JAMB 2014 Physics · Question 20
A circuit has a resistance of 200Ω. The resistance of the circuit can be reduced to 120Ω when
Options
- A.A 300Ω resistor is connected to it in parallel
- B.An 80Ω resistor is conneceted to it in series
- C.A 150Ω resistor is connected to it in parallel
- D.A 240Ω resistor is connected to it in series
Answer: A. A 300Ω resistor is connected to it in parallel
Explanation written with AI and checked against the answer key
Analyze the question: We have a 200Ω resistance and want to reduce it to 120Ω. Adding a resistor in series always increases resistance, so options B and D are out. Only a parallel connection can reduce it. Concept: For two resistors in parallel: R = (R₁ × R₂) / (R₁ + R₂) Solution: Option A: 200Ω parallel with 300Ω R = (200 × 300) / (200 + 300) = 60000 / 500 = 120Ω ✔ Option C: 200Ω parallel with 150Ω R = (200 × 150) / (200 + 150) = 30000 / 350 ≈ 85.7Ω ✘ Option B: series gives 200 + 80 = 280Ω ✘ Option D: series gives 200 + 240 = 440Ω ✘ Answer: Option A. Tip: Series always increases resistance, and parallel always gives a value smaller than the smallest resistor. So when a question asks for a reduction, go straight to the parallel options. Also, the product-over-sum formula is a quick way to check two resistors in parallel.AI can make mistakes. Report anything that looks wrong.
Past questions © the examination body (JAMB/WAEC/NECO). DeeepSolve is not affiliated with them. Rights holders: see /copyright.