Q. A battery is charged at a potential of $15\ V$ in 8 hours when the current flowing is $10\ A.$ The battery on discharge supplies a current of $5\ A$ for 15 hours. The mean terminal voltage during discharge is $14\ V.$ The $"Watt - hour"$ efficiency of battery is
A) 80%
B) 90%
C) 87.5%
D) 82.5%
Explanation
(c) The power of the battery, when charged, is given by
$P_{1} = V_{1}I_{1}$
The electrical energy dissipated is given by $E_{1} = P_{1}t_{1}$
$i.e.,\ E_{1} = V_{1}I_{1}t_{1} = 15 \times 10 \times 8 = 1200\ Wh$
Similarly, the electrical energy dissipated during the discharge a battery is given by,
$E_{2} = V_{2}I_{2}t_{2} = 14 \times 5 \times 15 = 1050\ Wh$
Hence, watt-hour efficiency of the battery is given by
$\eta = \frac{E_{2}}{E_{1}} \times 100 = 0.875 \times 100 = 87.5\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.