EKT212/4 Principles of Measurement and Instrumentation
(Sem2 2014/15)
Tutorial 2
Part 1:
(a) Explain with a circuit diagram the operation of a basic dc ammeter using PPMC
instrument.
(b)
What the value of shunt resistance is required for using 50 µA meter movement having
an internal resistance of 250 Ω for measuring current in the range of 0-500 mA?
Rs 
(c)
I m Rm
(50 )( 250)

 25m
I  I m 500m  50
Design a multirange ammeter with ranges of 0-1 A, 5 A, 25 A, 125 A employing
individual shunts for each range. A D’ Arsonval movement with an internal resistance of
730 Ω and a full scale current of 5 mA is available.
I m Rm
(5m)(730)

 3.67
I1  I m
1  5m
I R
(5m)(730)
Rs 2  m m 
 0.73
I2  Im
5  5m
I R
(5m)(730)
Rs 3  m m 
 0.15
I3  I m
25  5m
I R
(5m)(730)
Rs 4  m m 
 0.03
I 4  I m 125  5m
Rs1 
EKT212/4 Principles of Measurement and Instrumentation
(d)
(Sem2 2014/15)
Design an Ayrton shunt to provide an ammeter with current ranges of 0 -1 mA, 5 mA, 20
mA, and 50 mA, using a D’Arsonval movement having internal resistance of 50 Ω and a
full scale current of 100 µA.
EKT212/4 Principles of Measurement and Instrumentation
(e)
(Sem2 2014/15)
Figure 1 shows an Aryton shunt of an ammeter with a current range of 0 – 1 mA, 10 mA,
50 mA and 100 mA. A D’ Arsonval movement with an internal resistance of 100 Ω and
full scale current of 50 µA is used. Propose the value of Ra, Rb, Rc and Rd for this
design.
Figure 1
EKT212/4 Principles of Measurement and Instrumentation
(f)
(Sem2 2014/15)
A current meter that has an internal resistance of 100 Ω is used to measure the current
through resistor Ra in Figure 2. Determine the percentage of error of the reading due to
ammeter insertion.
Figure 2
Req 
(1k )(1k )
 500
1k  1k
I a (calculated) 
V
3

 2mA
R1  Req 1k  500
V
3

 1.875mA
R1  Rm  Req 1k  500  100
I
I
2m  1.875m
E %  a ( calculated) a ( measured) x100% 
x100%  6.25%
I a ( calculated)
2m
I a ( measured) 
EKT212/4 Principles of Measurement and Instrumentation
(Sem2 2014/15)
Part 2:
Question 1
(a) Explain with a circuit diagram the operation of a basic dc voltmeter using PPMC
instrument.
multiplier is to limit the current through movement so that it does not exceed full scale
deflection.
(b) Define sensitivity of voltmeters. What is the significance of sensitivity in voltmeters?
Sensitivit y 
1
(/V)
I fs
Sensitivity is based on the full scale current (fsd) results whenever resistance is present in the meter circuit
for each voltage applied.
The ideal voltmeter should have an infinite internal resistance so that it reacts almost like an open circuit.
High internal resistance does not contribute to the measured voltage drop.
Low sensitivity means high fsd current, subsequently low internal resistance. Low sensitivity voltmeter
usually contributes to slightly high error compared to high sensitivity voltmeter.
EKT212/4 Principles of Measurement and Instrumentation
(Sem2 2014/15)
(c) A basic D’Arsonval movement with a full scale deflection of 50µA and having an
internal resistance of 1800 Ω is available. It is to be converted into a 0 – 1V, 0-5V, 020V, 0-100V multirange voltmeter using individual multipliers for each range. Calculate
the values of the individual resistor.
Step 1 for 1V (V1)
R1 
V1
1
- Rm 
 1800  18.2k
I fsd
50µA
Step 2 for 5V (V2)
R2 
V2
5
- Rm 
 1800  98.2k
I fsd
50µA
Step 3 for 20V (V3)
R3 
V3
20
- Rm 
 1800  398.2k
I fsd
50µA
Step 4 for 100V (V4)
R4 
V4
100
- Rm 
 1800  1998.2k
I fsd
50µA
EKT212/4 Principles of Measurement and Instrumentation
(Sem2 2014/15)
Question 2
(a) State the effect of using a voltmeter of low sensitivity.
A low sensitivity voltmeter may give a correct reading when measuring voltage in a low resistance circuit, but it
may produce unreliable reading in a high resistance circuit.
(b) Convert a basic D’Arsonval movement with an internal resistance of 100 Ω and a full
scale deflection of 1mA into a multirange dc voltmeter with the voltage ranges of a 0 –
1V,0-10V, 0-50V.
Step 1 for 1V (V3)
V3
1
 Rm 
 100  0.9k
I fsd
1m
R3 
Step 1 for 10V (V2)
R2 
V2
10
 ( Rm  R3 ) 
 (100  0.9k )  9k
I fsd
1m
Step 3 for 50V (V1)
R1 
V1
50
 ( Rm  R3  R2 ) 
 (100  0.9k  9k )  40k
I fsd
1m
EKT212/4 Principles of Measurement and Instrumentation
(Sem2 2014/15)
Question 3
Figure 3
Two different voltmeters are used to measure the voltage across R2 in the circuit in Figure 3.
The meters are as follows:
(i)
5 V range, S = 20 kΩ/V.
(ii)
10 V range, S = 20 kΩ/V.
Let Vs = 40V, R1 = 20 kΩ, R2 = 1 kΩ and R3 = 10 kΩ. Identify which voltmeter introduces
least error due to loading.
EKT212/4 Principles of Measurement and Instrumentation
(Sem2 2014/15)
EKT212/4 Principles of Measurement and Instrumentation
(Sem2 2014/15)
Question 4
Figure 4
(a) The circuit diagram of Figure 4 shows a full wave rectifier ac voltmeter. The meter
movement has an internal resistance of 250 Ω and required 1 mA for full scale
deflection. The diodes each have a forward resistance of 50 Ω and infinite reverse
resistance. Calculate:
(i) The ohms per volt rating of the ac voltmeter.
(ii) The series resistance required for full scale meter deflection when 25V rms is
applied to the meter terminals.
Solution
(i)
Rm = 250 Ω, Ifsd =1 mA, RD=50Ω.
Ohms per volt rating =
(ii)
Series resistance,
S dc 
0.9 0.9

