THE COPPERBELT UNIVERSITY
SCHOOL OF ENGINEERING
ELECTRICAL ENGINEERING DEPARTMENT
EG 232-Measurements and Instruments, End of Term 2 Test Solution, 2017
Duration: 2 Hour
Instructions: Answer all the four questions, Each question has equal marks.
Question One
1. Identify the bridge in Fig.Q1 and derive the equation for determining the unknown
impedance. If R2=400Ω, R1=200 Ω C1=100 pF, C3=0.68 μF, calculate for Zx. What
type of load is Zx?
B
C1
Z2
R1
V
A
R2
D
C3
Z3
C
Zx
D
Fig.Q1
[7 Marks]
Solution
The bridge shown above is a Schering bridge which is used extensively in ther
measure ment of capacitance.
1
𝑌1 =
+ 𝑗𝜔𝐶1
𝑅1
𝑍2 = 𝑅2
𝑗
𝑍3 = −
𝜔𝐶3
𝑗
𝑍𝑋 = 𝑅𝑥 −
𝜔𝐶𝑥
𝐵𝑎𝑙𝑎𝑛𝑐𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛
𝑍1 ∙ 𝑍𝑥 = 𝑍2 ∙ 𝑍3
𝑍2 ∙ 𝑍3
𝑍𝑥 =
𝑍1
1
Since 𝑍1 = 𝑌 For simplicity we replace the impedance with the admittance since the
1
capacitor and resistor are connected in parallel.
𝑍2 ∙ 𝑍3
∴ 𝑍𝑥 =
= 𝑍2 ∙ 𝑍2 ∙ 𝑌1
𝑍1
𝑅𝑥 −
𝑗
𝑗
1
−𝑗𝑅2
𝑅2 𝜔𝐶1 𝑅2 𝐶1
𝑅2 𝑗
= 𝑅2 ∙ (−
) ∙ ( + 𝑗𝜔𝐶1 ) =
+
=
−
𝜔𝐶𝑥
𝜔𝐶3
𝑅1
𝜔𝐶3 𝑅1
𝜔𝐶3
𝐶3
𝜔𝐶3 𝑅1
the above equation has a real and imaginary parts. So we separate the two parts and
determine 𝑅𝑥 and 𝐶𝑥 independently.
𝑅2 𝐶1
𝑅𝑥 =
,
𝐶3
𝑗
𝑅2 𝑗
𝐶3 𝑅1
−
=−
→ 𝐶𝑥 =
𝜔𝐶𝑥
𝜔𝐶3 𝑅1
𝑅2
Now, If R2=400Ω, R1=200 Ω C1=100 pF, C3=0.68 μF,
400 ∙ 10−11
𝑅𝑥 =
= 5.9 ∙ 10−3 Ω
6.8 ∙ 10−7
200 ∙ 6.8 ∙ 10−7
𝐶𝑥 =
= 3.4 ∙ 10−7 𝐹 = 0.34 𝜇𝐹
400
𝑗
𝑍𝑥 = 𝑅𝑥 − 𝜔𝐶 = 𝑅𝑥 − 𝑗𝑋𝐶𝑥 If the frequency is 50 Hz,
𝑥
𝑗
𝑍𝑥 = 5.9 ∙ 10−3 − 2𝜋∙50∙3.4∙10−7 = 5.9 ∙ 10−3 − 𝑗9362.1 Ω
𝑍𝑥 = √𝑅𝑥2 + 𝑋𝐶2𝑥 = √(5.9 ∙ 10−3 )2 + (9362.1)2 = 9362.1 Ω
𝑍𝑥 is highly capacitive meaning it’s a capacitive load as the resistive part is negligible and
only accounts for the leakages resistance of a capacitor
2. With the aid of illustrations show how current transformers can be connected in Delta to
help in the measurement transmission line currents. Clearly show the polarity sides of the
CTs and relay terminals, and the currents flowing in the various branches of the circuit.
[5 Marks]
Solution
3. Starting with differentiating a transducer from a sensor, explain the operation principle of
a piezoelectric transducer and its main application in our day to day life.
[3 Marks]
Solution
The main difference between a transducer and a sensor is that the transducer converts
a physical quantity into electrical quantities (current / voltage) and starts the actuator,
while a sensor is a device that detects change in a physical quantity and sent this
information to a computer to start the actuator.
