EE 3203: ELECTRICAL AND ELECTRONIC
MEASUREMENTS
ASSIGNMENT 1
NAME
: KUMARASINGHE K.A.G.T.V
REG NO.
: EG/ 2019/3645
SUBMITTED DATE : 18/07/2022
Q1.
I.
State 5 static and dynamic performance characteristics of measuring instruments.
Statics performance characteristics
• Sensitivity
• Dead zone
• Tolerance
• Stability
• Accuracy
Dynamic performance characteristics
• Speed of response
• Lag
• Bandwidth
• Delay time
• Dynamic error
Select one from both static and dynamic characteristics and explain its practical application
in taking measurements. You may use figures, graphs, etc.
• Sensitivity
Sensitivity can be denoted the instrument response due to the smallest change in the
measured variable. It is defined as the ratio of the changes in the output of an
instrument to a change in the value of the quantity being measured.
We have to get most efficient instrument for measurement. That mean if we going to
be measured some small physical measurement, we have to get most sensitive
instrument for the purpose. Otherwise, the reading may be wrong than actual values.
Instrument Reading
II.
Sensitivity of Measurement
Measured Quantity
Figure 1 : Sensitivity of Measurement
High sensitivity instruments may cause drifts due to thermal or other effects, and
indications may be less repeatable or precise than those of lower sensitivity
instruments.
• Speed of response
The rapidity with which an instrument or measurement system responds to changes
in measured quantity is defined as speed of response. The response time it takes for
an instrument or system to settle to its final steady position after receiving input.
There are some measurements to get very quickly. So, we need to select highest
speed response instrument to fulfil the purpose.
III.
Calculate the maximum percentage error in the sum of 2 voltage measurements when V1 =
100 V ± 1% and V2 = 80 V ± 5%.
let the sum S for nominal case,
S
=
=
V1 + V2
180
Absolute error for V1
=
=
( 0.01 x 100 ) V
1V
Absolute error for V2
=
=
( 0.05 x 80 ) V
4V
Total Absolute error
(∆𝑆𝑡𝑜𝑡𝑎𝑙 )
=
=
1V+4V
5V
Therefore,
Maximum percentage error =
=
=
∆𝑆𝑡𝑜𝑡𝑎𝑙
5
𝑆
180
∗ 100%
2.778%