NABTEB 2023 BASIC ELECTRICITY ESSAY AND OBJ ANSWERS

BASIC ELECTRICITY-OBJ
Answers loading…
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
BASIC ELECTRICITY ESSAY- ANSWERS;-ExamgrandComNg
INSTRUCTION: ANSWER FIVE QUESTIONS ONLY
(1a)
(i) Capacitor: A capacitor is an electronic component that stores electrical energy in an electric field. It consists of two conductive plates separated by a non-conductive material.

(ii) Insulator: An insulator is a material that does not allow the flow of electrical charges easily. It has high electrical resistance and low conductivity.

(iii) Conductor: A conductor is a material that allows the flow of electrical charges or current through it easily. It has low electrical resistance and high conductivity.

(iv) Resistance: Resistance is a property of a material or component that impedes the flow of electrical current. Resistance opposes the flow of electrons and converts electrical energy into heat.

(v) Potential Difference: Potential difference is the electrical potential energy difference between two points in an electric circuit. It represents the work done per unit charge in moving an electric charge from one point to another.

(1b)
(i) Blue, grey, orange, silver:
The color codes for the given resistor are: Blue: 6, Grey: 8, Orange: 3, Silver: 10% tolerance
Maximum value = (68+10)x10³
= 680,000Ω
= 680kΩ
Minimum value = (60-10)x10³
= 58,000Ω
= 58kΩ

(ii) Red, yellow, brown, gold:
The color codes for the given resistor are: Red: 2, Yellow: 4, Brown: 1, Gold: 5% tolerance
Maximum value = (24+1)x10⁰
= 25Ω
Minimum value = (24-1)x10⁰
= 23Ω

(iii) Brown, black, red, no band:
The color codes for the given resistor are: Brown: 1, Black: 0, Red: 2
Maximum value = (10+2)x10⁰
= 12Ω
Minimum value = (10-2)x10⁰
= 8Ω

(1c)
Total emf = 1.5V+2.0V+3.5V
= 7.0V
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

(2a)
Ohm’s law is a fundamental law in electronics that establishes the relationship between current voltage and resistance in an electrical circuit. It states that the current in a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them.

(2b)
(i) In Parallel:
1/Req = 1/R1 + 1/R2 + 1/R3 + 1/R4
Where R1 = 2Ω, R2 = 3Ω, R3 = 4Ω, R4 = 6Ω
1/Req = 1/2 + 1/3 + 1/4 + 1/6
1/Req = 1.5
Req = 0.67ohms
The total current flowing through the circuit is:
I = V/Req
I = 12/0.67
I = 17.91A

See also  2023 NABTEB MAY/JUNE NBC/NTC EXAMINATION TIMETABLE;-Examgrand

(ii) In Series:
Req = R1 + R2 + R3 + R4
Req = 2 + 3 + 4 + 6
Req = 15Ω
The total current flowing through the circuit is:
I = V/Req
I = 12/15
I = 0.8A

(iii) The current through the 4ohms resistor:
I = V/R
I = 12/4
I = 3A
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

(3a)
(i) Soldered joint
(ii) Crimped Joint
(iii) Screw Terminal Joint
(iv) Wire Nut Joint
(v) Welded Joint
(vi) Compression Joint

(3b)
(i) Helps in promoting the wetting ability of the solder which allows for better flow and adhesion to the surfaces being joined.
(ii) Prevents oxidation by creating a protective layer over the solder ultimately resulting in better quality and reduced rework.
(iii) Allows for the efficient transfer of heat energy thereby reducing the possibility of a cold solder joint.
(iv) Helps in cleaning the surface by dissolving contaminants and allowing for proper adhesion.

(3c)
(i) Clean the surfaces to be soldered by removing any dirt, grease, oxidation, or contaminants.
(ii) Apply an appropriate flux to the surfaces being soldered.
(iii) Select a suitable soldering iron or soldering station with the appropriate wattage for the soldering task.
(iv) Apply a small amount of solder to the soldering iron tip to improve heat transfer and ensure good thermal contact.

(3d)
(PICK TWO ONLY)
(i) Use Personal Protective Equipment (PPE).
(ii) Ensure proper ventilation in the soldering area to remove fumes.
(iii) Keep wires away from the heat source to prevent melting or damage.
(iv) Ensure that the soldering iron is properly grounded
(v) Avoid soldering near flammable materials or substances.
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

(4ai)
(PICK FOUR ONLY)
(i) Constant voltage charging
(ii) Constant current charging
(iii) Trickle charging
(iv) Pulse charging
(v) Solar charging
(vi) Fast charging

(4aii)
(PICK TWO ONLY)
(i) Ensure proper ventilation in the charging area to disperse any gases or fumes that may be generated during charging.
(ii) Keep the charging area free from flammable materials, liquids, or gases.
(iii) Install fire detection and suppression systems
(iv) Avoid overloading electrical circuits and use appropriate circuit breakers or fuses to protect against overcurrent situations
(v) Maintain clear segregation between charged and discharged batteries to avoid accidental mixing or short circuits.
(vi) Monitor and control the temperature in the charging area to prevent overheating.