Examples of transducers:
 Linear variable differential transformer
 Piezoelectrical
 Photovoltaic
 Thermister
 Photo conductor ( light dependant resistor)
Examples of sensors:
 Smoke detector
 Motion detector
 Infrared (IR) sensor
 Proximity sensor
 Level sensor
The Piezoelectric transducer is used to convert mechanical tress, force and pressure
exited on it into electrical voltage. Its main applications in our daily life include:
 Automobile seat belts
 Automatic doors in airports e.t.c
 In microphone
Total [15 Marks]
Question Two
1. With the aid of an illustration, clearly explain the operation principles of a CRO and its
main applications.
[4 Marks]
Solution
Fig, S 2.1 showing the layout of and CRO.
An oscilloscope is a display device that converts electrical signals to visual signals. Its
main component is the cathode ray tube, whose main components are shown if the
fig.S2.1 above. The Filament that is connected to a low tension battery heats up the
cathode, which in turn releases the electrons through the thermionic process. The
anodes places in front of a cathode and kept at High voltage perform the functions of
focusing the electronic to form an electronic beam and accelerate it to the screen. The
deflection system is made up of two pairs of plates, X and Y. Y-plate are placed
horizontally to bring about Vertical deflection while X-plates are placed vertically to
bring about horizontal deflection. The end of the tube is made flate and coated with
phosphorus of Zink-sulphite to glow when an electron beam hits it.
Applications of CRO
 Acoustic research
 Transmission and receiving units of TV to convert electrical impulses to visuaf
signals
 In radar systems to detect planes
 In electronics for circuit studies.
2. With the help of illustrations, prove that the power measured using a two wattmeter
method in a three phase system with a delta connected leading load is 𝑃 = √3 ∙ 𝑉𝐿 ∙ 𝐼𝐿 ∙
cos 𝜙 .
[10 marks]
Solution
Fig S 2.2 showing the connection of two wattmeters to measure power by a delta
connected load.
Wattmeter Readings
Wattmeter 1 (W1)
Wattmeter 2 (W2)
Current coil= IR
Current coil=IY
Voltage coil=VRB
Voltage coil=VYB
𝑃1 = 𝑉𝑅𝐵 ∙ 𝐼𝑅 ∙ cos(𝑉𝑅𝐵 ∧ 𝐼𝑅 )
𝑃2 = 𝑉𝑌𝐵 ∙ 𝐼𝑌 ∙ cos(𝑉𝑌𝐵 ∧ 𝐼𝑌 )
Angles between VRB and IR, and VYB and IY using a phasor diagram of a delta connected
capacitive load show below.
Fig. S2.3 showing the phasor diagram of a balanced three phase capacitive ( leading)
delta connected load.
The total power consumed by the load is equal to the sum of the reading of 𝑊1 and 𝑊2 .
i.e.
𝑃 = 𝑊1 (P1) +𝑊2 (P2)
𝑊1 = 𝑃1 = 𝑉𝑅𝐵 ∙ 𝐼𝑅 ∙ cos(𝑉𝑅𝐵 ∧ 𝐼𝑅 ) = 𝑉𝐿 ∙ 𝐼𝐿 ∙ cos(30° + 𝜙) … … … 𝐸𝑞. (1)
𝑊2 = 𝑃2 = 𝑉𝑌𝐵 ∙ 𝐼𝑌 ∙ cos(𝑉𝑌𝐵 ∧ 𝐼𝑌 ) = 𝑉𝐿 ∙ 𝐼𝐿 ∙ cos(30° − 𝜙) … … … 𝐸𝑞 (2)
After adding Eq. (1) and Eq. (2) we get:
𝑊1 + 𝑊2 = 𝑉𝐿 ∙ 𝐼𝐿 ∙ [cos(30° + 𝜙) + cos(30° − 𝜙)]
Formula required is cos 𝐶 + cos 𝐷 = 2 ∙ cos (
𝐶+𝐷
2
𝐶−𝐷
) ∙ cos (
2
)
∴ 𝑊1 + 𝑊2 = 𝑉𝐿 ∙ 𝐼𝐿
∙ [2 ∙ cos (
(30° + 𝜙) + (30° − 𝜙)
(30° + 𝜙) − (30° − 𝜙)
) ∙ cos (
)]
2
2
= 2 ∙ 𝑉𝐿 ∙ 𝐼𝐿 ∙ cos 30° ∙ cos 𝜙 → cos 30° =
√3
2
√3
∙ cos 𝜙 = √3 ∙ 𝑉𝐿 ∙ 𝐼𝐿 ∙ cos 𝜙
2
ℎ𝑒𝑛𝑐𝑒 𝑃 = √3 ∙ 𝑉𝐿 ∙ 𝐼𝐿 ∙ cos 𝜙
4. What’s the difference between digital and analogue meters?
[1 Marks]
Solution
 Digital meters show the measured value in numerical for while analogue ones
show by means of a moving pointer of a scale.
