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BASIC ELECTRICITY OBJ(TYPE A)
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BASIC ELECTRICITY ANSWERS (TYPE A)
INSTRUCTION: ANSWER FIVE(5) QUESTIONS ONLY
(2ai)
Coulomb (C): A coulomb is the SI unit of electric charge. It is defined as the amount of charge transported by a current of 1 ampere in 1 second.
(2aii)
Matter: Matter is anything that has mass and occupies space. All physical substances, including solids, liquids, and gases, are forms of matter.
(2aiii)
Electric charge: Electric charge is a property of matter that causes it to experience a force when placed in an electric or magnetic field. It can be positive (+) or negative (−).
(2b)
(DRAW THE DIAGRAM)
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(3a)
Work is the process of transferring energy by a force moving an object through a distance. Its unit of measurement is the joule (J).
Power is the rate at which work is done or energy is transferred over time. Its unit of measurement is the watt (W), which equals one joule per second.
(3b)
Given:
Mass (m)= 20 g = 0.02 kg
Acceleration due to gravity,(g)= 10m/s²
Distance fallen(h)= 25 m
The kinetic energy (KE) after falling height(h) equals the loss in potential energy:
KE = mgh
KE = 0.02 × 10 × 25 = 5
KE = 5J
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(4)
Given;
Primary voltage Vp = 1100V
Step-down voltage ratio = 5 : 1
Resistance of secondary load R = 10Ω
(i) Secondary voltage (Vs);
Vs = Vp/5
Vs= 1100V/5
Vs= 220V
(ii) Secondary current (Is);
Is = Vs/R
Is = 220/10
Is = 22A
(iii) Primary current (Ip);
Vp × Ip = Vs × Is
Ip = (Vs × Is)/Vp
Ip = (220 × 22)/1100
Ip= 4.4A
(iv) Power consumed by the load (P);
P = Vs × Is²
P = 220 × 22
P = 4840W
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(5ai)
Frequency:
Frequency is the number of complete cycles or oscillations an alternating current makes in one second. It is measured in hertz (Hz).
(5aii)
Average value:
Average value is the mean or arithmetic average of all the instantaneous values of an alternating current over one half cycle of the waveform.
(5aiii)
Instantaneous value:
Instantaneous value is the value of an alternating current or voltage at a specific moment or point in time during the cycle.
(5b)
Given;
Period (T) = 20ms
Convert to seconds (s);
T = 20ms = 20/1000
T = 0.02s
Frequency (f) = 1/T
f = 1/(0.02)
f = 50Hz
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(6a)
Capacitance of a capacitor is the ability of a capacitor to store electric charge for a given potential difference across its plates. It is measured in farads (F), where 1 farad equals 1 coulomb per volt. C = Q/V
(6b)
(i) The area of the plates.
(ii) The distance between the plates.
(iii) The dielectric material (permittivity) between the plates.
(6c)
Given
C₁ = 3 µF,
C₂ = 6 µF,
C₃ = 8 µF (parallel)
C₄ = 10 µF (series with the parallel group)
V = 50 V
(i) Total capacitance of parallel group (Cₚₐᵣₐₗₗₑₗ);
Cₚₐᵣₐₗₗₑₗ = C₁ + C₂ + C₃ = 3 + 6 + 8
Cₚₐᵣₐₗₗₑₗ = 17µF
(ii) Total capacitance of circuit (Cₜₒₜₐₗ);
1/Cₜₒₜₐₗ = 1/Cₚₐᵣₐₗₗₑₗ + 1/C₄
Cₜₒₜₐₗ = (Cₚₐᵣₐₗₗₑₗ × C₄)/ (Cₚₐᵣₐₗₗₑₗ + C₄)
Cₜₒₜₐₗ = (17 × 10) / (17 + 10) = Cₜₒₜₐₗ = 170/27
Cₜₒₜₐₗ = 6.30µF
(iii) Charge stored in circuit(Q);
Q = Cₜₒₜₐₗ × V
Q = 6.30× 50
Q = 315µC
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(7a)
Given:
‑ Inductance L = 0.5H
‑ Supply voltage V = 100V (rms)
‑ Frequency f = 50Hz
(i) Inductive reactance Xₗ;
Xₗ = 2πfL = 2 × 50 × 0.5 × 3.142
Xₗ = 157.1Ω
(ii) Current in the circuit I;
I = V / Xₗ = 100 /157.1
I = 0.637A
(7b)
(i) Primary cells are not rechargeable and become useless after use, while secondary cells are rechargeable and can be reused multiple times.
(ii) Primary cells have higher initial voltage and energy density but shorter life span, whereas secondary cells have lower initial voltage but longer life span.
(iii) Primary cells use non-reversible chemical reactions, but secondary cells use reversible reactions allowing recharging.
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