GATE 2020 IN Q.30: solution and answer

GATE 2020 · INQ.30 · 2 marks · MCQ

$I_{1}$ , $I_{2}$ and $I_{3}$ in the figure below are mesh currents. The correct set of mesh equations for these currents, in matrix form, is ___

GATE 2020 IN question 30: A planar circuit with three meshes, all mesh currents drawn clockwise. — Figure from the GATE 2020 IN paper. It is part of the question.

(A)
$3 - 1 - 2 - 1 3 - 1 - 2 - 1 3 I_{1} I_{2} I_{3} = V_{1} V_{2} - V_{3}$
(B)
$3 - 1 - 2 - 1 3 - 1 - 2 - 1 - 3 I_{1} I_{2} I_{3} = V_{1} V_{2} V_{3}$
(C)
$- 3 - 1 - 2 - 1 3 - 1 - 2 - 1 3 I_{1} I_{2} I_{3} = V_{1} V_{2} - V_{3}$
(D)
$1 - 1 - 2 - 1 2 - 1 - 2 - 1 3 I_{1} I_{2} I_{3} = V_{1} V_{2} V_{3}$

Answer and solution

ANSWER

The mesh equations are $3 - 1 - 2 - 1 3 - 1 - 2 - 1 3 I_{1} I_{2} I_{3} = V_{1} V_{2} - V_{3}$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2020 IN (IIT Delhi)

Self-resistances (diagonal entries). Mesh 1: $1 + 2 = 3 Ω$ . Mesh 2: $1 + 1 + 1 = 3 Ω$ (the right $1 Ω$ plus the two middle $1 Ω$ resistors). Mesh 3: $1 + 2 = 3 Ω$ .
Mutual resistances (off-diagonal entries). Meshes 1 and 2 share the L–M $1 Ω$ . Meshes 2 and 3 share the M–R $1 Ω$ . Meshes 1 and 3 share the $2 Ω$ . All three currents are clockwise, so in every shared branch the two currents flow in opposite directions, and every mutual term is negative: $R_{12} = - 1$ , $R_{23} = - 1$ , $R_{13} = - 2$ .
KVL in mesh 1 (clockwise). Going up through $V_{1}$ from − to + is a rise of $V_{1}$ : $V_{1} = 1 (I_{1} - I_{2}) + 2 (I_{1} - I_{3}) \Rightarrow 3 I_{1} - I_{2} - 2 I_{3} = V_{1}$
KVL in mesh 2 (clockwise). Going left to right along the top through $V_{2}$ runs from − to +, a rise of $V_{2}$ : $V_{2} = 1 I_{2} + 1 (I_{2} - I_{3}) + 1 (I_{2} - I_{1}) \Rightarrow - I_{1} + 3 I_{2} - I_{3} = V_{2}$
KVL in mesh 3 (clockwise). Going down the right side through $V_{3}$ runs from + to −, a drop of $V_{3}$ : $0 = 1 (I_{3} - I_{2}) + V_{3} + 2 (I_{3} - I_{1}) \Rightarrow - 2 I_{1} - I_{2} + 3 I_{3} = - V_{3}$
Matrix form. $3 - 1 - 2 - 1 3 - 1 - 2 - 1 3 I_{1} I_{2} I_{3} = V_{1} V_{2} - V_{3}$ This is option (A). Option (C) has the wrong sign on $R_{11}$ . Option (B) has the wrong sign on $R_{33}$ and on $V_{3}$ . Option (D) has the wrong self-resistances $R_{11}$ and $R_{22}$ , and the wrong sign on $V_{3}$ .

KEY IDEA

When every mesh current in a resistive network without dependent sources is drawn in the same direction, you can write the mesh equations by inspection. The resistance matrix is symmetric. Each diagonal entry is the total resistance around that mesh and is always positive. Each off-diagonal entry is minus the resistance the two meshes share. Each right-hand-side entry is the net source voltage rise met when you go around the mesh in the direction of its current.

WATCH OUT FOR

Writing $+ V_{3}$ instead of $- V_{3}$ . The clockwise current $I_{3}$ goes down through $V_{3}$ from its + terminal to its − terminal, so $V_{3}$ is a drop in that mesh and appears as $- V_{3}$ on the source side. Another mistake is accepting a negative diagonal entry (options B and C). A self-resistance can never be negative, and the matrix must be symmetric. Those two checks rule out B and C straight away.

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 25 Laws & node/mesh questions, with notes

GATE 2020 IN Q.30: solution and answer

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