GATE 2022 EC Q.13: solution and answer

GATE 2022 · ECQ.13 · 1 mark · MCQ

The current $I$ in the circuit shown is ________.

GATE 2022 EC question 13: Three-mesh ladder circuit with a common bottom (ground) wire. — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
$1.25 \times 1 0^{- 3} A$
(B)
$0.75 \times 1 0^{- 3} A$
(C)
$- 0.5 \times 1 0^{- 3} A$
(D)
$1.16 \times 1 0^{- 3} A$

Answer and solution

ANSWER

$I = 0.75 \times 1 0^{- 3} A$ , option (B).

Official answer: BSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Take the bottom wire as the reference (0 V). The source's longer plate (its positive terminal) is on top, so the top-left node is at $5 V$ . Let $V_{A}$ be the voltage of the middle node, where the three 2 kΩ resistors meet.
The right-hand 2 kΩ is in series with an ideal current source, so it carries exactly $1 0^{- 3} A$ into node A whatever its resistance. Its value does not affect $V_{A}$ .
Write KCL at node A, with current into the node equal to current out of it: $\frac{5 - V _{A}}{2 k Ω} + 1 mA = \frac{V _{A}}{2 k Ω}$
Multiply through by $2 k Ω$ : $5 - V_{A} + 2 = V_{A} \Rightarrow V_{A} = 3.5 V$
Then $I = \frac{5 - V _{A}}{2 k Ω} = \frac{1.5 V}{2 k Ω} = 0.75 \times 1 0^{- 3} A$
Check by superposition. With the 5 V source acting alone (current source open-circuited), the right-hand 2 kΩ carries no current and the source drives two 2 kΩ resistors in series: $I^{'} = \frac{5}{4 k Ω} = 1.25 mA$ . With the 1 mA source acting alone (5 V source short-circuited), node A sees two equal 2 kΩ paths to ground, so each takes 0.5 mA. The one through the left resistor flows against the arrow for $I$ , so $I^{''} = - 0.5 mA$ .
Adding the two: $I = 1.25 - 0.5 = 0.75 mA = 0.75 \times 1 0^{- 3} A$ , which confirms option (B).

KEY IDEA

A resistor in series with an ideal current source has no effect on the rest of the circuit, because the source fixes the branch current at 1 mA whatever the resistor's value. That leaves a single unknown node voltage. One KCL equation at node A gives $V_{A} = 3.5 V$ , and $I = (5 - V_{A}) /2 k Ω$ . Superposition gives the same answer: 1.25 mA from the voltage source minus 0.5 mA pushed back by the current source.

WATCH OUT FOR

Using only the 5 V source's contribution (5 V across 4 kΩ = 1.25 mA, option A), or only the current source's share (−0.5 mA, option C), instead of adding the two. Another mistake is putting the right-hand 2 kΩ in parallel with the middle resistor. It is in series with the current source, so it carries exactly 1 mA and can be ignored when finding $V_{A}$ .

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 25 Laws & node/mesh questions, with notes

GATE 2022 EC Q.13: solution and answer

Answer and solution

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