GATE 2023 EE Q.64: solution and answer

GATE 2023 · EEQ.64 · 2 marks · NAT

A balanced delta connected load consisting of the series connection of one resistor ( $R = 15 Ω$ ) and a capacitor ( $C = 212.21 μ F$ ) in each phase is connected to three-phase, 50 Hz, 415 V supply terminals through a line having an inductance of $L = 31.83 mH$ per phase, as shown in the figure. Considering the change in the supply terminal voltage with loading to be negligible, the magnitude of the voltage across the terminals $V_{A B}$ in Volts is ________ (Round off to the nearest integer).

GATE 2023 EE question 64: Three supply terminals on the left, labelled '3-phase, 50 Hz, 415 V (r.m.s.) supply terminals'. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

$∣ V_{A B} ∣ \approx 415 V$ .

Official answer: 414 to 416Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Reactances at $ω = 2 π (50) = 314.16 rad/s$ : $X_{L} = ω L = 314.16 \times 31.83 \times 1 0^{- 3} \approx 10 Ω$ , and $X_{C} = \frac{1}{ω C} = \frac{1}{314.16 \times 212.21 \times 1 0 ^{- 6}} \approx 15 Ω$ .
Impedance of each delta branch (R in series with C): $Z_{Δ} = R - j X_{C} = 15 - j 15 Ω$ .
The load is balanced, so replace the delta with an equivalent star: $Z_{Y} = \frac{Z _{Δ}}{3} = 5 - j 5 Ω$ per phase.
Per-phase equivalent circuit (supply neutral to the load's equivalent star point $N^{'}$ ): the line inductance is in series with $Z_{Y}$ , so $Z_{total} = j 10 + (5 - j 5) = 5 + j 5 Ω$ , and $∣ Z_{total} ∣ = 52 = 7.071 Ω$ .
The supply terminal voltage does not change with loading, so the supply phase voltage stays at $V_{p h} = \frac{415}{3} = 239.6 V$ . Line current: $I_{L} = \frac{239.6}{7.071} = 33.88 A$ .
Voltage across one phase of the equivalent star load: $∣ V_{A N^{'}} ∣ = I_{L} ∣ Z_{Y} ∣ = 33.88 \times 52 = 239.6 V$ . The voltage-divider ratio shows the same thing: $\frac{V _{A N^{'}}}{V _{A N}} = \frac{5 - j 5}{5 + j 5} = - j$ , whose magnitude is $1$ .
Line voltage at the load terminals: $∣ V_{A B} ∣ = 3 ∣ V_{A N^{'}} ∣ = 3 \times 239.6 \approx 415 V$ . In magnitude, the capacitive load exactly cancels the drop across the line inductance. Only the phase changes: the load voltage lags the supply voltage by $9 0^{\circ}$ .

KEY IDEA

Convert the balanced delta load to an equivalent star ( $Z_{Y} = Z_{Δ} /3 = 5 - j 5 Ω$ ) and solve a single-phase circuit: a stiff source, then $j 10 Ω$ of line reactance, then the star-phase load. The divider ratio $(5 - j 5) / (5 + j 5) = - j$ has magnitude 1. So the load line voltage has the same magnitude as the supply (415 V) and lags it by $9 0^{\circ}$ .

WATCH OUT FOR

Putting the line inductance in series with the delta branch impedance without converting to star first, i.e. $j 10 + 15 - j 15$ . Line elements carry line current, so they can only be combined with the per-phase star equivalent $Z_{Δ} /3$ . The second trap is assuming a series line inductance must lower the load voltage. With a capacitive load there can be no drop at all, or even a rise. Here the magnitude is unchanged.

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 11 Three-phase (EE) questions, with notes

GATE 2023 EE Q.64: solution and answer

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