Solutions of Examples for Practice - · PDF fileSolutions of Examples for Practice Example...

Solutions of Examples for Practice

Example 1.4.5

Solution : A m = 111.5 V, % e = 5.3 %

Now, % e =A A

A

t m

t

�� 100 i.e. 5.3 =

A 111.5A

100t

t

��

� 0.053 A t = A t – 111.5 i.e. At = 117.74 V

Example 1.4.6

Solution :

1 scale division =Full scale division

Number of divisions=

100200

= 0.5 V

� Resolution =12

� Scale division =12

0 5� . = 0.25 V

Example 1.4.7

Solution : The overshoot of 5.18 – 5.02 = 0.16 A when final reading is 5.02 A.

a) % overshoot =0.165.02

100� � 3.187 %

b) % e =A A

A100

5 5.025

100t m

t

��

�� 0.4 %

Example 1.8.4

Solution : The result is tabulated as shown below where di is the deviation of ith reading

from the mean.

Number (n) x di = x – x di2

1 3.5 – 0.02375 5.6406 × 10– 4

2 3.452 – 0.07175 5.14806 × 10– 3

3 3.620 0.09625 9.26406 × 10– 3

4 3.523 – 7.5 × 10– 4

5.625 × 10– 7

n = 4 �x = 14.095 � |di| = 0.1925 �di2 = 0.0149767

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM(1 - 1)

1 Measurement Errors and Standards

1. Arithmetic mean = x =�xn

=14.095

4= 3.52375

2. Average deviation = D =�|d |

ni =

0.19254

= 0.048125

3. Standard deviation = � =� d

1i2

n �for n < 20

� � =0.0149767

4 1�= 0.07065

Example 1.8.5

Solution : Refer example 1.8.4 for the procedure and verify the answers as 2.75 V, 1.25,1.5411.

Example 1.8.6

Solution : The result is tabulated as shown where di is the deviation from mean.

No.(n)

x d x – xi �d

i

2

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 x �� 1013 d 1.4i �� d 0.36i2 ��

i) Arithmetic mean, x =� x

n=

101310

= 101.3

ii) Deviation from mean = Average deviation =�| |d

ni =

1.410

= 0.14

iii) Standard deviation, � =� d

ni2

1�=

0.369

= 0.2 V

iv) Probable error of one reading = 0.6745 � = 0.6745 0.2� = 0.1349 V

Electronics Measurement and Instrumentation 1 - 2 Measurement Errors and Standards

TECHNICAL PUBLICATIONS - An up thrust for knowledgeTM

em = Probable error of mean =0.6745

1

�

n �=

0.1349

10 1�= 0.0449

Example 1.9.7

Solution : The limiting error at full scale is,

A = 1.51.5

% .of 10100

10 015� � �

For a reading of 2.5 A it is,

% e =0.152.5

� 100 = 6 % ... % Limiting error

The limiting values of current are,

I = 2.5 0.15 A = 2.35 to 2.65 A

Example 1.9.8

Solution : The unknown resistance is,

Rx =R R

R2 3

1=

2700 470

120

�= 10575 �

Now the relative limiting error in Rx is,

eT = [ ]e e e1 2 3

� � = [0.1 0.5 0.5]� � = 1.1 %

Now eT = R

Rx

x� 100 i.e. 1.1 =

Rx

10575� 100

Rx = 116.325 �

Hence the guaranteed values of the resistance is between,

Rx– Rx < Rx < Rx + Rx

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

Example 1.9.9

Solution : Limiting error for both is 1 % of full scale.

� a 1 �1

100250 2.5 V� � … For voltmeter

� a 2 �1

100500 5 mA� � … For ammeter

� e1 � a

A

2.5

125.40.019931

1� � … A 125.4 V1 �

� e2 � a

A

5 10

288.5 100.017332

2

3

3�

�

��

�

�… A 288.5 mA2 �

i) Rindicated =A

A125.4

288.5 10

1

2 3�

��

�434.662 �



ii) For division of V and I, the resultant error is

eT = � � � � [e e ] [0.01993 0.01733] 0.037261 2

i.e. eT = 3.726 %

� eT � R

Rx

x

where Rx � ��

� �� R

434.662x

� Rx � 16.1955 �

Hence the limits within which the result can be guaranteed is,

434.662 – 16.1955 to 434.662 + 16.1955 i.e. 418.4665 � to 450.8575 �

Example 1.9.10

Solution : Refer example 1.9.6 for the procedure and verify the answers as :

Limiting error = 21.66 %, Limiting error in ohms = 721.982 �, Range = 2.611 k� to4.055 k�.

