GATE 2018 IN Q.6: solution and answer

GATE 2018 · INQ.6 · 1 mark · MCQ

A series R-C circuit is excited by a $1∠0 V$ sinusoidal ac voltage source. The locus diagram of the phasor current $\underline{I} = (x + j y) A$ , when $C$ is varied, while keeping $R$ fixed, is

GATE 2018 IN question 6: Each option is a sketch on complex-plane axes: horizontal axis labelled x (arrows both ways), vertical axis labelled +jy… — Figure from the GATE 2018 IN paper. It is part of the question.

(A)
Figure — semicircle in the first quadrant: starts at the origin, bulges upward (+jy), ends on the positive x-axis.
(B)
Figure — semicircle in the fourth quadrant: starts at the origin, bulges downward (−jy), ends on the positive x-axis.
(C)
Figure — semicircle in the second quadrant: from a point on the negative x-axis, bulges upward (+jy), ends at the origin.
(D)
Figure — semicircle in the third quadrant: from a point on the negative x-axis, bulges downward (−jy), ends at the origin.

Answer and solution

ANSWER

(A): the current phasor traces the upper semicircle of diameter $1/ R$ on the positive real axis, from the origin (as $C \to 0$ ) to $\frac{1}{R} ∠ 0^{\circ} A$ (as $C \to \infty$ ).

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

Impedance of the series R-C branch: $Z = R - j X_{C}$ , where $X_{C} = \frac{1}{ω C}$ . As $C$ goes from $0$ to $\infty$ , $X_{C}$ falls from $\infty$ to $0$ .
Ohm's law in phasor form with $V = 1∠ 0^{\circ} V$ : $\underline{I} = \frac{1}{R - j X _{C}} = \frac{R + j X _{C}}{R ^{2} + X _{C}^{2}}$ so $x = \frac{R}{R ^{2} + X _{C}^{2}}$ and $y = \frac{X _{C}}{R ^{2} + X _{C}^{2}}$ .
Signs: $R > 0$ and $X_{C} \geq 0$ , so $x \geq 0$ and $y \geq 0$ . The current leads the voltage (the circuit is capacitive), so the whole locus lies in the first quadrant. That rules out (B), (C) and (D).
Shape: $x^{2} + y^{2} = \frac{R ^{2} + X _{C}^{2}}{( R ^{2} + X _{C}^{2} ) ^{2}} = \frac{1}{R ^{2} + X _{C}^{2}} = \frac{x}{R}$ . Completing the square gives $(x - \frac{1}{2 R})^{2} + y^{2} = (\frac{1}{2 R})^{2}$ This is a circle with centre $(\frac{1}{2 R}, 0)$ and radius $\frac{1}{2 R}$ , so it passes through the origin.
End points: as $C \to 0$ ( $X_{C} \to \infty$ ), $\underline{I} \to 0$ , the origin. As $C \to \infty$ ( $X_{C} \to 0$ ), $\underline{I} \to \frac{1}{R} ∠ 0^{\circ} A$ , a point on the positive x-axis. The top of the arc is at $X_{C} = R$ , where $\underline{I} = \frac{1 + j}{2 R} = \frac{1}{2 R} ∠4 5^{\circ} A$ .
So the locus is the upper semicircle of diameter $1/ R$ standing on the positive real axis, running from the origin to $1/ R$ . That is option (A).

KEY IDEA

With $R$ fixed and $C$ varied, the impedance $Z = R - j X_{C}$ moves along the vertical half-line $Re Z = R$ below the real axis. Inverting a straight line that does not pass through the origin gives a circle that does pass through the origin, so $\underline{I} = V / Z$ traces a circle through the origin with diameter $1/ R$ on the real axis. Because the circuit is capacitive, the current leads the voltage, so only the upper half ( $y > 0$ ) is traced.

WATCH OUT FOR

Picking (B) by mixing up leading and lagging. The fourth-quadrant semicircle is the current locus of a series R-L circuit with $L$ varied, where the current lags. A related slip is to sketch the impedance locus, which for R-C lies below the real axis, and take it for the current locus: inverting $Z$ flips the sign of the angle. Options (C) and (D) need a negative real part of the current, which a positive resistance cannot give.

From the GATE 2018 IN paper. Every question of this paper · Download the full paper · All 34 AC, power & resonance questions, with notes

GATE 2018 IN Q.6: solution and answer

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