GATE 2020 EC Q.9: solution and answer

GATE 2020 · ECQ.9 · 1 mark · MCQ

In the circuit shown below, the Thevenin voltage $V_{T H}$ is

GATE 2020 EC question 9: A ladder circuit with a common bottom (reference) rail. — Figure from the GATE 2020 EC paper. It is part of the question.

(A)
2.4 V
(B)
2.8 V
(C)
3.6 V
(D)
4.5 V

Answer and solution

ANSWER

$V_{T H} = 3.6 V$ , option (C).

Official answer: CSource: Official answer key, GATE 2020 EC (IIT Delhi)

The output is open-circuited, so the $4 Ω$ resistor carries no current and drops no voltage: $V_{T H} = V_{B}$ , the voltage of the node joining the $2 A$ source and the right-hand $2 Ω$ resistor.
Source transformation on the left: $1 A$ in parallel with $1 Ω$ becomes a $1 \times 1 = 1 V$ source in series with $1 Ω$ . The current source pushes current up into node A, so the $1 V$ source has its positive terminal towards node A.
Going from the bottom rail up through the $1 V$ source and then left to right through the $2 V$ source ( $-$ on the left, $+$ on the right), both are rises, so they add: $1 + 2 = 3 V$ . The resistances in series are $1 + 2 = 3 Ω$ .
So node B is fed by a $3 V$ source through $3 Ω$ , plus the $2 A$ source into the node, with $2 Ω$ from the node to the reference rail.
KCL at node B (current leaving = current entering): $\frac{V _{B} - 3}{3} + \frac{V _{B}}{2} = 2$
Multiply out: $\frac{V _{B}}{3} + \frac{V _{B}}{2} = 2 + 1 = 3$ , so $\frac{5}{6} V_{B} = 3 \Rightarrow V_{B} = 3.6 V$ .
Check by a second source transformation: $3 V$ behind $3 Ω$ is $1 A$ in parallel with $3 Ω$ . The total current into the node is $1 + 2 = 3 A$ , flowing through $3 Ω ∥ 2 Ω = \frac{3 \times 2}{3 + 2} = 1.2 Ω$ , giving $3 \times 1.2 = 3.6 V$ .
$V_{T H} = 3.6 V$ , which is option (C).

KEY IDEA

With the output open, no current flows in the $4 Ω$ resistor, so $V_{T H}$ is simply the voltage of the node before it. Repeated source transformations reduce the left side to a $3 V$ source behind $3 Ω$ (or a $1 A$ source in parallel with $3 Ω$ ). One KCL equation, or current times parallel resistance, then gives the node voltage.

WATCH OUT FOR

Three mistakes are common. (1) Counting a voltage drop across the $4 Ω$ resistor, or adding $4 Ω$ into the node equation, even though it carries no current on open circuit. (2) Subtracting the $2 V$ source from the transformed $1 V$ source: both rise in the same direction ( $-$ to $+$ going left to right), so they add to $3 V$ . (3) Adding the two current sources directly ( $1 + 2 = 3 A$ ) without first moving the $1 A$ source past the $2 V$ source and the $2 Ω$ resistor by source transformation.

From the GATE 2020 EC paper. Every question of this paper · Download the full paper · All 16 Theorems questions, with notes

GATE 2020 EC Q.9: solution and answer

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