Example 1.9.11

Solution : Refer example 1.9.9 for the procedure and verify the answers as :

i) 435.2733 �

ii) 418.778 � to 451.7684 � .

��



Solutions of Examples for Practice

Example 2.2.5

Solution : For the bridge shown,

R = 500 � and �R = 20 �

Using approximate result,

VTH

=E

4 R

�R=

10 20

4 500

�

�= 0.1 V

while Req = R = 500 � and Rg = 125 � given

� Ig =V

R R

TH

eq g�=

01500 125

.�

= 160 �A

Example 2.2.6

Solution : For bridge balance,

R R1 3 = R R2 4

Here R4 = RR

RR

1010

10V1

23� � � = 10 k �

The R V is 10 k � when the temperature is 80C from the Fig. 2.2.11. Thus bridge isbalanced at 80C. At 60C, R V is 9 k� from the Fig. 2.2.11. Thus the bridge is unbalanced.

The error voltage is the voltage across galvanometer i.e. Thevenin’s voltage.

e = ER

R R

R

R R10

910 9

1010 10

4

3 4

1

1 2�

�

�

�

�

��

�

�

�

� �

��

= – 0.2631 VThe negative sign indicates opposite polarity of the error voltage. Thus error voltage is

0.2631 V.

Example 2.2.7

Solution : From given bridge, R1 = R2 = 100 �, R3 = 230 �

R4 =R R

R1 3

2=

100 230

100

�= 230 �


(2 - 1)

2 Bridge Measurements

Relative limiting error is,� R

R4

4=

� � �R

R

R

R

R

R1

1

3

3

2

2� � = (� 0.02 � 0.01 � 0.02)% = � 0.05 %

So limiting values of unknown resistance are,

= 230 � 0.05 % = 230 �0 05

100230

.�

�

�

�

�� = 230 � 0.115 �

= 229.885 � to 230.115 �

Example 2.3.3

Solution : A Kelvin double bridge can be drawn as shown in the Fig. 2.1.

By using the expression of unknown resistance we can write,

Rx =R

RR

bR

a b R

R

Rab

1

23

y

y

1

2�

� �

�

�

�

��

� Rx = � �100.24

200100.03 10 6� +

� �� 200 700 10

100.31 200 700 10

6

6

�

� � �

100.24200

100.31200

�

� �

��

� Rx = 49.97 � 10– 6� = 49.97 ��

Hence the magnitude of error in the measurement is given by,

error = Actual value of resistance Measured value of resistance

= 50 10 49 97 106 6� � . = 0.03 � 10– 6

Example 2.5.3

Solution : The bridge described above is as shown in the Fig. 2.2.

In general, the bridge balance condition is given by,

Z Zx1 = Z Z2 3

Electronics Measurement and Instrumentation 2 - 2 Bridge Measurements


R2 R

1

Ry

Rx

ab

1 2

3

4

5 6

G

+ –

E

R3

Standard resistor = R = 100.03

Outer ratio arms : R = 100.24 , R = 200

nner ratio arms : a = 100.31 , b = 200

Value of low resistance link = 700

3

1 2

��

� �

� �

��

I

Fig. 2.1

� Zx=Z Z

Z1

2 3 ...(1)

From Fig. 2.2 the impedance of arm ab is given by,

Z1 = � �R j X R j L R j f LL1 1 1 1 1 12� � � � ��

= � � � �50 2 50� �j � 0.1

= � �50 j 31.4159� �

The impedance of arm cd is given by,

Zx = R jX R jCx Cx x

x �

1�

Substituting values of all impedances inequation (1), we get,

Rj

Cxx

�

=� �� 100 1000

50 � j 31.4159

� Rj

Cxx

�

=100000 0º

59.0504 32.14º

�

�= 1693.4686 32.14º�

� R jCx

x

1�

= 1433.9457 – j 900.9081

Comparing real and imaginary terms we get,

Rx = 1433.9457 �

And,1

�Cx= 900.9081

� Cx =� ��

1 12� �

�900.9081 50 900.9081

��

� 3.5332 F.

