Electrical and Electronics Measurements MCQ Quiz - Objective Question with Answer for Electrical and Electronics Measurements - Download Free PDF
Last updated on Jun 23, 2025
Latest Electrical and Electronics Measurements MCQ Objective Questions
Electrical and Electronics Measurements Question 1:
Which of the following instruments can be used to generate a signal of various frequency contents?
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 1 Detailed Solution
The instrument used to generate a signal of various frequency contents is an Arbitrary waveform synthesizer.
Explanation:
- Arbitrary waveform synthesizer: This instrument can generate virtually any waveform, which inherently means it can create signals with various and precisely controlled frequency components. It can synthesize complex waveforms by combining different frequencies, amplitudes, and phases, as specified by the user.
- Spectrum Analyzer: This instrument is used to measure and display the frequency spectrum of an electrical signal, showing its component frequencies and their amplitudes. It does not generate signals.
- Logic Analyzer: This instrument is used to capture and display digital signals and their timing relationships, primarily for debugging digital circuits. It does not generate analog signals with various frequency content.
- Frequency Counter: This instrument is used to measure the frequency of a periodic electronic signal. It does not generate signals.
Electrical and Electronics Measurements Question 2:
A reference frequency of 10 kHz is applied to the horizontal input of a conventional cathode ray oscilloscope. When a test frequency is applied to the vertical input, the following stable Lissajous pattern is obtained. What is the test frequency applied?
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 2 Detailed Solution
Concept:
When a Lissajous pattern is formed on a Cathode Ray Oscilloscope (CRO), the ratio of frequencies of the vertical and horizontal signals can be determined from the pattern.
The general formula is:
Given:
Reference frequency (horizontal)
Lissajous pattern shows 2 vertical loops and 1 horizontal loop ⇒
Calculation:
Answer:
Option 3) 20 kHz
Electrical and Electronics Measurements Question 3:
An ammeter of range 1 A shows a change in reading of 1 mA for every mA change in input current and does not respond to smaller current change at its input. Its resolution when referred to full scale is
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 3 Detailed Solution
Detailed Solution:
The resolution is calculated as the ratio of the smallest measurable change in current to the full-scale current range of the ammeter. This can be mathematically expressed as:
Resolution = (Smallest detectable change in current) ÷ (Full-scale current range)
From the problem:
- The smallest detectable change in current = 1 mA
- The full-scale current range = 1 A = 1000 mA
Substituting the values into the formula:
Resolution = 1 mA ÷ 1000 mA = 0.001 = 0.1%
Electrical and Electronics Measurements Question 4:
Match the following sources of error in a CT with their corresponding causes
Error Type |
Cause |
||
1. |
Ratio Error |
A. |
Leakage flux in windings |
2. |
Phase Angle Error |
B. |
Magnetising current requirement |
3. |
Hysteresis Error |
C. |
Non-ideal core material properties |
4. |
Leakage Reactance Error |
D. |
Residual magnetism in the core |
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 4 Detailed Solution
Matching the Error Types and Causes in Current Transformers (CTs):
In a Current Transformer (CT), various sources of errors arise due to the physical and electrical characteristics of the transformer. These errors are classified into different types, and each error type has specific causes associated with it. Understanding the relationship between these errors and their causes is crucial for ensuring the accurate performance of CTs in electrical systems.
The correct matching of error types with their causes is:
Option 3: 1 → B, 2 → C, 3 → D, 4 → A
Let us analyze each error type and its corresponding cause:
1. Ratio Error (1 → B):
Explanation: Ratio error occurs when there is a discrepancy between the actual transformation ratio of the CT and its nominal or designed transformation ratio. This discrepancy primarily arises due to the magnetizing current requirement of the core. Magnetizing current is the current needed to establish the magnetic flux in the CT core, and it does not contribute to the secondary current. As a result, it causes a deviation in the actual ratio of primary to secondary currents, leading to ratio error.
Key Cause: Magnetizing current requirement (B).
2. Phase Angle Error (2 → C):
Explanation: Phase angle error is the angular displacement between the primary current and the secondary current of a CT. Ideally, the primary and secondary currents should be in phase (for a purely resistive burden). However, due to non-ideal properties of the core material (such as hysteresis and eddy current losses), there is a phase shift between the two currents. These non-ideal core properties affect the accuracy of the CT and result in phase angle error.
