50999516W23C Electronic Measurements and Instrumentation Solution Manual (1) PDF

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1 Units and Dimensions Solutions of Selected Examples for Practice Example 1.8.7 Solution : On right hand side, [f] [ML2 T -2 I -1 ] –2 –1 [B] = = = [MT I ] [A ] [L2 ] [L] –1 [l] = [L] and [v] = = [LT ] [T] \ [B] [l] [v] = [MT – 2 I – 1] [L] [LT –1 ] = [ML T 2 –3 I –1 ] These are dimensions of righ...

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EE ELECT RICAL MEASUREMENT S INST RUMENTAT ION Praveen P CLASS NOT ES ON ELECT RICAL MEASUREMENT S & INST RUMENTAT ION 2015 CLASS NOT ES ON Md Ift ekhar CALIBRAT ION AND T EST ING OF SINGLE PHASE ENERGY MET ER sat ya narayana

1

Units and Dimensions

Solutions of Selected Examples for Practice Example 1.8.7 Solution : On right hand side, [B] =

[] [ML2 T 2 I 1 ] = [MT = [A ] [L2 ]

[l] = [L]

and [v] =

[L] = [LT [T]

[B] [l] [v] = [MT – 2 I – 1] [L] [LT  These are dimensions of right hand side. [W] [ML2 T 2 ] = [e] = [Q] [T I] 2

–3

–1

–2

–1

I

–1

]

] 2

] = [ML T

–3

I

–1

]

–1

= [ML T I ] These are dimensions of left hand side, same as that of right hand side. This proves that the equation is dimensionally correct. Example 1.8.8 Solution :

Let us obtain the dimensions of each quantity on R.H.S. of the equation, [B] =

[] [M L2 T 2 I 1 ] = [MT = [A ] [L2 ]

–2 –1

I

]

2

[l] = [L], [b] = [L], A = [L ] 

[2b + l] = Length = [L]



[ ] =



[ ] =

Resistivity =

RA l

[R][A] [M L2 T 3 I 2 ] [L2 ] = [L] [l] 3

= [M L T

–3

I

–2

]

(1 - 1) TM

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R.H.S. =

1-2

Units and Dimensions

[MT 2 I 1 ] [L] [L] [L2 ] [L][M L3 T 3 I 2 ]

= [T I] and L.H.S. = [I] To satisfy the equation dimensionally, R.H.S. must be divided by [T] i.e. multiplied by [T 1 ]. The dimensions of  are cycles/sec. i.e. [T 1 ]. Hence there must be presence of  in the numerator of R.H.S. Hence the correct expression for Ie can be written as, Ie =

K B l b A (2b  l) 

Example 1.8.9 Solution : Let us find the dimensions of R.H.S. [E] [W] / [Q] = [R1] = [R2] = [R3] = [I] [I] [ML2 T 2 ] / [TI] 2 –3 –2 = [ML T I ] [I] The dimension of all resistances is same. 1 –1 [] = cycles/second = = [T ] … Cycles are dimensionless [T] = R

[C] =

[Q] [TI] [TI] = = [E] [W] [ML2 T 2 ] [Q] [TI]

= [M 

[R.H.S] =

–1

L

–2

4 2

T I]

[R]{1  [] 2 [R 3 ] 2 [C 3 ] 2 } 2

2

[] [R] [C 3 ]

2

where [R] = [R1] [R3] 2

Now as 1 is dimensionless, the dimensions of 1 +  R 23 C 23 is same as the dimensions of 2

 R 23 C 23 . 

[R.H.S] =

[ML2 T 3 I 2 ] [T 1 ] 2 [ML2 T 3 I 2 ] 2 [M1 L2 T 4 I 2 ] 2 [T 1 ] 2 [ML2 T 3 I 2 ] 2 [M1 L2 T 4 I 2 ]

= [T] 2

–3

–2

while [L.H.S.] = [R] = [ML T I ] So dimensionally equation is not correct. So on R.H.S., a quantity must be present to make the dimensions of R.H.S. same as L.H.S. 2 –3 –2 [T] [x] = [ML T I ]  

2

[x] = [ML T

–4

I

–2

] TM

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1-3

Units and Dimensions –1

Thus R.H.S. must be multiplied by x, i.e. divided by x –1

2

–4

–2 –1

–1

–2

.

4 2

x = [ML T I ] = [M L T I ] But these are the dimensions of capacitor C3 in the given equation. Hence for having dimensional correctness of the equation, R.H.S. must be divided by one more C3. Hence the equation becomes, R4 =

R 2 (1  2 R 23 C 23 ) 2 R 1 R 3 C 23

Example 1.8.10 Solution :

Now 

Let us find the dimensions of quantities involved on R.H.S. of the equation. [] [B] = [A ] E = N

d dt

[ ] = [E] [T] =

… As N is dimensionless.

