GATE 2018 IN Q.7: solution and answer

GATE 2018 · INQ.7 · 1 mark · MCQ

The Thevenin equivalent circuit representation across terminals p-q of the circuit shown in the figure is a

GATE 2018 IN question 7: A 4 V independent voltage source sits on the left, + terminal at the top and - terminal at the bottom. — Figure from the GATE 2018 IN paper. It is part of the question.

(A)
1 V source in series with 150 kΩ
(B)
1 V source in parallel with 100 kΩ
(C)
2 V source in series with 150 kΩ
(D)
2 V source in parallel with 200 kΩ

Answer and solution

ANSWER

The Thevenin equivalent across p-q is a 2 V source in series with 150 kΩ, option (C).

Official answer: CSource: Official answer key, GATE 2018 IN (IIT Guwahati)

Open-circuit voltage. With p-q open, no current flows in the 100 kΩ resistor leading to p, so there is no voltage drop across it. $V_{t h}$ is therefore the voltage across the shunt 100 kΩ resistor, with p positive with respect to q.
Voltage divider. The 4 V source drives the first series 100 kΩ resistor and the shunt 100 kΩ resistor in series: $V_{t h} = 4 \times \frac{100}{100 + 100} = 2 V$
Thevenin resistance. Set the independent 4 V source to zero, which means replacing it with a short circuit. Looking in from p-q, the first 100 kΩ resistor is now in parallel with the shunt 100 kΩ resistor: $100 ∥ 100 = \frac{100 \times 100}{100 + 100} = 50 k Ω$
That 50 kΩ is in series with the 100 kΩ resistor leading to p: $R_{t h} = 50 + 100 = 150 k Ω$
Form of the equivalent. A Thevenin equivalent is always an ideal voltage source $V_{t h}$ in series with $R_{t h}$ . A voltage source in parallel with a resistor, as in (B) and (D), behaves at its terminals like the bare source ( $R_{t h} = 0$ ), so neither of those can be right. The equivalent is a 2 V source in series with 150 kΩ, which is option (C).

KEY IDEA

$V_{t h}$ is the open-circuit voltage. No current flows in the resistor leading to the open terminal, so it is just the divider voltage across the shunt resistor, $4 \times \frac{1}{2} = 2 V$ . For $R_{t h}$ , short the voltage source: $(100 ∥ 100) + 100 = 150 k Ω$ . A Thevenin equivalent is always a source in series with a resistance.

WATCH OUT FOR

Three slips are common. (1) Treating the resistor leading to p as part of the divider, which gives $V_{t h} = 4/3 V$ or a similar wrong value, even though it carries no current when p-q is open. (2) Forgetting that same series resistor when finding $R_{t h}$ , which gives 50 kΩ. (3) Opening the voltage source instead of shorting it, which gives $100 + 100 = 200 k Ω$ and makes (D) look plausible. Separately, any option with a voltage source in parallel with a resistor has the wrong form for a Thevenin equivalent.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 16 Theorems questions, with notes

GATE 2018 IN Q.7: solution and answer

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