Set of gyro numerical questions with their solutions for the benefit of students appearing for their phase II MMD examinationsFull description
compilation of IIT JEE numericals on resistors and capacitors
SET-NET problems
Q1. A Linear Resistance potentiometer is 50mm long & is uniformly wound with a wire having a resistance of 10,000Ω. Under normal conditions, the slider is at the center of the potentiometer. Find the linear displacement when the resistances of the potentiometer as measured by Wheatstone bridge for two cases are; (1).3850Ω (2).756 (2).7560Ω. 0Ω.
Are the two displacements in same direction If it is possible to measure a minimum value of 10Ω resistance with the above arrangement, find the resolution. Ans. The resistance at normal position =
Resistance of potentiometer per unit length =
= 5000Ω
= 200Ω/mm
(1) Change of resistance form normal position = 5000 – 3850 =1150Ω Therefore, Displacement of slider from its normal position =
= 5.75mm
(2) Displacement =
6−
Resolution = minimum ×
= 12.8mm
= 0.05mm
Q2. In a proximity inductive transducer, the coil has an inductance of 2mH (Henry) = (Kgm2s-2A-2) when the target mode of Ferro-magnetic material is 1mm away from the core. Calculate the value of inductance when a displacement of 0.02mm is applied to the target in a direction moving it towards the core. Show that the change in inductance is linearly proportional to the displacement? Ans. Inductance with air gap length 1m, L=2mH
Length when 0.05mm displacement is done = 1-0.02 = 0.98mm Now inductance is inversely proportional to the length of air gap. ΔL is increase in inductance L+ΔL = 2×
= 2.04mH
.98
ΔL = 2.04 – 2.00 = 0.04mH Ratio ∆ ∆
=
.4
= 0.02 &
= 0.02
Example 3.The output voltage of a LVDT is 1.5 V at minimum displacement. At a load of 0.5 M Ω, the deviation from linearity is maximum and it is ± 0.003 V from a straight line through origin. Find the linearity at the given load. Solution
Linearity = ±
. .
×100 = ±0.2%.
Example 4 A steel cantilever is 0.25mm long, 20mm wide and 4mm thick. (a) Calculate the value of deflection at the free end for the cantilever when a force of 25 N is applied at this end. The modulus of elasticity for steel is 200 GN/m 2.
(b) An LVDT with a sensitivity of 0.5 V/mm is used. The voltage is read on a 10 V voltmeter having 100 divisions. Two – tenths of a division can be read with certainty. (c) Calculate the minimum and maximum value of force that can be measured with this arrangement. Solution:
(a)
Moment of area of cantilever l = bd3 =
×(0.02)×(0.004)3 =
0.107×10−9 m4, Deflection (b)
x=
=
×(.) ×× ×.×
6.8
Deflection per unit force =
= 6.08 mm.
= 0.2432 mm/N.
Overall sensitivity of measurement system: = (0.2432 mm/N)×(0.5 V/mm) = 0.1216 V/N. 1scale division = (10/1000) = 0.1 V, Sincetwo – tenths of a scale division can be read with certainty, Resolution = (2/10)×0.1 = 0.02 V. (c)
Minimum force that can be measured = 0.02/0.1216 = 0.1645 N. Maximum force that can be measured = 10/0.1216 = 82.2 N
Example 5 Fig shows a capacitive transducer using five plates. The dimensions of each plate are 25×25mm and the distance between plates is 0.25mm. This arrangement is to be used for measurement of displacement by observing the change in capacitance with distance x. calculate the sensitivity of the device. Assume that the plates are separated by air. The permittivity of air is 8.85×10 -12F/m. Solution: The five plate transducer forms a combination of four capacitors connected in parallel. The movable plate is moved through a distance x on the right side,
Capacitance of each capacitor
Cʹ =
∈˳(−)
Where w = width of each plate. Capacitance of transducer
C = 4Cʹ =
−4∈˳
Sensitivity of transducer =
=
4∈˳(−)
4×8.8× ×× .×
= -3540pF/m.
= 3.54pF/mm. (Disregarding the sign)
Example 6. A capacitive transducer uses two quartz diaphragms of area 750 mm2 separated by a distance of 3.5 mm. a pressure of 900 KN/m 2 when applied to the top diaphragm produces a deflection of 0.6 mm. the capacitance is 370 pF when no pressure applied to the diaphragms. Find the value of capacitance after the application of a pressure of 900kN/m 2. Solution: Suppose C1 and C2 are respectively the values of capacitance before and after application of pressure. Let d 1 and d2 be the values of
distances between the diaphragms for the corresponding pressure conditions. C1 = ∈ / and C2 = ∈ /
=
∴ C2 = C1×
But
d1 = 3.5mm and d2 = 3.5-0.6 = 2.9mm
∴ Value of capacitance after application of pressure C 2 = 370×3.5/2.9 = 446.5 pF Example 7 A capacitive transducer is made up of two concentric cylindrical electrodes. The outer diameter of the inner diameter is 3mm and the dielectric medium is air. The inner diameter of the outer electrode is 3.1 mm. Calculate the dielectric stress when a voltage of 100 V is applied across the electrodes. Is it within safe limits? The length of electrodes is 20mm. calculate the change in capacitance if the inner electrode is moved through a distance of 2mm. The breakdown strength of air is 3kV/mm.
Solution: Length of air gap between the two electrodes (3.13)/2=0.05mm.
∴ Dielectric stress
= 100/0.05 = 2000 V/mm = 2 kV/mm.
The breakdown strength of air is 3kV/mm and hence the dielectric is safe. Capacitance of the transducer C=
=
log ( )
×8.8× ×× .
log ( )
F = 33.9 Pf.
The moving is electrode is shifted through a distance of 2mm.
∴
l = 20-2 = 18mm
New value of capacitance
=
×8.8× ×8× .
log ( )
F = 30.5 pF
∴ Change in value of capacitance ΔC = 33.9-30.5 = 3.4 Pf.
Example 8. A quartz Piezo-electric crystalhaving a thickness of 2 mm and voltage sensitivity of 0.055 V-m/N is subjected to a pressure of 1.5 MN/m2. Calculate the voltage output. If the permittivity of quartz is
40.6×10− F/m, calculate its charge sensitivity. Solution: Voltage output E0 = g t p = 0.055×2×10− ×1.5×106 = 165 V
Charge sensitivity
d = ∈ º ∈ ʳ g = 40.6× 10− × 0.055 = 2.23 × 10− C/N
Example 9 A piezo - electric crystal having dimensions of 5mm ×
5 × 1.5 and a voltage sensitivity of 0.055 V-m/N is used for force measurement. Calculate the force if the voltage developed is 100 V. Solution: The applied pressure is:
P= Force
˳
=
N/m2 = .×.×
1.2 MN/m2
F = PA = 1.2 × 106 × 5 × 5 × 1 0−6 = 30N
Example 10 A barium titanate pickup has dimensions of 5mm ×
5 × 1.25. The force acting on it is 5 N. The charge sensitivity of barium titanate is 150pC/N and its permittivity is 12.5×10 -9 F/m. If the modulus of elasticity of barium titanate is 12×10 6 N/m2, calculate the strain. Also calculate the charge and the capacitance. A = 5×5×10-6 = 25×10-6 m2,
Solution: Area of plates
Pressure Voltage sensitivity Voltage generated 106 = 3 V.
