NABTEB Basic Electricity 2024 Obj & Essays Answers

NABTEB BASIC ELECTRICITY OBJ
01-10: DACBDAAADA
11-20: BCBBBCBAAB
21-30: ACBBCBABDC
31-40: DBDDCACAAB

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NABTEB 2024 BASIC ELECTRICITY ANSWERS

ANSWER FIVE QUESTIONS ONLY

(1a)
The purpose of colouring carbon resistors is to indicate their resistance value through a colour code system.

(1b)
(i) Brown Yellow Red and Gold
Brown = 1, Yellow = 4, Red = Multiplier (10²), Gold = ±5%
Resistance value = 14 x 10²
= 1400Ω
Maximum value = 1400 x 1.05
= 1470Ω
Minimum value: 1400 x 0.95
= 1330Ω

(ii) Red Black Green and Silver
Red = 2, Black = 0, Green = Multiplier (10⁵), Silver = ±10%
Resistance value = 20 x 10⁵
= 2,000,000Ω
= 2 MΩ
Maximum value = 2,000,000 x 1.10
= 2,200,000Ω
Minimum value = 2,000,000 x 0.90
= 1,800,000Ω

(iii) Orange Brown Blue and Gold
Orange = 3, Brown = 1, Blue = Multiplier (10⁶), Gold = ±5%
Resistance value = 31 x 10⁶
= 31,000,000Ω
= 31 MΩ
Maximum value = 31,000,000 x 1.05
= 32,550,000Ω
Minimum value = 31,000,000 x 0.95
= 29,450,000Ω

(iv) Blue Black Black and Silver
Blue = 6, Black = 0, Black = Multiplier (10⁰), Silver = ±10%
Resistance value = 60 x 10⁰
= 60Ω
Maximum value = 60 x 1.10
= 66Ω
Minimum value = 60 x 0.90
= 54Ω

(v) Brown Blue Red and Gold
Brown = 1, Blue = 6, Red = Multiplier (10²), Gold = ±5%
Resistance value = 16 x 10²
= 1600Ω
Maximum value = 1600 x 1.05
= 1680Ω
Minimum value = 1600 x 0.95
= 1520Ω
===========================

(2a)
(i) Electromotive force (EMF): Electromotive force, denoted as EMF represents the electrical energy per unit charge that a source of electrical energy can provide. It is measured in volts (V).

(ii) Potential difference: Potential difference is the difference in electric potential between two points in an electric circuit, representing the work done per unit charge in moving a charge between those two points. It is measured in volts (V).

(iii) Current: Current is the flow of electric charge through a conducting medium, measured as the rate of flow of charge through a given cross-sectional area. It is measured in amperes (A).

(2b)
Given:
Voltage (V) = 240V
Current (I) = 2.4A
Total Resistance (RT) = V/I
RT = 240/2.4
RT = 100Ω
Total Resistance (RT):
= Sum of parallel resistance (RP) + Sum of series resistance (RS)
RP = RT – RS
RS = 70Ω
RP = 100 – 70
RP = 30Ω
1/RP = 1/90 + 1/B
1/30 = 1/90 + 1/B
1/B = 1/30 – 1/90
1/B = (3 -1)/90
1/B = 2/90
B = 90/2
B = 45Ω
===========================

(3a)
(i) Voltage reaches its maximum level
(ii) Charging current drops to minimum
(iii) Battery temperature stabilizes

(3b)
Given:
Number of identical cells = 9
Electromotive force (EMF) = 2v
Internal resistance = 0.1Ω
Resistance Load = 7.5Ω

(i) All In Series:
Total EMF = 9 x 2
= 18v
Total internal resistance = 9 x 0.1
= 0.9Ω
Total resistance = 0.9 + 7.5
= 8.4Ω
Current = 18/8.4
= 2.14A
Potential difference across the load:
= 2.14 x 7.5
= 16.05Ω

(ii) All Cells in Parallel:
Total EMF = 2v
Total internal resistance = 0.1/9
= 0.0111Ω
Total resistance = 0.0111 + 7.5
= 7.5111Ω
Current = 2/7.5111
= 0.266A
Potential difference across the load:
= 0.266 x 7.5
= 1.995Ω