[W] [ML2 T 2 ] [T] = [T] [Q] [TI] 2

– 2 –1

= [ML T 

[B] =

I

[ML2 T 2 I 1 ] 2

[L ]

]

= [MT

–2

I

–1

]

2

[A] = [L ] Now

B = 0 H

… For air gap

H =

NI [I] –1 = [L I] = [L] l



0 =

[B] [MT 2 I 1 ] +1 = = [M L T 1 [H] [L I]



[R.H.S.] = [MT a

–2

= [M T

I

–1 a

2 b

–2a

–a

2b

a+c

… As N is dimensionless.

+1

T

c

+c

] [L ] [ML

I

] [L ] [M L

2b + c

– 2a – 2c

–2c

–2

T

I

I

–2

]

–2 c

– 2c

]

I

– 2c

]

– a – 2c

= [M L T I ] The pull on electromagnet is nothing but a force measured in newtons. So dimensions of L.H.S. are dimensions of force. –2 [L.H.S.] = [F] = [MLT ]  Equating the dimensions of L.H.S. and R.H.S. and to satisfy the given relation dimensionally all powers of L,M,T and I on both sides must be same. TM

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1-4

Units and Dimensions

a+c = 1

… (1)

2b + c = 1

… (2)

– 2a – 2c = – 2

… (3)

– a – 2c = 0 Adding (1) and (4), c = – 1 From (1), a = 2

… (4)

From (2), b = 1 Thus the given expression can be written as, P = K B A  0 1 = 2

where

KB 2 A 0

K = Constant of proportionality.

Example 1.8.11 Solution : Use M.K.S.A. system. L.H.S. = [I] = [I] For right hand side,

... In M.K.S.A. system

[] = rad/sec = [T  1 ]

... Radians is dimensionless

[M] = [L] = [ML2 T  3 I  2 ]

... Inductance

[E] = [ML2 T  3 I  1 ] [R] =

... Voltage or e.m.f.

[E] [ML2 T  3 I  2 ] [I]

... Resistance

 ME = [T  1 ] [ML2 T  2 I  2 ] [ML2 T  3 I  1 ] [M2 L4 T  6 I  3 ] 2 M2 = [T  1 ] 2 [ML2 T  2 I  2 ] 2 [M2 L4 T  6 I  4 ] R 1 R 2 = [R] 2 [ML2 T  3 I  2 ] 2 [M2 L4 T  6 I  4 ] 

[2 M2  R 1 R 2 ] 2 = {[M2 L4 T  6 I  4 ]} 2 [M4 L8 T  1 2 I  8 ] 2 L 1 R 21 = [T  1 ] 2 [ML2 T  2 I  2 ] [M2 L4 T  6 I  4 ] [M 3 L6 T  10 I  6 ] Now

2 M2  R 1R 2 

2

and 2 L 1 R 21 can be added only if these two are

dimensionally same. As the dimensions are not same, there is error in the equation. To make the dimensions equal, the dimensions of 2 L 1 R 2 must be multiplied by [ML2 T  2 I  2 ]. These are the dimensions of inductance L.

TM

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1-5

Units and Dimensions

Hence the term 2 L 1 R21 must be 2 L 1 L 2 R21 . [2 M2  R 1 R 2 ] 2  2 L 1 L 2 R 21 = [M4 L8 T  12 I  8 ]

 

R.H.S. =

[M2 L4 T  6 I  3 ] M4

L8

T  12

I 8



[M2 L4 T  6 I  3 ] 2

4

[M L T

6

I

4

]



1 [I

1

]

= [I] = L.H.S. Thus, now the equation is dimensionally correct. The correct equation is,



I =

 ME

2 M2  R 1R 2 

2

 2 L 1 L 2 R 21

qqq

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Basics of Measurement and Error Analysis

2

Solutions of Examples for Practice Example 2.4.8 Solution : A t = true value = 8.5 A A m = measured value = 8.3 A Absolute error = A t – A m = 8.5 – 8.3 = 0.2 A %e =

0.2 At - Am ´ 100 = ´ 100 = 2.35 % At 8 .5

Example 2.4.9 Solution :

= 111.5 V

Am

% e = 5.3 % Now \ \

%e =

At - Am ´ 100 At

5.3 =

A t - 111.5 ´ 100 At

0.053 A t = A t – 111.5 A t = 117.74 V

\

Example 2.4.10 Solution : 1 scale division = \

Resolution =

full scale division 100 = = 0.5 V number of divisions 200 1 2

´ scale division =

1 ´ 0. 5 2

= 0.25 V Example 2.7.4 Solution : The result is tabulated as shown where di is the deviation from mean.