Key Cause: Non-ideal core material properties (C).
3. Hysteresis Error (3 → D):
Explanation: Hysteresis error is caused by the residual magnetism or remanence in the CT core. When the core is magnetized by the primary current, some residual magnetism remains in the core even after the current is removed. This residual flux affects the accuracy of the CT and leads to hysteresis error. It is particularly significant in applications where the CT is subjected to varying primary currents.
Key Cause: Residual magnetism in the core (D).
4. Leakage Reactance Error (4 → A):
Explanation: Leakage reactance error arises due to the leakage flux in the CT windings. Leakage flux is the portion of the magnetic flux that does not link both the primary and secondary windings. This flux creates reactance in the windings, which impacts the accuracy of the CT. The leakage reactance causes a voltage drop in the windings, thereby introducing errors in the secondary current.
Key Cause: Leakage flux in windings (A).
Important Information
To further understand the analysis, let’s evaluate the other options:
Option 1: 1 → A, 2 → B, 3 → D, 4 → C
This option incorrectly associates the causes with the error types. For example, it matches ratio error with "leakage flux in windings" (A), which is incorrect because ratio error is primarily caused by the magnetizing current requirement (B). Similarly, other mismatches make this option incorrect.
Option 2: 1 → B, 2 → A, 3 → D, 4 → C
This option also contains incorrect associations. For instance, it matches phase angle error with "leakage flux in windings" (A), which is incorrect because phase angle error is caused by non-ideal core material properties (C). The other pairings are also mismatched.
Option 4: 1 → B, 2 → A, 3 → C, 4 → D
In this option, hysteresis error is incorrectly matched with "non-ideal core material properties" (C) instead of "residual magnetism in the core" (D). Leakage reactance error is also mismatched with "residual magnetism in the core" (D) instead of "leakage flux in windings" (A).
Conclusion:
Understanding the sources of errors in CTs and their respective causes is essential for ensuring the accuracy and reliability of these devices. The correct matching of error types with their causes is as follows:
- Ratio Error: Magnetizing current requirement (B).
- Phase Angle Error: Non-ideal core material properties (C).
- Hysteresis Error: Residual magnetism in the core (D).
- Leakage Reactance Error: Leakage flux in windings (A).
Option 3 correctly identifies these relationships, making it the correct choice.
Electrical and Electronics Measurements Question 5:
Calculate the ripple factor (in percentage) if the dc and ac voltmeter used to measure the output signal from a rectifier circuit gives the readings of 25 V dc and 1.5 V rms.
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 5 Detailed Solution
Explanation:
Ripple Factor Calculation:
The ripple factor is a measure of the residual AC component present in the output of a rectifier circuit in relation to the DC component. It quantifies the effectiveness of the rectifier in converting AC to DC. The formula for the ripple factor (ρ) is given by:
Ripple Factor (ρ) = (RMS value of the AC component) / (DC value)
To express the ripple factor in percentage:
Ripple Factor (%) = [(RMS value of the AC component) / (DC value)] × 100
From the problem statement:
- DC voltage (VDC) = 25 V
- RMS voltage of the AC component (VAC) = 1.5 V
Substituting the given values into the formula:
Ripple Factor (%) = [(1.5) / (25)] × 100
Ripple Factor (%) = (0.06) × 100
Ripple Factor (%) = 6%
Thus, the ripple factor is 6%, which matches Option 4.
Important Information:
The ripple factor is a crucial parameter in rectifier design and operation. It provides insights into the quality of the DC output and the extent of filtering required in the circuit. A lower ripple factor indicates a smoother and more stable DC output.
Analysis of Other Options
To further understand the analysis, let’s evaluate the other options:
- Option 1 (0.17): This value is significantly lower than the calculated ripple factor of 6%. It may have resulted from an incorrect calculation or misunderstanding of the formula.
- Option 2 (0.34): This value is also incorrect and does not match the calculated ripple factor. It suggests a calculation error or incorrect substitution of values.
- Option 3 (0.25): This value is closer to 25% and does not correspond to the calculated ripple factor of 6%. Again, it might result from an incorrect computation.