 Digital meters are made up of only electronic circuits whereas analogue meters
are mostly made up of mechanical systems. Though some may have some
electronic components working together with the mechanical system.
Total [15 Marks]
∴ 𝑊1 + 𝑊2 = 2 ∙ 𝑉𝐿 ∙ 𝐼𝐿 ∙
Question Three
1. A 3 phase 20 kVA load has a PF of 0.34. The power was measured by two wattmeter
method. Find the reading of each wattmeter (i) when the PF is leading, (ii) when PF is
lagging.
[4 Marks]
Solution
PF leading
𝑺 = 𝟐𝟎 𝒌𝑽𝑨 = 𝟐𝟎𝟎𝟎𝟎 𝑽𝑨
𝐜𝐨𝐬 𝝓 = 𝟎. 𝟑𝟒 → 𝝓 = 𝐜𝐨𝐬 −𝟏 𝟎. 𝟑𝟒 = 𝟕𝟎°
𝑺 = √𝟑 ∙ 𝑽𝑳 ∙ 𝑰𝑳
𝑺
𝟐𝟎𝟎𝟎𝟎
𝑽𝑳 ∙ 𝑰 𝑳 =
=
= 𝟏𝟏𝟓𝟒𝟕 𝑽𝑨
√𝟑
√𝟑
𝑾𝟏 = 𝑽𝑳 ∙ 𝑰𝑳 ∙ 𝐜𝐨𝐬(𝟑𝟎° + 𝝓) = 𝟏𝟏𝟓𝟒𝟕 ∙ 𝐜𝐨𝐬(𝟑𝟎° + 𝟕𝟎°) = − 𝟐𝟎𝟎𝟓. 𝟏 𝑾
𝑾𝟐 = 𝑽𝑳 ∙ 𝑰𝑳 ∙ 𝐜𝐨𝐬(𝟑𝟎° − 𝝓) = 𝟏𝟏𝟓𝟒𝟕 ∙ 𝐜𝐨𝐬(𝟑𝟎° − 𝟕𝟎°) = 𝟖𝟖𝟒𝟓. 𝟓 𝑾
PF lagging
𝑾𝟏 = 𝑽𝑳 ∙ 𝑰𝑳 ∙ 𝐜𝐨𝐬(𝟑𝟎° − 𝝓) = 𝟏𝟏𝟓𝟒𝟕 ∙ 𝐜𝐨𝐬(𝟑𝟎° − 𝟕𝟎°) = 𝟖𝟖𝟒𝟓. 𝟓 𝑾
𝑾𝟐 = 𝑽𝑳 ∙ 𝑰𝑳 ∙ 𝐜𝐨𝐬(𝟑𝟎° + 𝝓) = 𝟏𝟏𝟓𝟒𝟕 ∙ 𝐜𝐨𝐬(𝟑𝟎° + 𝟕𝟎°) = − 𝟐𝟎𝟎𝟓. 𝟏 𝑾
2. Fig. Q3, shows a megga.
Fig. Q3
a) Explain what happens when the testing leads are open circuited, short circuited
and have an unknown resistance connected between them.
Solution
Coil A tends to move the pointer counter clockwise while the coil B moves it
clockwise. Coil B is in series with R’ and the unknown resistance Rx . The
combination of the Coil B, R’ and Rx forms a series path between the positive and
negative brushes of a DC generator. Coil A is connected in series with R 1 and
connected across the generator.
 When the testing leads are open circuited
When the testing leads are open circuited, No current flows through coil
B but current flows through coil A and this deflects the pointer to
infinity, which indicate to large resistance to be measured.
 When the testing leads are short circuited
In this case the pointer deflects to zero as the current flowing through
coil B is large. The resistor R’ protects the instrument by limiting the
current that flows.

When the testing leads are have an unknown resistance connected
between them
When Rx the current that flow through coil B makes the pointer to
deflect in clockwise direction while that in coil a try to deflect it in
counter clockwise direction, the pointer comes to rest at a point where
the two torques are equal, this position depends on the value of
resistance.
b) Explain the procedure for using a megga.
