EPM1016 (06/07) – Tut 1

TUTORIAL 1: INSTRUMENTATION & MEASUREMENT SYSTEMS

1)Absolute error, A=Am – At = 1.46 – 1.50 = - 0.04VAbsolute correction, C = -A = +0.04V

Relative error,

Relative error (expressed as a percentage of f.s.d)

2)1 scale division = 500/100 = 5V

Resolution =

3)Let:

L Length of capillary tube which would be occupied by mercury contained in the bulb when it is not heated.

A Area of capillary tubeL+L Length of capillary tube which would be occupied by

mercury contained in the bulb when it is heated.V Change of volume.T Change of temperature.

It should be noted that there will be a change in the length of mercury but no change in the area and length of the capillary tube.

Given sensitivity,

Coefficient of volumetric expansion =

On resolving

since V = A . L

Since the capillary tube remain unchanged, the area of the tube and hence the column of mercury remains unaltered. Hence, one may writeV = A.(L)

Equating above two equations, one gets:

T. H. Oh 1

EPM1016 (06/07) – Tut 1

Volume of mercury in the capillary bulb:

4)Pressure measured, range: –10 Pa to 28 Pa.

Resolution = 0.25

Output voltage range: –10mV to 120 mV.

Division of the sensor =

Resolution of output voltage =

T. H. Oh 2

EPM1016 (06/07) – Tut 1

5)a) By using voltage divider,

b) Parallel combination of RB and RX (internal resistance of meter X)

c) Parallel combination of RB and RY (internal resistance of meter Y)

d) Voltmeter – X error = (3.53-5)/5 X 100% = -29.4% Voltmeter – Y error = (4.9-5)/5 X 100% = -2%

6)R = Ro [1 + (t-20)]

Ro =4 Ro = 4x 0.2%= 8 m

= 0.004 = 0.004 x 1% =

t = 25 oC t = 25 x 1% = 0.25 oC

= + + = + + = 0.01296

R = 4[1 + 0.004(25 - 20)] = 4.08

T. H. Oh 3

E = 30 V

RA = 25

RB = 5 RX

EPM1016 (06/07) – Tut 1

= 3.2x10-3 = 0.3%

R = 4.08 0.3%

T. H. Oh 4

EPM1016 (06/07) – Tut 1

7)

Inductance, ,

E = work done/chargeCharge = I*TWork done = F LForce, F = mass acceleration = M*(L/T2)

Base units, L = kgm2A-2s-2

8)Absolute error = 27 – 30 = -3mA,

Relative accuracy = 27/30 = 0.9

% accuracy = 0.9 X 100% = 90%

T. H. Oh 5