- Option 4 (0.06): This value is correct. When expressed as a percentage (multiplied by 100), it gives 6%, which aligns with the accurate calculation.
Conclusion:
The ripple factor is a key parameter for evaluating the performance of a rectifier circuit. For the given voltmeter readings, the ripple factor was calculated to be 6%, corresponding to Option 4. This value indicates the quality of the rectified output and helps in designing the appropriate filtering mechanism for further smoothing of the DC signal.
Top Electrical and Electronics Measurements MCQ Objective Questions
The type of damping use for moving iron instruments is ______.
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 6 Detailed Solution
Download Solution PDF
M.C. Instruments |
M.I. Instruments |
1. MC type instruments are more accurate. |
1. MI type are less accurate than MC type. |
2. Manufacturing cost is high. |
2. Cheap in cost. |
3. Reading scale is uniformly distributed. |
3. Non-uniform scale (scale cramped at beginning and finishing) |
4. Very sensitive in construction & for input. |
4, Robust in construction. |
5. Low power consumption |
5. Slightly high-power consumption. |
6. Eddy current damping is used. |
6. Air friction damping is used. |
7. Can be used only for D.C measurements. |
7. Can be used for A.C as well as for D.C measurements. |
8. Controlling torque is provided by spring. |
8. Controlling torque is provided by gravity or spring |
9. Deflection is proportional to current. (θ α l). |
9. Deflection is proportional to the square of the current. (θ α l2). |
10. Errors are set due to the ageing of control springs. Permanent magnet (i.e. No Hysteresis loss). |
10. Errors are set due to hysteresis and stray fields. (i.e. hysteresis loss takes place). |
Ammeter provides the path for:
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 7 Detailed Solution
Download Solution PDFAmmeter:
- It is used to measure the current.
- An ideal ammeter has zero internal resistance and thus it provides the path for maximum current.
- It is always connected in series as it measures current.
- The range of ammeter can be extended by using a low shunt resistance.
Voltmeter:
- It is used to measure the voltage.
- An ideal voltmeter has infinite resistance and thus it provides the path for minimum current.
- It is always connected in parallel as it measures voltage.
- The range of voltmeter can be extended by using a high series resistance.
A meter reads 125 V and the true value of the voltage is 125.5 V. Find the static error of the instrument.
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 8 Detailed Solution
Download Solution PDFConcept:Absolute Error:
The deviation of the measured value from the true value (or) actual value is called error. It is also known as a static error.
Static error (E) = Am – At
Am = Measured value
At = True value
Relative Static Error: The ratio of absolute error to the true value is called relative static error.
Limiting Error:
The maximum allowable error in the measurement is specified in terms of true value, is known as limiting error. It will give a range of errors. It is always with respect to true value, so it is a variable error.
Guaranteed Accuracy Error:
The allowable error in measurement is specified in terms of full-scale value is known as a guaranteed accuracy error. It is a variable error seen by the instrument since it is with respect to full-scale value.
Application:
Given-
Am = 125 V, At = 125.5 V
∴ Static error (E) = 125 - 125.5
E = 0.5 V
When the pointer of an indicating instrument is in motion, then the deflecting torque is opposed by:
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 9 Detailed Solution
Download Solution PDFThe essential features are possessed by an indicating instrument deflecting, controlling, and a damping device.
- Deflecting device: The deflection device produces deflecting torque which causes the moving system to move from its zero position.
- Controlling device: The controlling device produces the controlling torque (Tc) which opposes the deflecting torque and increases with the deflection of the moving system. It also brings the pointer back to zero when the deflecting torque is removed.
- Damping device: This device produces damping torque this torque is necessary to bring the pointer to rest quickly. This damping torque (Td) is used to reduce the oscillation.
A galvanometer is used to ________
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 10 Detailed Solution
Download Solution PDF- A Galvanometer is used to detect the direction of the current.
- This device is used for detecting and measuring a small amount of electric current.
- Johann Schweigger invented the device in 1820.
Measuring Instrument |
Quantity |
Photometers |
Light |
Magnetometer |
Magnetic induction |
Sound Level Meter |
sound |
More power is required to operate a moving-iron meter than a PMMC meter because of the magnetic circuit's high _____.
