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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

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