ISRO EC 2023 Q.69: solution and answer

ISRO 2023 · ECQ.69 · 1 mark · MCQ

I_1 = 10 A. Calculate the value of V_s.

ISRO EC 2023 question 69: Two separate loops sharing a common bottom wire. — Figure from the ISRO EC 2023 paper. It is part of the question.

(A)
-5 V
(B)
10 V
(C)
15 V
(D)
20 V

Answer and solution

ANSWER

$V_{s} = - 5 V$ , option (a).

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

The left loop contains only the source $V_{s}$ and the box, so the box voltage equals the source voltage: $V_{1} = V_{s}$ (the + marks of both are at the top).
The dependent source and the right-hand box form a separate loop. They share only the bottom wire with the left loop, because the top wire of the left loop is not joined to the top wire of the right part. No current can flow between the two parts, so the right part is a closed loop of its own.
The dependent source carries $2 V_{1}$ downward, and the right-hand box carries $I_{1}$ downward.
Apply KCL at the top node of the right loop. Both currents leave this node downward, and no other branch is attached to it, so $2 V_{1} + I_{1} = 0$ .
Hence $I_{1} = - 2 V_{1}$ . With $I_{1} = 10 A$ , $V_{1} = - \frac{10}{2} = - 5 V$ .
Therefore $V_{s} = V_{1} = - 5 V$ , which is option (a).

KEY IDEA

A circuit made of parts joined only through one common wire has no current between them. The right-hand loop therefore obeys KCL on its own, and the control voltage $V_{1}$ is fixed by $V_{s}$ alone.

WATCH OUT FOR

Treating $I_{1}$ as the current supplied by the dependent source, so that $I_{1} = 2 V_{1}$ and $V_{s} = + 5 V$ . Both $I_{1}$ and the source current point downward from the same node, so they must add to zero.

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.69: solution and answer

Answer and solution

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