Electrical measurement and instrumentation worksheet II
1. The resistance Rθ KΩ of a thermistor at θ K is given by:

1 
1
R  1.68 exp3050 

  298 

The thermistor is incorporated into a deflection bridge circuit given in the following figure.
Calculate the range of the output voltage corresponding to an input temperature range of 0
to 50oC. Assume that the output voltage is measured by an infinite impedance voltmeter.
Solution
for 0 0 C
T K   0  273.15  273.15K


1 
1 
1
 1
R  1.68 exp3050 

  1.68 exp3050

  298 
 273.15 298 


R273.15  4.2627 K
 R
R2 

Vout  Vs 

 R  R4 R2  R3 
4.2627 K
1K


Vout  2.56V 


 4.2627 K  1.22 K 1K  0.29 K 
Vout  0V
for 50 0 C
T K   50  273.15  323.15


1 
1 
1
 1
R  1.68 exp3050 

  1.68 exp3050

  298 
 323.15 298 


R323.15  0.7574K
0.7574K
1K


Vout  2.56V 


 0.7574K  1.22 K 1K  0.29 K 
Vout  1.0039V
 the range is
0 to 1V
2. In the Maxwell bridge shown in the following figure, let the fixed-value bridge components
have the following values: R3 =5Ω; C=1 mF. Calculate the value of the unknown
impedance (Lu, Ru) if R1 D 159=Ω and R2 =10Ω at balance.
Solution :
since the products of the resistance of opposite arms are equal
Ru R1  R2 R3
Ru 
R2 R3 10 * 5

 0.3145
R1
159
Lu  R2 R3 C  10 * 5 *1mF  50mH
3. A 2 mA meter with an internal resistance of 100 Ω is to be converted to 0-150 mA
ammeter. Calculate the value of the shunt resistance required in two ways.
Solution: Given values are,
Rm = 100 Ω, Im = 2 mA and I = 150 mA
Rsh 
Rsh =
Rsh 
I m Rm
I  I m 
2 x10 3 Ax100 
150 x10 3 A  2 x10 3 A

200 x10 3 A
148 x10 3 A
Rsh 1.351

4. Design and draw a multirange d.c milliammeter with a basic meter having a resistance of 75 Ω
and a full scale deflection for the current of 2mA if the required ranges are: 0 -10mA, 0 – 50mA
and 0 – 100mA
Solution: The first range is 0 – 10 mA and hence I1 = 10 mA while Im = 2 mA and Rm = 75 Ω
I Rm
So, Rsh1  m
=
I1  I m
2 mA x75 150mA

18.75
10 mA 2mA
8mA
The second range is 0 – 50 mA and hence I2 = 50 mA while Im =
2 mA and Rm = 75 Ω
Rsh 2 
I m Rm
2 mA x75 150mA


 3.125
I 2  I m 50 mA 2mA
48mA
The second range is 0 – 100 mA and hence I3 = 100 mA while Im = 2 mA and Rm = 75 Ω
Rsh 3 
I m Rm
2 mA x75 150mA


1.53
I 3  I m 100 mA 2mA
98mA
5. Design and draw an Ayrton shunt to provide an ammeter with current ranges 1A, 5A and
10A. A basic meter resistance is 50Ω and its full scale deflection current is 1mA.
Solution:
In position '1', R1 is shunt with R2 +R3 + Rm
Thus, I1 R1   I m R2  R3  Rm  where I1= 10A, Im
=1mA and Rm = 50Ω,
Then, 10A R1 = 1x10-3A (R2 +R3 + Rm)
R1 = 10 -4 ((R2 +R3 + 50Ω) ----------------------- (a)
In position '2', R1 + R2 is shunt with R3 + Rm
Therefore,
where I2 = 5A,
I 2 R1  R2   I m R3  Rm 
5A (R1 + R2) = 1x10-3A (R3 + 50Ω)
Thus, R1 + R2 = 2x10- 4 (R3 + 50Ω) ----------------- (b)
In position '3', R1 + R2 + R3 is shunt with Rm
So, I 3 R1  R2  R3   I m Rm where I3= 1A,
1A (R1 + R2 + R3) = 1x10-3A x 50Ω
R1 + R2 + R3 = 0.05Ω ------------------------------------------- (c)
From equation (c), R1 + R2 = 0.05 Ω - R3
Substituting in equation (b) we get,
0.05 Ω - R3 = 2x10- 4 (R3 + 50Ω)
0.05 Ω - R3 = 2x10- 4 R3 + 0.01Ω
0.05 Ω - 0.01Ω = R3 (1 +2x10- 4)
Thus, R3 =
0.04
 0.0399 .
1.0002
So, R1 + R2 = 0.05 Ω – 0.0399 Ω = 0.01Ω
Then R2 = 0.01Ω - R1
Substituting R2 in equation (a),
R1 = 10 - 4 (0.01Ω - R1 + 0.0399Ω + 50Ω)
R1 = 10 - 6 Ω – 10 - 4 R1 + 3.99 – 6 Ω + 5- 3 Ω
1.0001 R1 = 5.00499 - 3 Ω
R1 = 0.005 Ω
And
R2 = 0.01 Ω - 0.005 Ω = 0.005 Ω
From the above result, the shunt resistances of the designed Ayrton shunt are:
R1 = 0.005 Ω
R2 = 0.005 Ω and
R3 = 0.0399Ω
6. A moving coil instrument gives a full scale deflection with a current of 40 µA, while the
internal resistance of the meter is 500Ω. It is to be used as a voltmeter to measure a range
of 0 – 10V. Calculate the multiplier resistance needed.
Solution: Given values are: Rm = 500Ω, Im = 40 µA and V = 10V
Thus RS 
VS
10V
 Rm 
 500  249.5 K
Im
40 x 10 6 A
7. Calculate the value of the multiplier resistance on the 500 V range of a d.c. voltmeter, that
uses 50µAmeter movement with an internal resistance of 200 Ω. Using the sensitivity
method of calculating the multiplier (series string) resistances.
Solution: The sensitivity of the meter is:
S
Now
1
1

 20000  V  20 K V
I m 50 x10 6 A
RS  SV  Rm  20000  / V x 500V   200
RS 10000000  200
RS  9998000   9.998 M  10 M
8. A galvanometer has a current sensitivity of 1µA/mm and a critical damping resistance of
1KΩ. calculate:
i.
the voltage sensitivity
ii.
megohm sensitivity
Voltage sensitivity= critical damping resis tan ce* a current sensitivity
i.
Voltage sensitivity  1kΩΩ1μA/mm
Voltage sensitivity  1mV/mm
voltage sensitivity
current sensitivity
ii.
1V / mm
megohm sensitivity 
1A / mm
9. A PMMC instrument with FSD=50 µA and Rm=1700Ω is to be employed as a volt meter
with ranges of 10V, 50V, and 100V. calculate the required values of multiplier resistors
for universal multiplier resistances.
megohm sensitivity 
Rm  R1 
R1 
V1
Im
V1
10V
 Rm 
 1700
Im
50A
R1  198.3K
Rm  R1  R2 
R2 
V2
Im
V2
50V
 ( Rm  R1 ) 
 (1700  198.3K)
Im
50A
R2  800K
Rm  R1  R2  R3 
R3 
V3
Im
V3
100V
 ( Rm  R1  R2 ) 
 (1700  198.3K  800 K)
Im
50A
R3  1M