See also  NABTEB 2023 OFFICE PRACTICE ESSAY AND OBJ ANSWERS

(4bi)
Average Value:
= (2/π) × (maximum value of the voltage)
= (2/π) × (200V)
= 127.32V

(ii) Root Mean Square (RMS) Value:
Vrms = Vmax/√2
Vrms = 200/√2
= 141.42V
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

(6a)
(PICK THREE ONLY)
(i) Heating Effect – Example: Electric heaters.
(ii) Magnetic Effect – Example: Electric motors generators.
(iii) Chemical Effect – Example: Electroplating processes.
(iv) Luminescent Effect – Example: Fluorescent.
(v) Physiological Effect – Example: Electric shocks

(6b)
(PICK SIX ONLY)
(i) Copper
(ii) Aluminum
(iii) Silver
(iv) Gold
(v) Brass
(vi) Bronze
(vii) Iron
(viii) Carbon
(ix) Stainless Steel
(x) Plasma

(6c)
(PICK SIX ONLY)
(i) Rubber
(ii) Glass
(iii) Plastic
(iv) Ceramic
(v) Porcelain
(vi) Wood
(vii) Mica
(viii) Fiberboard

(6d)
Given:
R1 = 6Ω, R2 = 9Ω, R3 = 12Ω
1/Requ = 1/R1 + 1/R2 + 1/R3
1/Requ = 1/6 + 1/9 + 1/12
1/Requ = (6/36) + (4/36) + (3/36)
1/Requ = 13/36
Requ = 36/13
Requ = 2.77Ω
I = V/Requ
I = 50/2.77
I = 18.04A
••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••

(7a)
A transformer is a static device that transfers electrical energy between two or more circuits through the principle of electromagnetic induction. It consists of two or more coils of wire, known as windings, which are electrically insulated from each other but magnetically linked by a common iron core.

(7b)
(i) Power Transformers: These transformers are designed for high-power applications and are used to step up or step down voltage levels in power distribution systems.

(ii) Distribution Transformers: These transformers are used for the distribution of electrical power at lower voltage levels, typically between the power distribution network and end consumers.

(iii) Isolation Transformers: These transformers are used to isolate electrical circuits or equipment from the power source. They provide galvanic isolation, protecting sensitive devices from electrical noise, voltage spikes, or grounding issues.

See also  NABTEB 2023 FINANCIAL ACCOUNTING ESSAY AND OBJ ANSWERS;-ExamgrandComNg

(7c)
(i) Copper Loss or Ohmic Loss: Copper losses occur due to the resistance of the transformer windings. When current flows through the windings, resistance causes a power loss in the form of heat. Copper losses increase with the square of the current and are proportional to the resistance of the windings.

(ii) Iron Core Loss or Core Loss: Core losses occur due to the magnetization and demagnetization of the transformer’s core during each AC cycle. These losses include hysteresis loss (energy lost due to the reversal of magnetization in the core) and eddy current loss (energy lost due to circulating currents induced in the core). Core losses result in heat dissipation.

(7d)
(i) Core Material Selection: Choosing a core material with low hysteresis and eddy current losses helps reduce core losses. Materials like silicon steel are commonly used for transformer cores due to their favorable magnetic properties.

(ii) Laminated Core: Constructing the transformer core using laminations of thin insulated metal sheets reduces eddy current losses. The insulation between laminations prevents the formation of large eddy currents.

(iii) Efficient Winding Design: Optimizing the design of the transformer windings reduces copper losses. This includes using appropriate wire sizes, minimizing the length of the windings, and employing techniques to reduce resistance, such as using larger conductors or stranded wires.

(iv) Cooling Systems: Efficient cooling methods, such as forced air or liquid cooling, can help dissipate heat generated by losses and prevent the transformer from overheating.

(7e)
Given:
Step-down ratio = 20:1
Primary voltage = 240V
Secondary Voltage = Primary Voltage/Step-down Ratio
Secondary Voltage = 240V/20
Secondary Voltage = 12V

••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••
COMPLETED

Be the first to comment

Leave a Reply

Your email address will not be published.


*