ISRO EC 2023 Q.70: solution and answer

ISRO 2023 · ECQ.70 · 1 mark · MCQ

Find V_A

ISRO EC 2023 question 70: A 8 A ideal current source (arrow pointing up) on the far left, with an 8 Ω resistor in parallel across it. — Figure from the ISRO EC 2023 paper. It is part of the question.

(A)
24 V
(B)
12 V
(C)
10 V
(D)
8 V

Answer and solution

ANSWER

$V_{A} = 24 V$ , option (a).

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

Replace the $8 A$ current source with its parallel $8 Ω$ by its Thevenin equivalent: $8 A \times 8 Ω = 64 V$ in series with $8 Ω$ .
The series $2 Ω$ resistor adds to this, so the left part is a $64 V$ source behind $8 + 2 = 10 Ω$ feeding the node whose voltage is $V_{A}$ .
At this node the other two branches are the $16 V$ battery (positive plate towards the node) in series with $5 Ω$ , and the $10 Ω$ resistor to the bottom rail. Write KCL with all currents leaving the node:
$\frac{V _{A} - 64}{10} + \frac{V _{A} - 16}{5} + \frac{V _{A}}{10} = 0$
Multiply by $10$ : $(V_{A} - 64) + 2 (V_{A} - 16) + V_{A} = 0$ , so $4 V_{A} = 96$ .
$V_{A} = 24 V$ .
Check: the left part supplies $\frac{64 - 24}{10} = 4 A$ ; the battery branch takes $\frac{24 - 16}{5} = 1.6 A$ and the $10 Ω$ takes $2.4 A$ , which total $4 A$ .

KEY IDEA

Turn the current source with its parallel resistor into a voltage source with a series resistor, then write one KCL equation at the single remaining node.

WATCH OUT FOR

Reversing the battery polarity (the longer plate is the positive terminal, and it faces the node), which gives $8 V$ instead of $24 V$ , or forgetting to add the series $2 Ω$ to the $8 Ω$ after the source transformation.

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.70: solution and answer

Answer and solution

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