Example 2.5.4

Solution : From the given information, the Maxwell's capacitance bridge is as shown in

the Fig. 2.3.

The equation for balance is,

Z Z1 4� = Z Z2 3�

i.e. � �R j LR

1+ j C R1 14

4 4�

�

�

�

��

�= ( ) ( )R R2 3

� R R j L R1 4 1 4� � = R R j R R R C2 3 2 3 4 4� �

Equating real terms, we get,

R R1 4 = R R2 3



A.C.

Source

50 Hz

Z3

Z2

0.1H

Zx

a

bd

c

Rx

Z1

50 �

100

�

G

1000

�

Cx

Fig. 2.2

a

b

c

dA.C.

supply

f Hz

D

L1

1000 �

R2

1000

�

Z2

C4

Z4

R1

Z1

1000

�

R3

Z3

0.5 F�

R4

Fig. 2.3

� R1 =R R

R2 3

4�

( ) ( )1000 1000

1000= 1000 �

Equating imaginary terms, we get

�L R1 4 = �R R R C2 3 4 4

� L1 = R R C (0.5 10 )2 3 46� � ( ) ( )1000 1000 = 0.5 H

Example 2.6.2

Solution : The bridge is as shown in the

Fig. 2.4.

Under balance condition, we have,

R2 = 2410 �

R3 = 750 �

R4 = 64.5 �

RC4 = 0.4 �

C4 = 0.35 �F

The bridge shown is Hay's bridge. In general, the bridge balance condition is given by,

Z Z1 4 = Z Z2 3

i.e. � �� R j R R j X1 L 1 4 C4 C4� � X = R2R3

� � �R j 64.5 0.4j

C1 L 1 4� �

�

��

�

� X

�= (2410) (750)

�� R j 64.9 j1

2 500 0.35 101 L 1 6� �

� � � �

�

� X

��

�

� = 1807500

� � �� R j 64.9 j 909.451 L 1� X = 1807500

� � �� R j 911.7627 85.91º1 L 1� � X = 1807500

� R j1807500

911.7627 85.91º1 L 1� �

�X = 1982.42 85.91º 141.3928 j 1977.37� � �

Comparing real term, R1 = 141.3928 �

Similarly comparing imaginary term, XL 1= 1977.37

� L1 =1977.37

2 500� ��

�0.6294 H



A.C.

supply

500 Hz

Z2

R2

R4

R3

C4

Z3

R1

L1

Z4

D

A

BD

C

Rc4

Z1

Fig. 2.4

Example 2.6.3

Solution : From given data, Hay's bridge can be

drawn as shown in the Fig. 2.5.

At f = 50 Hz, reactance offered by C4 is,

jXC4=

�

j

C

j

2 f C4 4� �

=

� � � �

j

2 50 0.38 10 6�

� j XC4= – j 8376.5759 �

� Z4 = R j X4 C4 = 833 – j 8376.5759

� Z4 = 8417.8924 �– 84.32º

Now in general for a.c. bridge, the condition of

balance is given by,

Z Z1 4 = Z Z2 3

� Z1 =Z Z

Z

(16800) (100)

8417.8929 84.32º2 3

4�

�= 199.5749 � + 84.32º

� Z1 = R jX R j L R j 2 f L1 L 1 1 1 1 1� � � � �� = 19.7524 + j 198.595 �

Comparing imaginary terms on both the sides,

2 f L1� = 198.595

L1 =198.595

2 f198.595

2 50� ��

� �= 0.6321 H

Comparing real terms on both the sides,

R1 = 19.7524 �

Thus, under balanced condition at 50 Hz, arm AB consists pure resistance R 19.75241 � � in

series with inductance L1 = 0.6321 H.

