GATE 2020 EE Q.19: solution and answer

GATE 2020 · EEQ.19 · 1 mark · NAT

The Thevenin equivalent voltage, $V_{T H}$ , in V (rounded off to 2 decimal places) of the network shown below, is ________

GATE 2020 EE question 19: The circuit has a common bottom (reference) rail and a top rail, reading left to right. — Figure from the GATE 2020 EE paper. It is part of the question.

Answer and solution

ANSWER

$V_{T H} = 14.00 V$ .

Official answer: 13.80 to 14.20Source: Official answer key, GATE 2020 EE (IIT Delhi)

Open circuit at the output. Nothing is connected across the output terminals, so no current flows in the right-hand $3 Ω$ . With no current it drops no voltage, so $V_{T H} = V_{B}$ , the voltage of the node where the $2 Ω$ , the current source and the output $3 Ω$ meet, measured from the bottom rail.
Voltage source in parallel with a resistor. The left $3 Ω$ is connected directly across the ideal $4 V$ source. It draws current from the source but cannot change the source's terminal voltage, so $V_{A} = 4 V$ whatever that resistor's value is.
KCL at node B. The $5 A$ source pushes current into B. The output branch is open, so the only way out of B is through the $2 Ω$ to node A: $\frac{V _{B} - 4}{2} = 5$
Solve. $V_{B} = 4 + 5 \times 2 = 14 V$ .
Result. $V_{T H} = V_{B} = 14.00 V$ . The sign is positive because the + output terminal connects to B through the current-free $3 Ω$ , and the − terminal is the bottom rail.

KEY IDEA

Only two facts are needed. An ideal voltage source fixes its node voltage no matter what resistor sits across it, and a resistor that carries no current (the open-circuited output $3 Ω$ ) drops no voltage. The whole $5 A$ therefore flows through the $2 Ω$ , so $V_{T H} = 4 + 5 \times 2 = 14 V$ .

WATCH OUT FOR

Letting the left $3 Ω$ carry part of the $5 A$ (treating it as a divider with the $2 Ω$ , or using source transformation on it). It is directly across an ideal voltage source, so it has no effect on $V_{A}$ or $V_{B}$ . Another slip is to count a voltage drop across the output $3 Ω$ . With the terminals open it carries no current, so it drops nothing.

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

GATE 2020 EE Q.19: solution and answer

Answer and solution

More Theorems questions