GATE 2023 EE Q.32: solution and answer

GATE 2023 · EEQ.32 · 1 mark · NAT

For the circuit shown in the figure, $V_{1} = 8 V$ , DC and $I_{1} = 8 A$ , DC. The voltage $V_{ab}$ in Volts is ________ (Round off to 1 decimal place).

GATE 2023 EE question 32: A rectangular circuit with four distinct nodes. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

$V_{ab} = 6.0 V$ .

Official answer: 5.9 to 6.1Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Simplify the topology first. The middle wire runs from the junction under $a$ straight across and up to $b$ with no element in it, so that junction and $b$ are one node. $V_{1}$ therefore sits directly between the top-left node $P$ and $b$ . Its long plate ( $+$ ) faces $P$ , so $V_{P} - V_{b} = 8 V$ .
Combine the parallel elements. Both $3 Ω$ resistors join $a$ to $b$ , so $R_{ab} = 3 ∥ 3 = 1.5 Ω$ . Both $2 Ω$ resistors join $M$ to $b$ , so together they are $1 Ω$ . Node $M$ touches only these resistors and $I_{1}$ , so the $1 Ω$ is in series with $I_{1}$ , making one branch from $b$ through $M$ to $P$ .
An ideal voltage source holds its terminal voltage whatever current flows through it. The $I_{1}$ + $1 Ω$ branch is connected across $V_{1}$ (from $b$ to $P$ ), so it can only change the current in $V_{1}$ . $V_{P b}$ stays at $8 V$ , and $I_{1}$ has no effect on $V_{ab}$ . (Superposition gives the same result: with $V_{1}$ replaced by a short, $P$ and $b$ are one node, so $I_{1}$ circulates through the short and adds nothing to $V_{ab}$ .)
The only other path from $P$ to $b$ is the $0.5 Ω$ resistor in series with $R_{ab} = 1.5 Ω$ . Nothing else connects to $a$ , so the voltage divider rule applies: $V_{ab} = V_{P b} \frac{1.5}{0.5 + 1.5} = 8 \times \frac{1.5}{2} = 6 V$
Check with KCL, taking $V_{b} = 0$ , so $V_{P} = 8 V$ and $V_{a} = 6 V$ . The $0.5 Ω$ resistor carries $(8 - 6) /0.5 = 4 A$ from $P$ to $a$ , and this splits into $2 A + 2 A$ through the two $3 Ω$ resistors ( $6/3 = 2 A$ each). At $P$ : $8 A$ arrives from $I_{1}$ , $4 A$ leaves through the $0.5 Ω$ , and the other $4 A$ leaves through $V_{1}$ into $b$ . At $b$ : $4 + 4 = 8 A$ comes in and returns to $M$ through the $1 Ω$ pair, so $V_{M} = - 8 V$ . KCL holds at every node, so $V_{ab} = 6.0 V$ .

KEY IDEA

An ideal voltage source fixes the voltage across its own terminals. Once you see that the junction under $a$ is wired straight to $b$ , $V_{1}$ sits directly across $P$ and $b$ . The $I_{1}$ branch (in series with the two $2 Ω$ resistors) is connected across $V_{1}$ , so it cannot change any voltage in the rest of the circuit. What is left is an $8 V$ source driving $0.5 Ω$ in series with $3 ∥ 3 = 1.5 Ω$ , and the voltage divider rule gives $6 V$ .

WATCH OUT FOR

Missing the plain wire from the junction under $a$ to $b$ . A student who misses it treats the vertical $3 Ω$ , the middle $2 Ω$ and $V_{1}$ as meeting at a separate node and solves a much harder, wrong circuit. A second trap is letting $I_{1}$ push current through the divider: it only changes the current through the ideal source $V_{1}$ . Reading the battery backwards (long plate towards $b$ ) gives $- 6 V$ . The long plate is the positive terminal, and here it faces the top-left node.

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

GATE 2023 EE Q.32: solution and answer

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