(2 - 1) TM

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Basics of Measurement and Error Analysis

2

No. (n)

x

di = x – x

1

101.2

– 0.1

0.01

2

101.4

0.1

0.01

3

101.7

0.4

0.16

4

101.3

0

0

5

101.3

0

0

6

101.2

– 0.1

0.01

7

101.0

– 0.3

0.09

8

101.3

0

0

9

101.5

0.2

0.04

10

101.1

– 0.2

0.04

n = 10

i) Arithmetic mean, x =

å x = 1013

å

d i = 1.4

di

å d 2i = 0.36

Sx 1013 = = 101.3 n 10

ii) Deviation from mean = Average deviation =

S|d i| 1.4 = 0.14 = n 10

S di 2 0.36 = 0.2 V = 9 n -1 iv) Probable error of one reading = 0.6745 s = 0.6745 ´ 0.2 = 0.1349 V

iii) Standard deviation, s =

e m = Probable error of mean = =

0.6745 s n -1

0.1349 = 0.0449 10 - 1

Example 2.7.5 Solution : The results are tabulated as shown, TM

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Basics of Measurement and Error Analysis

No. (n)

x

d i =x -x

d i2

1

29.6

– 1.975

3.9006

2

32.4

0.825

0.6806

3

39.4

7.825

61.2306

4

28.9

– 2.675

7.1556

5

30.0

– 1.575

2.4806

6

33.3

1.725

2.9756

7

31.4

– 0.175

0.0306

8

29.5

– 2.075

4.3056

9

30.5

– 1.075

1.1556

10

31.7

0.125

0.0156

11

33.0

1.425

2.0306

12

29.2

– 2.375

5.6406

n = 12

å x = 378.9

å |d 2i | = 23.85

å |d 2i | = 91.6022

i)

Mean value = Arithmetic mean = x = =

Sx n

378.9 = 31.575 12

ii) When values are arranged in ascending order then, Median value = x(n + 1)/ 2 for odd values = midway between centre two values for even values For given set in ascending order x 1 = 28.9, x 2 = 29.2, x 3 = 29.5, x 4 = 29.6, x5 = 30.0, x 6 = 30.5, x7 = 31.4 x 8 = 31.7, x 9 = 32.4, x 10 = 33.0, x 11 = 33.3, x 12 = 39.4 As n = 12 is even, centre two values are x 6 and x7 x + x7 30.5 + 31.4 = = 30.95 \ x median = 6 2 2 iii) Standard deviation, s =

å di 2 n -1

=

91.6022 12 - 1

= 2.8857

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Basics of Measurement and Error Analysis

Example 2.7.6 Solution : The values obtained from the measurements are tabulated as follows. Sr. No.

x

d i = x1 – x

d 2i

1.

147.2

– 0.39

0.1521

2.

147.4

– 0.19

0.0361

3.

147.9

+ 0.31

0.0961

4.

148.1

+ 0.51

0.2601

5.

147.7

+ 0.11

0.0121

6.

147.5

– 0.09

0.0081

7.

147.6

+ 0.01

0.0001

8.

147.4

– 0.19

0.0361

9.

147.6

+ 0.01

0.0001

10.

147.5

– 0.09

0.0081

å x = 1475.9 a) Arithmetic mean,

x=

b) Standarad deviation,

å

å

x

n

=

1475.9 = 147.59 10

d 2i n -1

å

s =

å d 2i = 0.609

d i = 1.9

as

n < 20

0.609 = 0.2601 9

=

c) The probable error of average of the ten readings, em = 0.6745

s 0.2601 = 0.6745 n -1 10 - 1

= 0.058479 Example 2.8.13 Solution : The thousands place is 4 and its value is 4000. 0 . 025 =±1W \ error = ± 4000 ´ 100 The hundred place is 3 and its value is 300. \

error = ± 300 ´

0 . 075 = ± 0.225 W 100

The tens place is 5 and its value is 50. TM

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error = ± 50 ´

2-5

Basics of Measurement and Error Analysis

0 .1 = ± 0.05 W 100

The units place is 2 and its value is 2. \ \

error = ± 2 ´

0 . 15 = ± 0.003 W 100

Total error = ± [1 + 0.225 + 0.05 + 0.003] = ± 1.278 W

\

% limiting error = ±

± 1 . 278 ´ 100 = ± 0.0293 % 4352

Example 2.8.14 Solution : The limiting error at full scale is, 1.5 dA = 1.5 % of 10 = ´ 10 = 0.15 100 For a reading of 2.5 A it is, %e =