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 11 Detailed Solution
Download Solution PDFMoving iron meter has large magnetic reluctance as compared to PMMC meter. That’s why more power is required to operate the moving iron meter.
Advantages of moving iron:
- It is a universal instrument which can be used for the measurement of AC and DC quantities
- These instruments can withstand large loads and are not damaged even under severe overload conditions
- It is very cheap due to the simple construction
Disadvantages of moving iron:
- These instruments suffer from error due to hysteresis, frequency change and stray losses
- The reading of the instrument is affected by temperature variation
Note: In terms of accuracy PMMC meter has the highest accuracy. The order of accuracy is given below.
Induction
If a 100 watts bulb is used for 10 hours, then the amount of consumed electrical energy will be -
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 12 Detailed Solution
Download Solution PDFThe correct answer is 1 KWh.
Concept:
Electric bulb: It is an electric device that converts electric energy into heat and light energy.
- Resistance of the bulb
- If we consider the bulb as a resistor then, we can easily find the current and voltage drops.
- Power consumed ∝ Brightness
The rate of work done by the electric current is called as electric power.
The difference of potential between two points is called a potential difference.
- Electric energy (E) = electric power (P) × time (t)
Unit of electric energy = unit of electric power × unit of time = kilowatt × hour = kilowatt-hour
Key Points
- Given that
- Power of the bulb = 100 W
- Time = 10 hours
We know that,
Energy = Power × Time
⇒ Energy consume = 100 w × 10 hrs = 1000 watt - hr = 1 KWh.
For balanced lagging power factor, according to two wattmeter method,
W1 = VL IL cos (30 - ϕ)
W2 = VL IL cos (30 + ϕ)
The total reactive volt-amperes Q is given by
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 13 Detailed Solution
Download Solution PDFIn a two-wattmeter method,
The reading of first wattmeter (W1) = VL IL cos (30 – ϕ)
The reading of second wattmeter (W2) = VL IL cos (30 + ϕ)
Total power in the circuit (P) = W1 + W2 = √3 VLIL cos ϕ
Total reactive power in the circuit
Power factor = cos ϕ
The braking torque of induction type single-phase energy meter is:
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 14 Detailed Solution
Download Solution PDFEnergy meter:
Energy meter or Watt-hour meter is used to measure the energy in kWh.
It is an integrating type instrument.
Its working principle is similar to the transformer.
There are three essential mechanisms required in the energy meter named Driving torque, Braking torque, and registered mechanism.
Driving torque:
This torque is required to revolve the disc or rotate the disc.
It is obtained by the electromagnetic induction effect.
Braking torque:
It is required to rotate the disc at a constant speed.
It is obtained by using a permanent magnet placed inside the energy meter near the Aluminum disc.
Eddy currents have induced in the magnet due to the movement of the rotating disc through the magnetic field. This eddy current reacts with the flux and exerts a braking torque which opposes the motion of the disk. The speed of the disk can be controlled by changing flux.
Breaking torque of induction type single-phase energy meter is:
K = constant
ϕ = flux
N = speed in rpm
R = radius of the disc
Re = resistance in path of current (i.e. disc)
The braking torque of induction type single-phase energy meter is directly proportional to the square of the flux.
Registered mechanism:
It registers the no. of rotations or revolutions of the disc which is proportional to the energy consumed in kWh.
Meter constant = (No. of revolutions / kWh)
Points to remember:
Creeping:
Sometimes the disc of the energy meter makes the slow but continuous rotation at no load i.e. when the potential coil is excited but with no current flowing in the load called creeping error
This error may be caused due to overcompensation for friction, excessive supply voltage, vibrations, stray magnetic fields, etc
It can be reduced by making two opposite holes on the disc.
Null type recorders are __________ recorders.
Answer (Detailed Solution Below)
Electrical and Electronics Measurements Question 15 Detailed Solution
Download Solution PDFNull type instrument: An instrument in which zero or null indication determines the magnitude of measured quantity, such type of instrument is called a null type instrument.
It uses a null detector which indicating the null condition when the measured quantity and the opposite quantity are same.
Examples: potentiometer, bridges, LVDT etc.