GATE 2022 EE Q.34: solution and answer

GATE 2022 · EEQ.34 · 1 mark · NAT

Two balanced three-phase loads, as shown in the figure, are connected to a $1003$ V, three-phase, 50 Hz main supply. Given $Z_{1} = (18 + j 24) Ω$ and $Z_{2} = (6 + j 8) Ω$ . The ammeter reading, in amperes, is ________. (round off to nearest integer)

GATE 2022 EE question 34: Three supply lines R (top), Y (middle) and B (bottom); — Figure from the GATE 2022 EE paper. It is part of the question.

Answer and solution

ANSWER

The ammeter reads 20 A.

Official answer: 20Source: Official answer key, GATE 2022 EE (IIT Kharagpur)

Phase voltage of the supply (balanced, star-equivalent): $V_{p h} = \frac{100 3}{3} = 100 V$ . The ammeter sits in line R ahead of both loads, so it reads the total line current drawn by the delta load and the star load together.
Delta-to-star conversion of the balanced delta: $Z_{Y} = \frac{Z _{1}}{3} = \frac{18 + j 24}{3} = (6 + j 8) Ω$ .
Both loads are balanced, so the star point of the $Z_{2}$ load and the equivalent star point of the converted delta are both at the supply neutral potential. The per-phase impedances are therefore in parallel: $Z_{e q} = \frac{( 6 + j 8 ) ( 6 + j 8 )}{( 6 + j 8 ) + ( 6 + j 8 )} = \frac{6 + j 8}{2} = (3 + j 4) Ω, ∣ Z_{e q} ∣ = 3^{2} + 4^{2} = 5 Ω$
Line current: $I_{L} = \frac{V _{p h}}{∣ Z _{e q} ∣} = \frac{100}{5} = 20 A$
Check by treating the loads separately. Star load: $\frac{100}{∣6 + j 8∣} = \frac{100}{10} = 10 A$ per line. Delta load: each $Z_{1}$ sees the full line voltage, so its phase current is $\frac{100 3}{∣18 + j 24∣} = \frac{100 3}{30} A$ , and its line current is $3 \times \frac{100 3}{30} = 10 A$ . Both line currents lag the R phase voltage by the same angle $tan^{- 1} (4/3) = 53.1 3^{\circ}$ (for the delta, $I_{R Y}$ lags $V_{R Y}$ by $53.1 3^{\circ}$ , the line current lags $I_{R Y}$ by $3 0^{\circ}$ , and $V_{R Y}$ leads $V_{R}$ by $3 0^{\circ}$ ), so they are in phase and add directly: $10 + 10 = 20 A$ .

KEY IDEA

Replace the balanced delta by its star equivalent ( $Z_{1} /3$ ). Two balanced star loads fed from the same supply have their star points at neutral potential, so the per-phase impedances are in parallel. Here $Z_{1} /3 = Z_{2} = (6 + j 8) Ω$ , so the per-phase impedance is $(3 + j 4) Ω$ , $∣ Z ∣ = 5 Ω$ , and the line current is $100/5 = 20 A$ .

WATCH OUT FOR

Three errors are common here. (1) Leaving out the factor of 3 in the delta-to-star conversion, or dividing the delta's phase current by $3$ instead of multiplying by it. Either slip gives the wrong delta line current. (2) Applying the line voltage $1003$ V across a star impedance, which gives 17.3 A per branch. (3) Reporting the current of only one load (10 A). Adding the two line-current magnitudes arithmetically is valid here only because both loads have the same impedance angle, $53.1 3^{\circ}$ . In general the currents must be added as phasors.

From the GATE 2022 EE paper. Every question of this paper · Download the full paper · All 11 Three-phase (EE) questions, with notes

GATE 2022 EE Q.34: solution and answer

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