JAMB 2024 Physics · Question 19
A cell of internal resistance of 2 supplies current through a resistor, X if the efficiency of the cell is 75%, find the value of X.
Options
- A.2
- B.4
- C.6
- D.8
Answer: C. 6
Explanation written with AI and checked against the answer key
Analyze the question: A cell with internal resistance r = 2 Ω drives current through an external resistor X. The efficiency is 75%. We need X. Concept: The efficiency of a cell is the fraction of the total power that is delivered to the external circuit: Efficiency = (power in external resistor) / (total power) = I²X / I²(X + r) = X / (X + r) (Equivalently, efficiency = terminal voltage / e.m.f. = V/E.) Solution: 0.75 = X / (X + 2) 0.75(X + 2) = X 0.75X + 1.5 = X 1.5 = 0.25X X = 6 Ω Check: 6 / (6 + 2) = 6/8 = 0.75 = 75% ✓ Answer: The value of X is 6 Ω, which is option C. Tip: Efficiency = R / (R + r). For quick work, rearrange it as R = r × η / (1 − η). Here, R = 2 × 0.75 / 0.25 = 6 Ω. Whenever the efficiency is 75%, the external resistance is 3 times the internal resistance.AI can make mistakes. Report anything that looks wrong.
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