NABTEB GCE 2024 Basic Electricity Answers

NABTEB GCE 2024 BASIC ELECTRICITY ANSWERS

NABTEB GCE BASIC ELECTRICITY OBJ
01-10: CABCADACBA
11-20: BCBBCCABAA
21-30: AABCBACBAB
31-40: DCBABACBBA

_ANSWER FIVE QUESTIONS ONLY_

(1ai)
The basic particles present in the nucleus of an atom are protons and neutrons.

(1aii)
An atom of an element is normally neutral because it contains an equal number of protons positively charged and electrons negatively charged. These charges cancel each other out, resulting in no net charge.

(1aiii)
An ion is a charged particle formed when an atom either loses or gains one or more electrons.

(1aiv)
(i) Cation: A positively charged ion formed when an atom loses electrons.
(ii) Anion: A negatively charged ion formed when an atom gains electrons.

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(2ai)
A conductor is a material that allows the flow of electrical current or heat through it easily due to the presence of free electrons. Examples include metals like copper and silver.

(2aii)
An insulator is a material that does not allow the flow of electrical current or heat through it easily, as it lacks free electrons. Examples include rubber and glass.

(2b)
(i) Mica – Insulator
(ii) Brass – Conductor
(iii) Air – Insulator
(iv) Silver – Conductor
(v) Copper – Conductor
(vi) Water – Conductor
(vii) Earth – Conductor
(viii) Dry wood – Insulator

(2c)
Solder is a metal alloy, usually made of tin and lead, used to join metal components together by melting it at a relatively low temperature.

(2d)
Soldering is the process of joining two or more metal components by melting solder and allowing it to flow between the components to create a strong electrical or mechanical connection.
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(4a)
Coulomb’s law of electrostatics states that the force between two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them.

(4bi)
Given:
q1 = 2×10⁻⁶C
q2 = 8.885×10⁻⁶C
r = 0.12m
k = 8.99×10⁹Nm²C²
F = (k x q1 x q2)/(r²)
F = (8.99×10⁹) x (2×10⁻⁶x8.885×10⁻⁶)/(0.12)²
F = (8.99×10⁹) x (1.777×10⁻¹¹)/(0.0144)
F = (8.99×10⁹)x(1.234×10⁻⁹)
F = 11.09N

(4bii)
Given:
r is reduced by 0.02m
New r = 0.12 – 0.02
r = 0.10m
F’ = (k x q1x q2)/(r²)
F’ = (8.99×10⁹)x(2x 10⁻⁶x8.885×10⁻⁶)/(0.10)²
F’ = (8.99×10⁹)x(1.777×10⁻¹¹)/(0.01)
F’ = (8.99×10⁹)x(1.777×10⁻⁹)
F’ = 15.98 N
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(5a)
Given:
Resistance (R) = 6Ω
Inductance (L) = 0.03H
Supply voltage (V) = 50V
Frequency (f) = 60Hz
Supply voltage (Vrms) = 50V
XL = 2πfL
XL = 2π x 60 x 0.03
XL = 11.31Ω
Impedance (Z) = √(R²+XL²)
Z = √(6²+11.31²)
Z = √(6 + 128.94)
Z = √(164.94)
Z = 12.83Ω
Current (I) = Vrms/Z
I = 50/12.83
I = 3.89A

(5b)
tan θ = XL/R
tan θ = 11.31/6
tan θ = 1.885
θ = tan⁻¹(1.885)
θ = 62.45°

(5c)
Power factor = cos θ
Power factor = cos(62.45°)
Power factor = 0.469

(5d)
P = Vrms x I x Power factor
P = 50 x 3.89 x 0.469
P = 91.14 W
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(6ai)
A shunt is a low resistance placed in parallel with a device like a galvanometer to bypass excess current and allow only a small fraction of the current to flow through the device.

(6aii)
A multiplier is a high resistance placed in series with a device like a galvanometer to limit the current passing through it and extend the range of measurement.

(6bi)
Given:
Resistance of galvanometer (Rg) = 5Ω
Full-scale deflection current (Ig) = 20mA
= 0.02A
Ammeter for 30 A:
Full-scale current for the ammeter (I) = 30A
Shunt resistance (Rs):
Ig = (V)/(Rg)
V = (Ig)x(Rg)
V = 0.02 x 5
V = 0.1V
Rs = (V)/(I – Ig)
Rs = (0.1)/(30 – 0.02)
Rs = (0.1)/(29.98)
Rs = 0.00334Ω

(6bii)
Voltmeter for 10 V:
Multiplier resistance (Rm):
V = Ig x Rg
Rm = (V/Ig) – Rg
Rm = (10/0.02) – 5
Rm = 500 – 5
Rm = 495Ω