Post on 10-Mar-2018
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
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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
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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 �
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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 � .
���
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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 �
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(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
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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
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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
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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�
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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)
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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� �
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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)
Electronics Measurement and Instrumentation 2 - 8 Bridge Measurements
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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.
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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
���
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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