0.89
16
0.00
04
0.24
86
= 376.12/15 = 25.07
D = ∑ ti- t/n = 0.89/15 = 0.059 mV
σ = √ (∑ d2 /n) = √ 0.2461/15 =
0.128
Temperature error probability = 0.128
x 0.6745 = 0.087
14