(iii) 3 Cells in Series and 3 Sets in Parallel:
Total EMF in series:
= 3 x 2
= 6v
Total internal resistance in series:
= 3 x 0.1
= 0.3Ω
Total EMF in parallel = 6v
Total internal resistance in parallel:
= 0.3/3
= 0.1Ω
Total resistance = 0.1 + 7.5
= 7.6Ω
Current = 6/7.6
= 0.789A
Potential difference across the load:
= 0.789 x 7.5
= 5.92Ω
===========================

(5a)
(i) Inductive reactance is the opposition that an inductor offers to the flow of alternating current. Its unit of measurement is the ohm (Ω).

(ii) Capacitive reactance is the opposition that a capacitor offers to the flow of alternating current. Its unit of measurement is the ohm (Ω).

(iii) Impedance is the total opposition that a circuit offers to the flow of alternating current. It is a combination of resistance, inductive reactance, and capacitive reactance, measured in ohms (Ω).

(5b)
Given:
Voltage (V) = 100 V
Frequency (f) = 50 Hz
Resistor (R) = 10 Ω
Inductance (L) = 50 mH
= 0.05 H
Capacitance (C) = 350 μF
= 350 × 10⁻⁶F

(i) Current (I):
Inductive reactance (XL) = 2πFL
XL = 2 x π x 50 x 0.05
XL = 31.42Ω
Capacitive reactance (XC)= 1/(2πFC)
XC = 1/(2 x π x 50 x 350 x 10⁻⁶)
XC = 91.09Ω
Total impedance (Z):
1/Z = 1/R + 1/XL + 1/XC
1/Z = 1/10 + 1/31.42 + 1/91.09
1/Z = 0.1+ 0.0318 + 0.01098
1/Z = 0.14278
Z = 1/0.14278
Z = 7.0Ω
Current (I) = V/Z
I = 100/7.0
I = 14.29A

(ii) Power Factor:
PF = R/Z
PF = 10/7.0
PF = 1.43

(iii) Angle of Lag (θ):
θ = Cos⁻¹(PF)
θ = Cos⁻¹(1)
θ = 0°

(iv) Power Consumed by the Circuit:
P = VI Cosθ
P = 100 x 14.29 Cos(0)
P = 100 x 14.29 x 1
P = 1429W
===========================

(6a)
(i) Accuracy
(ii) Resolution
(iii) Response time

(6b)
(i) 0 – 5 mA:
R = (50×10⁻³x50)/ (5×10⁻³)
R = 500Ω

(ii) 0 – 10 A:
R = (50×10⁻³x50)/ 10
R = 0.25Ω

(iii) 0 – 15 A:
R = (50×10⁻³x50)/ 15
R = 0.167Ω

(iv) 0 – 50 V:
R = 50/(50×10⁻³)
R = 1kΩ

(v) 0 – 100 V:
R = 100/ (50×10⁻³)
= 2kΩ

(vi) 0 – 150 V:
R = 150/ (50×10⁻³)
R = 3kΩ
===========================

(7a)
(i) Heating Effect: When an electric current flows through a conductor, it produces heat due to the resistance of the material. This is commonly used in devices such as electric stoves and toasters to generate heat.

(ii) Magnetic Effect: An electric current flowing through a conductor creates a magnetic field around it. This principle is the basis for electromagnets used in various applications such as electric motors and transformers.

(iii) Chemical Effect: Electric currents can cause chemical reactions in certain materials. This effect is utilized in processes like electroplating and electrolysis, where a chemical change is induced by passing an electric current through a solution.

(7b)
Given:
Resistance (R) = 30Ω
Voltage (V) = 240V
Time (t) = 1 hour = 3600 seconds
Power (P) = V²/R
P = (240)²/30
P = 1920W
Energy consumed (E):
E = Power x Time
E = 1920 x 3600
E = 6,912,000J
E = 6,912,000/3,600,000
E = 1.92kWh
===========================

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