GATE 2018 IN Q.30: solution and answer

GATE 2018 · INQ.30 · 2 marks · NAT

In the given circuit, superposition is applied. When $V_{2}$ is set to 0 V, the current $I_{2}$ is $- 6$ A. When $V_{1}$ is set to 0 V, the current $I_{1}$ is $+ 6$ A. Current $I_{3}$ (in A) when both sources are applied will be (up to two decimal places) ______.

GATE 2018 IN question 30: Two-mesh circuit with a common bottom (reference) wire. — Figure from the GATE 2018 IN paper. It is part of the question.

Answer and solution

ANSWER

$I_{3} = 1.00 A$ .

Official answer: 0.95 to 1.05Source: Official answer key, GATE 2018 IN (IIT Guwahati)

All three reference arrows point away from the top node A, so KCL at A gives $I_{1} + I_{2} + I_{3} = 0$ . Take the bottom wire as the reference, so $V_{A}$ is the voltage of node A. The source polarities are not needed, because the current given for each sub-circuit fixes that sub-circuit on its own.
Only $V_{1}$ acting ( $V_{2}$ shorted): the $2 Ω$ resistor now runs from A to the reference through the short, so $I_{2}^{'} = \frac{V _{A}^{'}}{2} = - 6 \Rightarrow V_{A}^{'} = - 12 V$ .
Then $I_{3}^{'} = \frac{V _{A}^{'}}{6} = \frac{- 12}{6} = - 2 A$ . As a check, KCL gives $I_{1}^{'} = - (I_{2}^{'} + I_{3}^{'}) = - (- 6 - 2) = 8 A$ .
Only $V_{2}$ acting ( $V_{1}$ shorted): the $3 Ω$ resistor now runs from A to the reference through the short, so $I_{1}^{''} = \frac{V _{A}^{''}}{3} = 6 \Rightarrow V_{A}^{''} = 18 V$ .
Then $I_{3}^{''} = \frac{V _{A}^{''}}{6} = \frac{18}{6} = 3 A$ . As a check, $I_{2}^{''} = - (I_{1}^{''} + I_{3}^{''}) = - (6 + 3) = - 9 A$ .
Superposition: $I_{3} = I_{3}^{'} + I_{3}^{''} = - 2 + 3 = 1.00 A$ This agrees with $V_{A} = V_{A}^{'} + V_{A}^{''} = - 12 + 18 = 6 V$ , which gives $I_{3} = 6/6 = 1 A$ .

KEY IDEA

When one source is set to zero, the resistor in that source's branch ends up connected between node A and the reference. The one branch current you are given then fixes the node voltage $V_{A}$ ( $I_{2}^{'} = V_{A}^{'} /2$ , $I_{1}^{''} = V_{A}^{''} /3$ ). The $6 Ω$ current in each sub-circuit is $V_{A} /6$ , and superposition adds the two. You never need the source values or their polarities.

WATCH OUT FOR

One mistake is to read each given current as a current flowing into node A. All three reference arrows point away from A, so $I_{2}^{'} = - 6 A$ makes $V_{A}^{'} = - 12 V$ , not $+ 12 V$ . Another is to try to find $V_{1}$ and $V_{2}$ first when their polarities are not shown. A third is to add the given currents $I_{2}^{'}$ and $I_{1}^{''}$ directly instead of first working out each sub-circuit's own $I_{3}$ .

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 16 Theorems questions, with notes

GATE 2018 IN Q.30: solution and answer

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