GATE 2023 EC Q.32: solution and answer

GATE 2023 · ECQ.32 · 1 mark · NAT

In the circuit shown below, the current $i$ flowing through $200 Ω$ resistor is ______ mA (rounded off to two decimal places).

GATE 2023 EC question 32: Three-node ladder with a common bottom (ground) rail. — Figure from the GATE 2023 EC paper. It is part of the question.

Answer and solution

ANSWER

$i = \frac{15}{11} mA \approx 1.36 mA$ .

Official answer: 1.30 to 1.40Source: Official answer key, GATE 2023 EC (IIT Kanpur)

Take the bottom rail as ground. Call the top-left node $A$ (voltage $V_{A}$ ) and the node at the right end of the $1 k Ω$ resistor $B$ (voltage $V_{B}$ ). Work in volts, $k Ω$ and $mA$ .
Source transformation on the left: the $2 V$ source (+ towards $A$ ) in series with $2 k Ω$ becomes a $\frac{2}{2} = 1 mA$ source into $A$ in parallel with $2 k Ω$ .
Combine everything between $A$ and ground: the currents add, $1 + 1 = 2 mA$ into $A$ (the transformed source plus the upward $1 mA$ source), and the resistances combine, $2 k Ω ∥ 2 k Ω = 1 k Ω$ . Back to Thevenin form, that is $2 \times 1 = 2 V$ behind $1 k Ω$ .
Source transformation on the $1 mA$ source in parallel with the $1 k Ω$ resistor between $A$ and $B$ : it becomes a $1 \times 1 = 1 V$ source in series with $1 k Ω$ . The current source pushes current towards $B$ , so the + terminal is on the $B$ side: a rise of $1 V$ going from $A$ to $B$ , which adds to the $2 V$ .
That leaves one loop: $2 + 1 = 3 V$ behind $1 + 1 = 2 k Ω$ , driving the $200 Ω = 0.2 k Ω$ load. $i = \frac{2 + 1}{1 + 1 + 0.2} mA = \frac{3}{2.2} mA = \frac{15}{11} mA \approx 1.364 mA$
Check by nodal analysis (currents leaving each node). KCL at $A$ : $\frac{V _{A} - 2}{2} + \frac{V _{A}}{2} - 1 + (V_{A} - V_{B}) + 1 = 0 \Rightarrow 2 V_{A} - V_{B} = 1$ . KCL at $B$ : $(V_{A} - V_{B}) + 1 = \frac{V _{B}}{0.2} \Rightarrow V_{A} = 6 V_{B} - 1$ . Substituting, $2 (6 V_{B} - 1) - V_{B} = 1$ , so $11 V_{B} = 3$ , $V_{B} = \frac{3}{11} V$ and $i = \frac{V _{B}}{0.2} = \frac{15}{11} mA$ .
Rounded to two decimal places, $i = 1.36 mA$ .

KEY IDEA

Use repeated source transformation. Everything between node $A$ and ground reduces to $2 V$ behind $1 k Ω$ . The $1 mA$ source across the $1 k Ω$ becomes a series $1 V$ source whose + terminal faces $B$ . That leaves a single loop: $3 V$ across $2.2 k Ω$ . Two-node nodal analysis gives the same result.

WATCH OUT FOR

Getting the polarity of the transformed $1 mA ∥ 1 k Ω$ source wrong. A current source pointing from $A$ to $B$ becomes a voltage source with its + terminal at $B$ , so it adds to the $2 V$ . Subtracting it gives $\frac{1}{2.2} \approx 0.45 mA$ . A second trap is dropping the upward $1 mA$ source when combining the branches at $A$ , which gives $\frac{2}{2.2} \approx 0.91 mA$ .

From the GATE 2023 EC paper. Every question of this paper · Download the full paper · All 25 Laws & node/mesh questions, with notes

GATE 2023 EC Q.32: solution and answer

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