NABTEB 2023 Basic Electricity Essays And Obj Questions & Answers

Basic-Elect-Obj
01-10 CCBCCBAABC
11-20 BAABDBADBC
21-30 ACDBBBBBAC
31-40 DCCBCBCACD

Completed

BASIC ELECTRICITY ESSAYS

*NUMBER ONE*

(1ai)
A capacitor is a device that stores electrical energy in an electric field. It consists of two conductive plates separated by an insulating material called a dielectric.

(1aii)
An insulator is a material that does not conduct electric current easily meaning that it has high electrical resistance.

(1aiii)
A conductor is a material that allows electric current to flow through it easily meaning that it has low electrical resistance.

(1aiv)
Resistance is the opposition to the flow of electric current in a conductor. It is measured in ohms (Ω).

(1av)
Potential difference also known as voltage is the energy per unit charge required to move a charge from one point to another in an electric field.

(1bi)
Blue-grey-orange-silver resistor has a value of 68 kΩ ± 10%
Maximum value = (68 kΩ + 10% of 68 kΩ) = 74.8 kΩ
Minimum value = (68 kΩ – 10% of 68 kΩ) = 61.2 kΩ

(1bii)
Red-yellow-brown-gold resistor has a value of 240 Ω ± 5%
Maximum value = (240 Ω + 5% of 240 Ω) = 252 Ω
Minimum value = (240 Ω – 5% of 240 Ω) = 228 Ω

(1biii)
Brown-black-red-no band resistor has a value of 1 kΩ ± 2%
Maximum value = (1 kΩ + 2% of 1 kΩ) = 1.02 kΩ
Minimum value = (1 kΩ – 2% of 1 kΩ) = 0.98 kΩ

(1c)
Total emf = 1.5V + 2.0V + 3.5V = 7.0V.
Therefore the total emf of the three cells is *7.0V* .

(2a)
Ohm’s Law states that the current passing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance between them.

(2bi)
When the resistors are connected in parallel the equivalent resistance (Req) is =
1/Req = 1/2Ω + 1/3Ω + 1/4Ω + 1/6Ω
Req = 1.2Ω
Using Ohm’s Law =
I = V/Req = 12/1.2 = 10 A

Therefore the resultant current when the resistors are connected in parallel is 10 A.

(2bii)
When the resistors are connected in series the equivalent resistance (Req) is =
Req = 2Ω + 3Ω + 4Ω + 6Ω
Req = 15Ω
Using Ohm’s Law =
I = V/Req = 12/15 = 0.8 A
Therefore the resultant current when the resistors are connected in series is *0.8 A*

(2biii)
The current in the 4Ω resistor can be calculated by using Ohm’s Law in both cases =

(i) In parallel =
I = V/R = 12/4 = 3 A
Therefore the current in the 4Ω resistor when the resistors are connected in parallel is *3 A.*

(ii) In series:
I = V/R = 12/15 x 4 = 3.2 A
Therefore the current in the 4Ω resistor when the resistors are connected in series is *3.2 A.*

(3a)
(PICK ANY SIX)
(i) Butt Joint
(ii) Lap Joint
(iii) T-Joint
(iv) Corner Joint
(v) Edge Joint
(vi) Scarf Joint
(vii) Tee Joint
(viii) Splice Joint

(3b)
(PICK ANY FOUR)
(i) Clean the surface of the materials
(ii) Prevent oxidation of the soldering material
(iii) Enhance wetting of the joint
(iv) Remove any impurities or residues
(v) Assist the flow of molten solder
(vi) Increase conductivity of the joint
(vii) Protect the joint from the atmosphere
(Viii) Ensure a stronger and more reliable joint

(3c)
(i) Clean and prepare the surface of the materials to be soldered
(ii) Apply an appropriate amount of flux to the surface
(iii) Heat the materials to the required temperature using a soldering iron or torch
(iv) Feed or apply solder to the joint while heating and then cool the joint gradually

(3d)
(PICK ANY TWO)
(i) Wear protective clothing gloves and glasses
(ii) Use a well-ventilated area to avoid inhaling hazardous fumes
(iii) Keep the workbench clear and organized to avoid accidents
(iv) Avoid touching a hot soldering iron to other materials or skin
(v) Unplug the soldering iron when not in use
(vi) Use a suitable soldering stand or holder for the iron
(vii) Keep solder and flux away from food drink and other things that may contaminate them
(viii) Dispose of used flux and solder in a secure manner to protect the environment.

(4ai)
(PICK ANY FOUR)
(i) Constant current charging
(ii) Constant voltage charging
(iii) Trickle charging
(iv) Fast charging
(v) Inductive charging
(vi) Solar charging.

(4aii)
(PICK ANY TWO)
(i) Ensure correct polarity while installing the cells in a device.
(ii) Keep the cells away from fire or any source of heat.
(iii) Do not mix different types of cells and batteries.
(iv) Use chargers designed for the specific type of cell or battery being charged.
(v) Avoid overcharging as it can cause the cell to explode or catch fire.
(vi) Store the cells in a cool and dry place.

(4bi)
The average value is calculated as:
= (2/π) × (peak value of the voltage)
= (2/π) × (200V)
= 127.32V
Therefore the average value of the alternating voltage is *127.32V*

(4bii)
The RMS value is calculated as:
RMS value = (peak value of the voltage) / √2
= 200V / √2
= 141.42V
Therefore the RMS value of the alternating voltage is *141.42V.*

(6a)
(PICK ANY THREE)
(i) eating effect: When current is passed through a resistor it produces heat. Example: An electric stove.
(ii) Magnetic effect: Electric current produces a magnetic field around it. Example: An electromagnet.
(iii) Chemical effect: Electric current can cause chemical reactions. Example: Electrolysis of water.
(iv) Illumination effect: Electric current produces light. Example: Incandescent light bulbs.
(v) Mechanical effect: Electric current produces mechanical motion. Example: Electric motors.
(vi) Shock effect: Electric current can cause shock or electrocution. Example: Touching a live wire.

(6b)
(PICK ANY SIX)
(i) Copper
(ii) Silver
(iii) Aluminum
(iv) Gold
(v) Iron
(vi) Brass
(vii) Bronze
(viii) Nickel

(6c)
(PICK ANY SIX)
(i) Rubber
(ii) Plastic
(iii) Glass
(iv) Air
(v) Porcelain
(vi) Dry wood
(vii) Bakelite
(viii) Ceramic

(6d)
1/R = 1/6 + 1/9 + 1/12 = 1/2
R = 2 ohms
I = V/R = 50/2 = *25 amperes*
Therefore the current in the circuit is *25 amperes.*