GATE 2018 EE Q.55: solution and answer

GATE 2018 · EEQ.55 · 2 marks · NAT

The voltage $v (t)$ across the terminals $a$ and $b$ as shown in the figure, is a sinusoidal voltage having a frequency $ω = 100$ radian/s. When the inductor current $i (t)$ is in phase with the voltage $v (t)$ , the magnitude of the impedance $Z$ (in $Ω$ ) seen between the terminals $a$ and $b$ is ________ (up to 2 decimal places).

GATE 2018 EE question 55: Terminals a (top, marked +) and b (bottom, marked -), with v(t) applied across them and an arrow labelled Z pointing into… — Figure from the GATE 2018 EE paper. It is part of the question.

Answer and solution

ANSWER

$∣ Z ∣ = 50.00 Ω$ . The series inductor ( $L = 0.5 H$ ) cancels the $- j 50 Ω$ reactance of the RC parallel branch, leaving only its resistive part.

Official answer: 50.0Source: Official answer key, GATE 2018 EE (IIT Guwahati)

Capacitor reactance at $ω = 100 rad/s$ : $X_{C} = \frac{1}{ω C} = \frac{1}{100 \times 100 \times 1 0 ^{- 6}} = 100 Ω$ , so the capacitor's impedance is $- j 100 Ω$ .
Parallel combination of R and C: $Z_{p} = \frac{100 ( - j 100 )}{100 - j 100} = \frac{- j 100}{1 - j} = \frac{- j 100 ( 1 + j )}{( 1 - j ) ( 1 + j )} = \frac{100 - j 100}{2} = 50 - j 50 Ω$
Total impedance seen from a-b (L in series with $Z_{p}$ ): $Z = j ω L + 50 - j 50 = 50 + j (100 L - 50) Ω$ .
In-phase condition: the inductor sits in series with the terminals, so the inductor current $i (t)$ is the terminal current. $i$ in phase with $v$ therefore means $Z = V / I$ is purely resistive: $Im (Z) = 0 \Rightarrow 100 L = 50 \Rightarrow L = 0.5 H$ .
Magnitude at that condition: $Z = 50 + j 0 Ω$ , so $∣ Z ∣ = 50.00 Ω$ .

KEY IDEA

Because the inductor carries the terminal current, "i in phase with v" means the whole input impedance is purely resistive. First reduce the parallel R-C branch to its series form, $50 - j 50 Ω$ . The inductor's $+ j ω L$ has to cancel the $- j 50$ , so the resistive part left over, $50 Ω$ , is $∣ Z ∣$ . You don't need to find L to get the answer.

WATCH OUT FOR

A common mistake is to resonate L with the capacitor alone, using $ω^{2} L C = 1$ , which gives $L = 1 H$ . That ignores the parallel resistor. With $L = 1 H$ you get $Z = 50 + j 50 Ω$ and $∣ Z ∣ = 70.71 Ω$ , and there the current is not in phase with the voltage. Another mistake is to quote $∣ Z_{p} ∣ = ∣50 - j 50∣ = 70.71 Ω$ as the answer, which leaves out the inductor.

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

GATE 2018 EE Q.55: solution and answer

Answer and solution

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