GATE 2022 EC Q.38: solution and answer

GATE 2022 · ECQ.38 · 2 marks · MCQ

For the circuit shown, the locus of the impedance $Z (j ω)$ is plotted as $ω$ increases from zero to infinity. The values of $R_{1}$ and $R_{2}$ are:

GATE 2022 EC question 38: Left: a two-terminal network seen from the input, labelled Z(j ω) with an arrow pointing into it. — Figure from the GATE 2022 EC paper. It is part of the question.

(A)
$R_{1} = 2 k Ω, R_{2} = 3 k Ω$
(B)
$R_{1} = 5 k Ω, R_{2} = 2 k Ω$
(C)
$R_{1} = 5 k Ω, R_{2} = 2.5 k Ω$
(D)
$R_{1} = 2 k Ω, R_{2} = 5 k Ω$

Answer and solution

ANSWER

$R_{1} = 2 k Ω$ and $R_{2} = 3 k Ω$ , option (A).

Official answer: ASource: Official answer key, GATE 2022 EC (IIT Kharagpur)

Write the impedance of $R_{1}$ in series with $R_{2} ∥ C$ : $Z (j ω) = R_{1} + \frac{R _{2}}{1 + j ω C R _{2}}$
Low-frequency end ( $ω = 0$ , the capacitor is an open circuit): $Z (0) = R_{1} + R_{2}$ . The plot marks $ω = 0$ at $Re (Z) = 5 k Ω$ , so $R_{1} + R_{2} = 5 k Ω$ .
High-frequency end ( $ω \to \infty$ , the capacitor is a short circuit across $R_{2}$ ): $Z (\infty) = R_{1}$ . The plot marks $ω \to \infty$ at $Re (Z) = 2 k Ω$ , so $R_{1} = 2 k Ω$ .
Therefore $R_{2} = 5 - 2 = 3 k Ω$ .
Check against the height of the arc. Since $Z - (R_{1} + \frac{R _{2}}{2}) = \frac{R _{2}}{2} \cdot \frac{1 - j ω C R _{2}}{1 + j ω C R _{2}},$ and the last fraction always has magnitude 1, the locus is a semicircle of radius $R_{2} /2$ centred at $R_{1} + R_{2} /2$ . Its imaginary part is negative, so it lies above the axis on a $- Im (Z)$ plot. Its highest point, reached at $ω = 1/ (R_{2} C)$ , is $- Im (Z) = R_{2} /2 = 1.5 k Ω$ , which matches the plot.
Hence $R_{1} = 2 k Ω$ and $R_{2} = 3 k Ω$ : option (A).

KEY IDEA

Read the two ends of the locus. At $ω = 0$ the capacitor is an open circuit, so $Z = R_{1} + R_{2}$ . As $ω \to \infty$ it shorts out $R_{2}$ , so $Z = R_{1}$ . The locus of $R_{2} ∥ C$ is a semicircle of diameter $R_{2}$ , so the arc's height, $R_{2} /2$ , gives an independent check.

WATCH OUT FOR

Reading the ends of the locus the wrong way round, so that $R_{1} = 5 k Ω$ and $R_{2} = 2 k Ω$ (option B). Another mistake is taking the arc's height of $1.5 k Ω$ as the radius of the whole locus measured from the origin. A third is treating the right-hand end (5) as $R_{2}$ on its own rather than $R_{1} + R_{2}$ , which gives option D. The capacitor removes $R_{2}$ at high frequency, so the high-frequency end (2) is $R_{1}$ alone.

From the GATE 2022 EC paper. Every question of this paper · Download the full paper · All 34 AC, power & resonance questions, with notes

GATE 2022 EC Q.38: solution and answer

Answer and solution

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