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:
330 330
± ±
5 % = 330 ± 16.5Ω 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 = 907.5 Ω
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. 100
I1
mA
accuracy
is measured as 79 mA on a analog of
instrument
±3%
of
full
with
an
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
V
and
=
V2
5
V.
The
accuracies are ±0.5 V for for
V2.
V1 = 12
measuring V1
and ±2%
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, 0.5V) – (5V
±
V 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 B)
% error X =
±
(% error A
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
±
R1
1 mA
25
±
6%
V 1.59% mA
25V
V R = ---I
=
±
6%
--------------63 ±
8
7.59% 8.
= 396.83
Ω
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
V)2 10%
(12V
V2
Ω
P = ---R
=
±
6%
--------------470 ±
9
0.306 W
V2 P = ----
144 ---------- =
=
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 9.
±
±
22% W
The output voltage from a precision 12
V power supply, monitored at intervals over a period of time, produced the following readings: 11.999 V, V5
V3
V1
= 12.001 V,
= 11.998 V,
= 12.002 V,
V6
= 12.003 V,
= 11.997 V,
12.002 V,
V8
and
= 11.997 V. Calculate the
V10
average
= 12.003 V,
V4
=
V2
voltage
V9
level,
= 11.998 V the
mean 10
V7
=
deviation, the standard deviation, and the
probable
error
in
the
measured
voltage at any time. d= VVi 1 12.001 0.001 Vi
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 0.003 0 ∑ 120.00 0.022 0
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
D=
∑
= 120.00/10 = 12.000 V
V- Vi /n = 0.022/10 = 2.2 mV
σ
= √ (∑ d2/n) = 2.32 mV
√ 0.000054/10=
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
= 25.04°C,
T13
= 25.02°C,
= 25.02°C, 25.05°C.
T14
T12
= 25.03°C and
Determine
temperature,
the
mean
the
T15
=
average
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
25.0 5 25.0 2 25.0 3 25.0 7 25.5 5 25.0 5 25.0 4 25.0 5 25.0 7 25.0 3 25.0 2 25.0 4 25.0 2 25.0
d= tit 0.02 0.05 0.04 0.00 0.48 0.02 0.03 0.02 0.00 0.04 0.05 0.03 0.05 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
D=
∑
σ
= 0.128
ti-
t
√ (∑
0.02 0.89
16 0.00 04 0.24 86
= 376.12/15 = 25.07 /n = 0.89/15 = 0.059 mV
d2/n) =
√ 0.2461/15=
Temperature error probability = 0.128 x 0.6745 = 0.087
14