ISRO EC 2023 Q.68: solution and answer

ISRO 2023 · ECQ.68 · 1 mark · MCQ

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

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Answer and solution

ANSWER

The battery delivers $26.4 kJ$ , option (a).

Official answer: ASource: ISRO's revised and final answer key of 16 February 2024

Energy delivered is $E = \int_{0}^{T} v (t) i (t) d t$ , with the current constant at $I = 2 A$ .
The discharge time is $T = 20 \times 60 = 1200 s$ .
The terminal voltage falls linearly from $12 V$ to $10 V$ , so its time average is $V_{a v g} = \frac{12 + 10}{2} = 11 V$ .
Because $i$ is constant, $E = I \int_{0}^{T} v (t) d t = I V_{a v g} T$ .
$E = 2 \times 11 \times 1200 = 26400 J = 26.4 kJ$ .
Using the end value $10 V$ throughout would give $2 \times 10 \times 1200 = 24 kJ$ , and using $12 V$ throughout would give $28.8 kJ$ . Neither is the energy for a linearly falling voltage, and the other two options match no correct calculation. The answer is option (a).

KEY IDEA

With constant current and a voltage that changes linearly with time, the energy is current times the average voltage times the time: $E = I \frac{V _{1} + V _{2}}{2} T$ .

WATCH OUT FOR

Taking the final voltage ( $10 V$ ) or the initial voltage ( $12 V$ ) for the whole discharge instead of the average of the two, or forgetting to convert $20$ minutes to $1200$ seconds.

From the ISRO EC 2023 paper. Every question of this paper · The full paper (PDF, ISRO) · All 5 ISRO Laws & node/mesh questions, with notes

ISRO EC 2023 Q.68: solution and answer

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