GATE 2020 EE Q.29: solution and answer

GATE 2020 · EEQ.29 · 2 marks · MCQ

Two buses, $i$ and $j$ , are connected with a transmission line of admittance $Y$ , at the two ends of which there are ideal transformers with turns ratios as shown. Bus admittance matrix for the system is:

GATE 2020 EE question 29: Single-line diagram. — Figure from the GATE 2020 EE paper. It is part of the question.

(A)
$[- t_{i} t_{j} Y t_{i}^{2} Y t_{j}^{2} Y - t_{i} t_{j} Y]$
(B)
$[t_{i} t_{j} Y - t_{i}^{2} Y - t_{j}^{2} Y t_{i} t_{j} Y]$
(C)
$[t_{i}^{2} Y - t_{i} t_{j} Y - t_{i} t_{j} Y t_{j}^{2} Y]$
(D)
$[t_{i} t_{j} Y - (t_{i} - t_{j})^{2} Y - (t_{i} - t_{j})^{2} Y t_{i} t_{j} Y]$

Answer and solution

ANSWER

The bus admittance matrix is $[t_{i}^{2} Y - t_{i} t_{j} Y - t_{i} t_{j} Y t_{j}^{2} Y]$ , which is option (C).

Official answer: CSource: Official answer key, GATE 2020 EE (IIT Delhi)

Each ideal transformer scales voltage by its tap ratio and, with power conserved, scales current inversely. In the figure the bus sits on the 1 side of each transformer (ratios $1 : t_{i}$ and $t_{j} : 1$ ). The line-side voltages are therefore $V_{i}^{'} = t_{i} V_{i}$ and $V_{j}^{'} = t_{j} V_{j}$ .
Current in the line from the $i$ end to the $j$ end: $I_{L} = Y (V_{i}^{'} - V_{j}^{'}) = Y (t_{i} V_{i} - t_{j} V_{j})$ .
Power balance at the $i$ transformer: $V_{i} I_{i} = V_{i}^{'} I_{L}$ , so $I_{i} = t_{i} I_{L} = t_{i}^{2} Y V_{i} - t_{i} t_{j} Y V_{j}$ .
Power balance at the $j$ transformer: $V_{j} I_{j} = - V_{j}^{'} I_{L}$ , so $I_{j} = - t_{j} I_{L} = - t_{i} t_{j} Y V_{i} + t_{j}^{2} Y V_{j}$ .
Collect the currents as $[I_{i} I_{j}] = Y_{b u s} [V_{i} V_{j}]$ :
$Y_{b u s} = [t_{i}^{2} Y - t_{i} t_{j} Y - t_{i} t_{j} Y t_{j}^{2} Y]$
Check: the matrix is symmetric, as it must be for a bilateral network with no phase shifter. The diagonals are positive, and the row sums are not zero unless $t_{i} = t_{j}$ . Only option (C) has this form, so the answer is (C).

KEY IDEA

An ideal transformer with ratio $1 : t$ multiplies the voltage on the line side by $t$ and the bus current by $t$ . Each diagonal entry therefore picks up $t^{2} Y$ and the off-diagonal entries pick up $t_{i} t_{j} Y$ .

WATCH OUT FOR

Scaling the voltage by the tap ratio but forgetting to scale the current as well, which gives $t Y$ terms instead of $t^{2} Y$ . Another slip is attaching the ratio to the wrong side of the transformer, so that the line-side voltage becomes $V / t$ .

From the GATE 2020 EE paper. Every question of this paper · Download the full paper · All 10 Per unit & load flow questions, with notes

GATE 2020 EE Q.29: solution and answer

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