Previous Year Question
A battery has a short-circuit current of 30 A and an open-circuit voltage of 12 V. If the battery is connected to a load resistance of 2Ω, then what will be the power delivered by the battery?
Correct answer: C
Explanation
The battery internal resistance is found from open-circuit voltage and short-circuit current: r=12/30=0.4 Ω. With a 2 Ω load the current is 12/(2+0.4)=5 A, so load power is I²R=25×2=50 W.
Formula
r=Voc/Isc=0.4Ω; I=12/2.4=5A; PL=I²RL=50W
Step-by-step solution
Correct Answer
C — 50 W
Concept and Reasoning
The battery internal resistance is found from open-circuit voltage and short-circuit current: r=12/30=0.4 Ω. With a 2 Ω load the current is 12/(2+0.4)=5 A, so load power is I²R=25×2=50 W.
Calculation / Key Relation
r=Voc/Isc=0.4Ω
I=12/2.4=5A
PL=I²RL=50W
Exam Takeaway
Short-circuit current reveals the internal resistance.
Common Mistake
Do not use 12 V directly across the load when the battery has internal resistance.
Common mistake
Do not use 12 V directly across the load when the battery has internal resistance.
Exam tip
Short-circuit current reveals the internal resistance.
Quick method
Short-circuit current reveals the internal resistance.