Solution
To used a megga to test the insulation of a cable, connect on testing lead to the
conductor (core) and the other to the insulation, After Insulating the wiring form
the meggas , crank the megga until the clutch slips and take note of the reading
NOTE: Normal insulation resistance should not read infinity and a small
resistance mean insulation breakdown.
c) How are excessive voltages avoided in meggas when the generator is cranked
faster than the rated speed?
[3+2+2 Marks]
Solution
To avoid excessive voltages when the generator is cranked faster than the rated
speed, most meggas are equipped with friction clutches, slip to prevent the
generator speed and out but voltage from exceeding their rated values.
3. Explain the general operation principle of all kinds of Induction types of Instruments.
[4 Marks]
Solution
Fig. S3
The principle of working and construction of induction type meter is very simple and
easy to understand that's why these are widely used in measuring energy in domestic as
well as industrial world. In all induction meters we have two fluxes which are produced
by two different alternating currents on a metallic disc. Due to alternating fluxes there is
an induced emf, the emf produced at one point (as shown in the fig, S3 given below)
interacts with the alternating current of the other side resulting in the production of the
deflection torque. Similarly, the emf produced at the point two interacts with the
alternating current at point one, resulting in the production of torque again but in opposite
direction. Hence due to these two torques which are in different directions, the metallic
disc moves.
Total [15 Marks]
Question Four
1. Name, make a sketch and explain the operation principle of the device you would
recommend for the measurements of the flow of fluids in a pipe.
[5 Marks]
Solution
There are different types of fluid flowmeters - Orifices, Venturies, Nozzles, Rotameters,
Pitot Tubes, Calorimetrics, Turbine, Vortex, Electromagnetic, Doppler, Ultrasonic,
Thermal, Coriolis
The most common principals for fluid flow metering are:





Differential Pressure Flowmeters
Velocity Flowmeters
Positive Displacement Flowmeters
Mass Flowmeters
Open Channel Flowmeters
Fig. Q 4.1
A venturi meter is used to measure the flow speed of a fluid in a pipe. The meter is
connected between two sections of the pipe (Fig.Q4.1 ); the cross-sectional area A of the
entrance and exit of the meter matches the pipe's cross-sectional area. Between the
entrance and exit, the fluid flows from the pipe with speed V and then through a narrow
''throat'' of cross-sectional area a with speed v. A manometer connects the wider portion
of the meter to the narrower portion. The change in the fluid's speed is accompanied by a
change Δp in the fluid's pressure, which causes a height difference h of the liquid in the
two arms of the manometer. (Here Δp means pressure in the throat minus pressure in the
pipe.) Suppose that the fluid is fresh water, that the cross-sectional areas are 71 cm2 in
the pipe and 34 cm2 in the throat, and that the pressure is 58 kPa in the pipe and 43 kPa
in the throat. What is the rate of water flow in cubic meters per second?
2. What is a multi-meter?
[2 Marks]
Solution
A multi-meter is a very popular measuring instrument. As its name suggest, it’s a meter
that is composed of multiple meters in one package, It can be used as a DC ammeter, DV
voltmeter, AC voltmeter and Ohmmeter. It is usually made of D’arsonval meter move
movement and a built in rectifiers.
3. State the main advantages of Instrument transformer.
[4 Marks]
Solution
i) They make it possible to measure large voltages and currents using a small rated
instrument rated 5A and 110V
ii) They provide electrical isolation from high voltage power systems, making it save for
the operator and reducing the insulation required on the meters.
iii) Instrument transformers makes to possible to standardise the meter that can be used to
measure various voltages, making it easier to replace damaged instruments with
standard health meters
iv) Instrument transformers consume less power because of the low voltages and currents
associated with them.
v) Several measuring instruments can be connected to one instrument transformer.
4. A 100:5-A CT is used to measure current of a motor using a 5-A ammeter. If the
ammeter reading were 4-A, what is the line (primary) current?
[2 Marks]
Solution
The rating of 100:5-A means that if I1 is 100-A, I2 will be 5-A.
I.e. CT ratio Kn = I1/I2 = 100/5 = 20:1.
∴ for I2 = 4-A
I1 = I2.Kn = 4(20) = 80-A.
I.e. primary (line) current is 80-A
5. What’s the difference between an ohmmeter and megga?
[2 Marks]
Solution
Both Ohmmeters and Meggas are used to measure resistance, but the main difference
between them is that an Ohmmeter only measure resistance up to kilo ohms, while a
mega can measure multi millions of resistance. For instance to measure resistance of
electronic components we can use an ohmmeter while to measure the resistance of a
cable we need to used a mega as the values expected are in mega ohms.
Total [15 marks]
The End