 0.9k / V
I fsd 1mA
RS  Sac  Vrms  ( Rm  2 RD )
RS  0.9k / V  25V  (250  2(50))  22.15k
(b) Sketch the circuit and waveforms for an ac voltmeter using PMMC instrument and a
half wave rectifier. Explain the circuit operation.
At +ve cycle, D1 is forward biased. The current is flowing through the movement that caused
deflection.
At -ve cycle, D1 is reversed biased. No current is flowing through the movement, therefore no
deflection.
EKT212/4 Principles of Measurement and Instrumentation
(Sem2 2014/15)
(c) A D’Arsonval meter movement with a full scale deflection current rating 0f 1mA and
an internal resistance of 500Ω is to be used in a half wave rectifier ac voltmeter.
Calculate the ac and dc sensitivity and the value of the multiplier resistor for a
30Vrms range.
S DC 
1
I fsd

1
 1k / V
1m
S AC  0.45S DC  0.45(1k )  0.45k / V
Rs  (S AC xVAC )  Rm  (0.45k )(30)  500  13k
Part 3:
(a) Explain with a circuit diagram the operation of a series type ohmmeter.
EKT212/4 Principles of Measurement and Instrumentation
(Sem2 2014/15)
(b) A 1mA full scale deflection current meter movement is to be used as an ohmmeter
circuit. The meter movement has an internal resistance of 500 Ω and a 5V battery will
be used in the circuit. Mark off the meter face (dial) for reading resistance with a
20%, 50% and 100% deflection.
Solution
Rm = 500 Ω, Ifsd =1 mA, E=5V
The value of Rs which limit current to FSD current,
Rs 
Rs 
E
I fsd
5V
 500  4.5k
1mA
The value of Rx with 20% deflection
Rx 
5V
 4.5  0.5k  20k
0.2mA
The value of Rx with 50% deflection
Rx 
5V
 4.5  0.5k  5k
0.5mA
The value of Rx with 100% deflection
Rx 
5V
 4.5  0.5k  0k
1mA
 Rm
EKT212/4 Principles of Measurement and Instrumentation
(Sem2 2014/15)
(c) With reference to Figure 5, state the ohmmeter scale when the current is 0A, 1/2 FSD,
1/4FSD and FSD
Figure 5
I x Rx  I m Rm
Rx 
I m Rm
I R
 m m
Ix
I  Im
The value of Rx at 0A scale(Im=0):
Rx 
I m Rm 0(500)

 0
I  I m 5  0
The value of Rx at ½ FSD (Im=0.5Ifsd):
Rx 
0.5I fsd Rm

I  0.5I fsd
0.5(5 )(500)
 500
5  0.5(5 )
The value of Rx at 1/4 FSD (Im=0.25Ifsd):
Rx 
0.25I fsd Rm
I  0.25I fsd

0.25(5 )(500)
 166.7
5  0.25(5 )
The value of Rx at FSD (Im=Ifsd):
Rx 
I fsd Rm
I  I fsd

(5 )(500)
 
5  5