GATE 2023 IN Q.61: solution and answer

GATE 2023 · INQ.61 · 2 marks · NAT

In the diagram shown, the frequency of the sinusoidal source voltage $V_{S}$ is 50 Hz. The load voltage is 230 V (RMS), and the load impedance is $\frac{230}{2} + j \frac{230}{2} Ω$ . The value of attenuator $A_{1} = \frac{1}{50 2}$ . The multiplier output voltage $V_{o} = \frac{V _{x} V _{y}}{1 V}$ , where $V_{x}$ and $V_{y}$ are the inputs. The magnitude of the average value of the multiplier output $V_{o}$ is ________ V (rounded off to one decimal place).

GATE 2023 IN question 61: A sinusoidal source V_S on the left. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

The magnitude of the average multiplier output is $∣ \overline{V_{o}} ∣ = 2.3 V$ .

Official answer: 2.1 to 2.5Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Load current, from Ohm's law with the given load voltage: $∣ Z ∣ = (\frac{230}{2})^{2} + (\frac{230}{2})^{2} = 230 Ω$ and $∠ Z = tan^{- 1} (1) = 4 5^{\circ}$ . Taking the load voltage as reference, $I = \frac{230∠ 0 ^{\circ}}{230∠4 5 ^{\circ}} = 1∠ - 4 5^{\circ} A (rms)$ , so the current lags the load voltage by $4 5^{\circ}$ . The 50 Hz figure is not needed, because the impedance is given directly.
Current-sensing input: the load current returns to the grounded source terminal through the series $1 Ω$ resistor, and $V_{x}$ is taken from the resistor's ungrounded end. So $V_{x} = I \times 1 Ω = 1∠ - 4 5^{\circ} V (rms)$ , in phase with the current.
Voltage-sensing input: $A_{1}$ scales the load voltage and the phase shifter adds $+ 9 0^{\circ}$ , so $V_{y} = \frac{230}{50 2} ∠9 0^{\circ} = 3.2527∠9 0^{\circ} V (rms)$ . The polarity of $A_{1}$ 's input leads is not marked, and reversing them would only flip the sign of $V_{y}$ .
The average of the product of two sinusoids at the same frequency is $\overline{v_{x} v_{y}} = V_{x, rms} V_{y, rms} cos θ$ , where $θ$ is the phase angle between them. Here $θ = 9 0^{\circ} - (- 4 5^{\circ}) = 13 5^{\circ}$ .
$\overline{V_{o}} = \frac{1 \times 3.2527 \times cos 13 5 ^{\circ}}{1 V} = \frac{4.6}{2} \times (- \frac{1}{2}) V = - 2.30 V$
Check through reactive power: the $9 0^{\circ}$ shift turns the multiplier into a varmeter, so $\overline{V_{o}} = \frac{Q \times 1 Ω \times A _{1}}{1 V}$ . With $Q = V I sin ϕ = 230 \times 1 \times sin 4 5^{\circ} = 162.63 VAR$ , this gives $\frac{162.63}{50 2} = 2.30 V$ in magnitude.
The sign depends only on the unmarked polarity of $A_{1}$ 's input and on the phase-shift convention. The question asks for the magnitude, so $∣ \overline{V_{o}} ∣ = 2.3 V$ .

KEY IDEA

The multiplier's $V_{x}$ input is proportional to the load current and its $V_{y}$ input is proportional to the load voltage shifted by $9 0^{\circ}$ . That makes the circuit a varmeter: the average output is $V_{x, rms} V_{y, rms} cos (9 0^{\circ} + ϕ)$ , whose magnitude is $V I sin ϕ \times R_{shunt} \times A_{1}$ for a load with power-factor angle $ϕ$ . Here that is $230 \times 1 \times sin 4 5^{\circ} \times 1 \times \frac{1}{50 2} = 2.3 V$ .

WATCH OUT FOR

Ignoring the $+ 9 0^{\circ}$ phase shifter and computing active power, $P \times R \times A_{1} = 162.63/70.71 = 2.3 V$ , reaches the right number only by accident, because $cos 4 5^{\circ} = sin 4 5^{\circ}$ . The method would fail for any other power factor. Two other slips: using peak values instead of rms (which gives $4.6 V$ ), and taking $∣ I ∣ = 230/ (230/ 2) = 2 A$ by dividing by one component of $Z$ instead of $∣ Z ∣ = 230 Ω$ .

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 8 V, I, power & energy questions, with notes

GATE 2023 IN Q.61: solution and answer

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