Question Statement:
If generator efficiency is 90%, the maximum electricity produced (from Q.48's engine), in kW, is:
Options:
-
(A) 2.85
-
(B) 3.00
-
(C) 3.15
-
(D) 3.50
Step-by-Step Solution & Explanation
• **Concept:** Maximum electrical output equals the engine useful shaft brake power multiplied by the generator electrical conversion efficiency.
• **Given:** Engine brake power from Q.48 $BP = 3.5\text{ kW}$, generator efficiency $\eta_g = 90\% = 0.90$.
• **Calculation:**
$P_{elec} = BP \times \eta_g = 3.5\text{ kW} \times 0.90 = 3.15\text{ kW}$
• **Conclusion:** The maximum electricity produced is $3.15\text{ kW}$. **Correct Answer: C**.
Practice This in Real CBT Exam Interface
Test yourself with the official TCS-iON palette, virtual calculator, and immediate performance analytics.
Launch CBT Mock Test