Example 2.7.2

Solution : Let us consider Wien bridge

as shown in the Fig. 2.6.

From bridge balance condition,

�2 =1

R C R C1 1 4 4

C4 =1

R C R21 1 4�

… (1)

NowR

R2

3=

R

R

C

C1

4

4

1�



D

Z1

Z4

Z3

R = 1003

�

R =

833

4

�

L1

R1

A

B

C

D

Z2

C = 0.38 F4

�

R

= 16800

2

�

a. c. supply @ 50 Hz

Fig. 2.5

R1

C4

R2

R3

R4

D

C1

A. C.

Supply

Fig. 2.6

Substituting value of C4 in above equation

R

R2

3=

R

R1

4�

1

R C R21 1

44�

…(2)

�30 10

1 10

3

3

�

�=

8 10 1

2 8 10 6 10

3

4 2 3 6 24

��

� � � � � � � �R R( ) ( )� 2.5 10 3

� 30 =8 10 3

4 4

��

R R0.01407

� R4 = 266.667 �

Putting R4 in equation (1), we get,

C4 =1

2 8 10 6 10 266 6672 3 6( ) .� � � � � � � �� 2.5 10 3

� C4 = 0.3166 nF

Example 2.7.3

Solution : Z1 = R1 – j1

1�C= R1 – j

1

1.5 10 6�� = R1 – j

666.667 10 3�

�

Z2 = R2 = 800 �

Z4 = R4 = 400 �

Z3 = R3 || C3

� Y3 =1

Z 3=

1R 3

+ j � C3

=1

330� � j 0.2 10 6 �

According to balance condition,

Z Z1 4 = Z Z2 3

i.e. Z2 =Z Z

Z1 4

3= Z1 Z4 (Y3)

� 800 = R j666.667 10

1

3

��

�

�

��

× 400 × [3.0303 × 10 – 3 + j 0.2 × 10 – 6�]

� 2 = 3.0303 × 10 3 R1 – j2020.203

�+ j 0.2 10 6� R1 � + 0.13334

� 2 = [3.0303 × 10 3 R1 + 0.13334] + j 0.2 10 R 202020.2036

1� �

� �

��

�

Equating real parts,

2 = 3.0303 × 10 3 R1 + 0.13334

� R1 = 616 � = R (unknown)



R1

C1

C3

R2

R4

R3

Detector

0.2 µF

A

1.5 µF

B

C

800 �

D

330 �400 �

Fig. 2.7

Equating imaginary parts,

0.2 × 10– 6 × R1 � –2020.203

�= 0

� �2 =2020.203

0.2 10 6166� �= 16397735.28

� � = 4.049411 krad/sec = 2� f

� f = 644.4838 Hz …Supply frequency

Example 2.8.2

Solution : The a.c. bridge is as

shown in the Fig. 2.8.

From circuit arrangement, it is

Anderson's bridge. The value of

unknown resistor is given by,

R1 =R R

Rr2 3

41

=( ) ( )250 100

200– 43.1 = 81.9 �

Unknown inductance is given by,

L1 = � �� CR

RR R r + R R3

42 4 2 4�

=1 10 100

200

6� �

[(250 + 200) (229.7) + (250) (200)]

= 0.0766 H = 76.6825 mH

Example 2.8.3

Solution : From the given data, an

Anderson's bridge can be drawn as shown

in the

Fig. 2.9.

By the standard formula, the unknown

resistance Rx in branch AB is given by,

Rx = R1 =� ��

� �

R R

R

600 800

6002 3

4� � 800 �

Similarly the unknown inductance inbranch AB is given by,

Lx = � �C R

RR r R r R R3

42 4 2 4� �



a

b

c

d

A.C.

voltage

source

1 HzR

4 R = 1003

�

D

229.7 �

r

r = 43.11

�

L , R1 1

=200 �

R = 2502

�

1 F�

E

C

Fig. 2.8

400 �

r

D

Rx

Lx

C

1 F�R = 800

3�R = 600

4�

A.C.

supply

at

100 Hz

C

EBD

R = 6002

�

A

Fig. 2.9

=� ��

� �� 1 10 8006�

� � �

600600 400 600 400 600 600 = 1.12 H

Example 2.10.4

Solution : According to the data, the

low voltages Schering bridge can be

drawn as shown in the Fig. 2.10.

