Basic Electronics and Electrical Engineering: Chapter 6: Measurement and Instrumentation

Measurement and Instrumentation: Important Solved Examples Problems

Anna University Important Solved Examples Problems with Formula, Equation, Explained Solution - Basic Electronics and Electrical Engineering: Chapter 6: Measurement and Instrumentation

Basic Electronics and Electrical Engineering


Chapter 6: Measurement and Instrumentation

 

SOLVED EXAMPLES

 

EXAMPLE 1

A moving coil instrument has the following data, number of turns 100, width of coil 2 cm, depth of coil 3 cm and flux density in the air gap is 0.1 wb/m2. Calculate the deflecting torque when carrying a current of 10 mA. Also calculate the deflection if the control spring constant is 20×10‒7 m/degree.

Solution

Deflecting torque,

Td=NBIA

Td = 100×0.1×10×10‒3×0.02×0.030

Td=6×10‒5 Nm

Controlling Torque,

Tc=KSθ

Tc=20×10‒7 × θ

At steady state position,

Tc = Td

20×10‒7 × θ = 6×10‒5

 θ = (6×10‒5) / (20×10‒7)

θ = 30°

 

EXAMPLE 2

The PMMC voltmeter has a coil of 120 turns wound on a rectangular former 35 mm depth and 20 mm wide. The flux density is 20 wh/m2. Calculate the deflecting torque for a coil current of 25 mA.

Solution

Deflecting torque, Td = NBIA

= 120 × 20×10‒3 × 25×10‒3 × 35×10‒3 × 20 ×10‒3

Td = 21×10‒6 Nm

 

EXAMPLE 3

A PMMC voltmeter has 150 turns wound on a rectangular former of size 30×20 mm. The flux density is 0.2 wb/m2. Calculate the deflecting torque for a coil current of 20 mA.

Solution

Deflecting torque, Td = NBIA

= 150 × 0.2 × 20×10‒3 × 30×10‒3 × 20×10‒3

Td = 36 × 10‒5 Nm

 

EXAMPLE 4

The change of inductance for a moving iron ammeter is 3 mH/degree. The control spring constant is 5×10‒7 N‒m/degree. The maximum deflection of the pointer is 100°. What is the current corresponding to maximum deflection?

Solution

Deflection, θ = ½ (I2/KS) × (dL/dθ)

current, I2 = (θ×2×KS) / (dL/dθ)

= (100×2×5×10‒7) / (3×10‒6)

I2 = 33.33

I = 5.77 A

 

EXAMPLE 5

It is required to convert a 5mA meter with 20Ω internal resistor into 5A ammeter. Calculate the value of shunt resistance required and multiply factor of the shunt.

Solution

Multiply factor, n = I / Ish

= 5 / 0.005 = 1000

Rsh = Rn / n‒1 = 20 / (1000‒1)

Rsh = 0.02 Ω

 

Basic Electronics and Electrical Engineering: Chapter 6: Measurement and Instrumentation : Tag: Basic Engineering : - Measurement and Instrumentation: Important Solved Examples Problems


Basic Electronics and Electrical Engineering: Chapter 6: Measurement and Instrumentation



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