GATE 2018 IN Q.32: solution and answer

GATE 2018 · INQ.32 · 2 marks · MCQ

In the given circuit, the mesh currents $I_{1}$ , $I_{2}$ and $I_{3}$ are

GATE 2018 IN question 32: A planar circuit with three meshes. — Figure from the GATE 2018 IN paper. It is part of the question.

(A)
$I_{1} = 1$ A, $I_{2} = 2$ A and $I_{3} = 3$ A
(B)
$I_{1} = 2$ A, $I_{2} = 3$ A and $I_{3} = 4$ A
(C)
$I_{1} = 3$ A, $I_{2} = 4$ A and $I_{3} = 5$ A
(D)
$I_{1} = 4$ A, $I_{2} = 5$ A and $I_{3} = 6$ A

Answer and solution

ANSWER

$I_{1} = 1 A$ , $I_{2} = 2 A$ and $I_{3} = 3 A$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2018 IN (IIT Guwahati)

Current-source constraint. The 2 A source is the only element in the branch shared by meshes 1 and 3. With both meshes clockwise, $I_{3}$ flows up this branch and $I_{1}$ flows down it. The source forces 2 A upward, so $I_{3} - I_{1} = 2 (1)$
KVL in mesh 2 (clockwise). Going left to right through the 5 V source takes us from − to +, which is a rise of 5 V. Along the bottom of mesh 2 we travel right to left. The leftward current in the 1 Ω is $(I_{2} - I_{3})$ and the leftward current in the 6 Ω is $(I_{2} - I_{1})$ : $- 5 + 1 (I_{2} - I_{3}) + 6 (I_{2} - I_{1}) = 0 \Rightarrow - 6 I_{1} + 7 I_{2} - I_{3} = 5 (2)$
Supermesh of meshes 1 and 3. KVL cannot be written for mesh 1 or mesh 3 on its own, because the voltage across the current source is unknown. Instead, go clockwise round the outer boundary of the two meshes and leave out the source: up the 2 Ω (current $I_{1}$ ), right through the 6 Ω (current $I_{1} - I_{2}$ ), right through the 1 Ω (current $I_{3} - I_{2}$ ), down the right 1 Ω (current $I_{3}$ ), then back along the bottom wire, which has no elements: $2 I_{1} + 6 (I_{1} - I_{2}) + 1 (I_{3} - I_{2}) + 1 \cdot I_{3} = 0 \Rightarrow 8 I_{1} - 7 I_{2} + 2 I_{3} = 0 (3)$
Substitute (1) into (2). Put $I_{3} = I_{1} + 2$ into (2): $- 6 I_{1} + 7 I_{2} - (I_{1} + 2) = 5$ , so $- 7 I_{1} + 7 I_{2} = 7$ and $I_{2} = I_{1} + 1$ .
Substitute into (3). $8 I_{1} - 7 (I_{1} + 1) + 2 (I_{1} + 2) = 0 \Rightarrow 3 I_{1} - 3 = 0 \Rightarrow I_{1} = 1 A$ .
Back-substitute. $I_{2} = I_{1} + 1 = 2 A$ and $I_{3} = I_{1} + 2 = 3 A$ . Check in (2): $- 6 + 14 - 3 = 5$ . Check in (3): $8 - 14 + 6 = 0$ .
Physical check with node voltages. Take the bottom wire as $0 V$ . The 2 Ω carries $I_{1} = 1 A$ upward, so $V_{L} = - 2 V$ . The right 1 Ω carries $I_{3} = 3 A$ downward, so $V_{R} = 3 V$ . Then $V_{R} - V_{L} = 5 V$ , which is exactly the source voltage, with + on the right. At node M, the 2 A from the source splits into 1 A leftward through the 6 Ω and 1 A rightward through the 1 Ω. Both branches give $V_{M} = 4 V$ , so KCL and KVL hold everywhere. The answer is option (A).
Why the supermesh equation decides it. All four options satisfy $I_{3} - I_{1} = 2$ and $I_{2} = I_{1} + 1$ , so equations (1) and (2) cannot separate them. Only (3) does. For example, option (B) gives $8 (2) - 7 (3) + 2 (4) = 3 \neq = 0$ .

KEY IDEA

When a current source is the only element in a branch shared by two meshes, it gives a constraint between their mesh currents, here $I_{3} - I_{1} = 2$ . The KVL equation that replaces those two meshes goes round their combined outer boundary (the supermesh) and leaves out the source, because the voltage across the source is unknown.

WATCH OUT FOR

There are two common mistakes. The first is to write KVL round mesh 1 or mesh 3 alone as if the current source had zero voltage across it. The second is to test the options only against the source constraint. Every option satisfies $I_{3} - I_{1} = 2$ , and every option also satisfies the mesh-2 equation, so only the supermesh equation $8 I_{1} - 7 I_{2} + 2 I_{3} = 0$ picks out (A). Watch the sign of the 5 V source too: going clockwise in mesh 2 you meet − then +, which is a rise.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 25 Laws & node/mesh questions, with notes

GATE 2018 IN Q.32: solution and answer

Answer and solution

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