Z1 = Impedance of branch AB

= R ||1

j C11�

�

��

�

�

Y1 =1

Rj C

11� �

Z2 = Impedance of branch

AD =1

j C2�

�

��

�

�

Y2 = j C2�

Z 3 = Impedance of branch BC

= R ||1

j C33�

�

��

�

�

� Y3 =1

Rj C

33� �

Z4 = Impedance of branch CD = R ||1

j C44�

�

��

�

�

� Y4 =1

Rj C

44� �

At balance,

Y1Y4 = Y2 Y3

�1

Rj C

1R

j C1

14

4��

�

�

��

�

�

�� = � �j C

1R

j C23

3� ��

�

�

��

�1

R RC C j

C

R

C

R1 4

21 4

4

1

1

4

�

�

�

��

�

�

�

�� =

j C

RC C2

3

22 3

�� … (1)

Comparing real terms, we get,

1R R

C C1 4

1 4 �2 = �2 C C2 3 … (2)



C3

A.C.

supply

C

BD

A

D

C1

R3

C2

C4

Detector

R4

R1

Specimen

Fig. 2.10

Comparing imaginary terms, we get,

C

R

C

R4

1

1

4� =

C

R2

3… (3)

Simplifying equations (2) and (3) for C1, we get,

C1 =

C R

RC C C R

1 C R

2 4

3

22 3 4 4

2

242

42

�

�

�

�

But it is clear that,

� �! C42 R 1 and C C C R

C R

R42 2

2 3 4 42 2 4

3""" """ , we can write,

C1 =C R

R2 4

3… (4)

Case 1 : Without specimen inserted between terminals AB.

#C1 =� ��

� �

150 10 500

500

12�

= 150 10 F 150 pF12� �

Case 2 : With specimen inserted between terminals AB,

##C1 =� ��

� �

C R

R

900 10 50000

50000900 10 F 900 pF2 4

3

1212�

��

But in general, the capacitance is given by,

C =$ $ $Ad

A

d0 r

�

where A = Area of plates, d = Distance between plates.

Without specimen inbetween electrodes, the capacitance is given by,

#C =$0 A

dbecause $r = 1 for free space … (5)

Similarly with specimen inbetween electrodes, the capacitance is given by,

##C =$ $0 r A

d…(6)

� Dividing equation (6) by equation (5), we get,

$r =##

��

��

CC

900 10

150 10

12

126.



Example 2.10.5

Solution : For Schering bridge the equations at balance are,

r1 =C

CR and C

R

RC4

23 1

4

32�

� r1 =C

CR

0.5 10 260

106 10

4

23

6

12�

� �

��

1.23 � 106 �

C1 =R

RC 106 104

32

12� � � �( / )1000

260

�129.77 pF

The power factor of the sheet = � �C r 2 50 130 10 1.1 112� � � � � � 23 106�

= 0.05

The capacitance C1 = $ $r 0Ad

Hence relative permittivity is given by,

$r =C d

A130 10 4.5 10

8.854

1

0

12 3

$�

� � �

�

102

0.12122

� ��

��

�

�

�= 5.9

Example 2.10.6

Solution : The bridge can be drawn as shown

in the Fig. 2.11.

The unknown capacitance in the Schering

bridge is given by,

C1 = CR

R0.5 10

72.63002

4

3

9� � � = 12 pF

The dielectric loss angle of capacitor is,

� = � �tan 1 � C R4 4

= � �tan � � � � �1 2 50 0.148 10 72.66�

= 0.1934

��



a

b

c

dA.C.

supply

50 Hz

C1

0.148 F�

72.6 �

C2

500

pF

C4

R1

300 �

R3

R4

Fig. 2.11

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