0.15 ´ 100 = 6 % 2.5

... % Limiting error

The limiting values of current are, I = 2.5 ± 0.15 A = 2.35 to 2.65 A Example 2.8.15 Solution : The unknown resistance is, Rx =

R2 R 3 2700 ´ 470 = = 10575 W R1 120

Now the relative limiting error in R x is, Now

e T = ± [e1 + e 2 + e 3 ] = ± [0.1 + 0.5 + 0.5] = ± 1.1 % dRx dRx eT = ± ´ 100 i.e. 1.1 = ± ´ 100 Rx 10575 d R x = ± 116.325 W

Hence the guaranteed values of the resistance is between, \

Rx – dRx < Rx < Rx + dRx 10575 – 116.325 < Rx < 10575 + 116.325 i.e. 10458.675 < Rx < 10691.325

Example 2.8.16 Solution : Limiting error for both is ± 1 % of full scale. \

da 1 =

1 ´ 250 = 2.5 V 100

… For voltmeter

\

da 2 =

1 ´ 500 = 5 mA 100

… For ammeter TM

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2-6



e1 =

da 1 2.5 = = 0.01993 A1 125.4

\

e2 =

da 2 5 ´ 10 - 3 = 0.01733 = A2 288.5 ´ 10 - 3

i)

R indicated =

Basics of Measurement and Error Analysis

… A 1 = 125.4 V … A 2 = 288.5 mA

A1 125.4 = = 434.662 W A2 288.5 ´ 10 - 3

ii) For division of V and I, the resultant error is e T = ± [e 1 + e 2 ] = ± [0.01993 + 0.01733] = ± 0.03726 i.e.

e T = ± 3.726 %

\

eT = ±

\

0.03726 = ±

\

dR x where R x = 434.662 W Rx dR x 434.662

± dR x = ± 16.1955 W

Hence the limits within which the result can be guaranteed is, 434.662 – 16.1955 to 434.662 + 16.1955 i.e. 418.4665 W to 450.8575 W Example 2.8.17 Solution : Refer example 2.8.5 for the procedure and verify the answers as : Limiting error = ± 21.66 %, Limiting error in ohms = ± 721.982 W, Range = 2.611 kW to 4.055 kW. Example 2.8.18 Solution : Refer example 2.8.16 for the procedure and verify the answer as : i) 435.2733 W ii) 418.778 W to 451.7684 W. Example 2.8.19 Solution : Refer example 2.8.15 for the procedure and verify the answers as : i) 200 W, ii) ± 5.5 %, iii) ± 11 W.

qqq

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Analog and Electronic Instruments

Solutions of Selected Examples for Practice Kept this unsolved example for student's practice.

Example 3.4.2 Example 3.6.2

Solution : R m = 100 W, Full scale deflection current I m = 2 mA, Current range required I = 0.150 mA,

m=

R sh =

I

=

Im

150 ´ 10 -3 2 ´ 10 -3

= 75

Rm 100 = = 1.3513 W m - 1 75 - 1

Power consumption of meter is, Pm =

(I m ) 2 R m = (2 ´ 10 -3 ) = 100 = 0.0004 W 2

Power consumption of shunt, Psh = Total

(I m ) 2 R sh = [(150 - 2) ´ 10 -3 ] ´ 1.3513 = 0.0295 W 2

P = Pm + Psh = 30 mW

Example 3.6.3 Solution : Rm = 100 W , Im = 1 mA, I = 100 mA Rm 100 Rsh = = = 1.0101 W é I - 1ù é100 - 1ù ûú ûú ëê 1 ëêI m Now the Rsh is doubled i.e. 2 ´ 1.0101 = 2.0202 W \

2.0202 =

\

100 I

ù é - 1ú ê -3 û ë1 ´ 10

1000 I - 1 = 49.5

i. e.

I = 50.5 mA

... New range of ammeter

Example 3.6.4 Solution : The arrangement is shown in the Fig. 3.1. (3 - 1) TM

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3-2

Ish

\

m =

Im

Rm = 25 W

Rsh

I m = 1 mA \

Im = 1 mA

I

i) Initially copper coil is directly across the shunt without 75 W resistance in series. I = 100 mA,

I

Analog and Electronic Instruments

75 W For Rx temperature compensation

R =1 + m R sh Fig. 3.1

25 100 = 1+ 1 R sh

\ R sh = 0.2525 W When temperature increases by 10 °C i.e. D t = 10 °C R 2 = R 1 [1 + a 1 D t] \

where a 1 = Resistance temperature coefficient at t 1...