Topic: Circuit Analysis
A 12V battery discharges for 20 minutes at a constant current of 2A and terminal voltage drops linearly from 12V to 10V. What is the total energy delivery by the battery during this time?
- A. 26.4 KJ
- B. 12.2 KJ
- C. 120.5 KJ
- D. 24 KJ
Answer: A. 26.4 KJ
Explanation: Energy $E = \int P(t) dt = \int V(t) I(t) dt$.
The current is constant at $I = 2A$. The voltage drops linearly from 12V to 10V, so the average voltage over the discharge period is $V_{avg} = \frac{12 + 10}{2} = 11V$.
Average Power $P_{avg} = V_{avg} \times I = 11V \times 2A = 22W$.
The time duration is $t = 20 \text{ minutes} = 20 \times 60 = 1200 \text{ seconds}$.
Total Energy $E = P_{avg} \times t = 22W \times 1200s = 26400 J = 26.4 kJ$.