PYQs / GATE EE / 2020 / Set 1 / Q39 GATE EE 2020 Set 1 — Question 39 Go beyond PYQs with Success Tracker AI-powered personalised practice and doubt support. Unlimited practice on eligible plans; AI usage limits apply. MCQ +2 / -0.67 Medium Y-Bus Formation Per-Unit System & Load Flow Power Systems
Power Systems → Per-Unit System & Load Flow → Y-Bus Formation
Last updated 5 September 2026
Question Two buses,
i i i and
j j j , are connected with a transmission line of admittance
Y Y Y , at the two ends of which there are ideal transformers with turns ratios as shown. Bus admittance matrix for the system is:
Correct answer (C) [figure]
Solution The system consists of a transmission line with admittance
Y Y Y connected between two ideal transformers.
1. Transformer at bus i i i : It has a turns ratio of 1 : t i 1:t_i 1 : t i i i i V i V_i V i i ′ i' i ′ V i ′ = t i V i V_i' = t_i V_i V i ′ = t i V i i i i I i I_i I i I i ′ = I i t i I_i' = \frac{I_i}{t_i} I i ′ = t i I i
2. Transformer at bus j j j : It has a turns ratio of t j : 1 t_j:1 t j : 1 j j j V j V_j V j j ′ j' j ′ V j ′ = t j V j V_j' = t_j V_j V j ′ = t j V j j j j I j I_j I j I j ′ = I j t j I_j' = \frac{I_j}{t_j} I j ′ = t j I j
3. Line Admittance : The current I i ′ I_i' I i ′ Y Y Y i ′ i' i ′ j ′ j' j ′ I i ′ = Y ( V i ′ − V j ′ ) = Y ( t i V i − t j V j ) I_i' = Y(V_i' - V_j') = Y(t_i V_i - t_j V_j) I i ′ = Y ( V i ′ − V j ′ ) = Y ( t i V i − t j V j ) 4. Relating to Bus Currents : Substituting I i ′ = I i t i I_i' = \frac{I_i}{t_i} I i ′ = t i I i I i t i = Y ( t i V i − t j V j ) ⟹ I i = t i 2 Y V i − t i t j Y V j \frac{I_i}{t_i} = Y(t_i V_i - t_j V_j) \implies I_i = t_i^2 Y V_i - t_i t_j Y V_j t i I i = Y ( t i V i − t j V j ) ⟹ I i = t i 2 Y V i − t i t j Y V j 5. Similarly for Bus j j j : The current I j ′ I_j' I j ′ j ′ j' j ′ i ′ i' i ′ I j ′ = Y ( V j ′ − V i ′ ) = Y ( t j V j − t i V i ) I_j' = Y(V_j' - V_i') = Y(t_j V_j - t_i V_i) I j ′ = Y ( V j ′ − V i ′ ) = Y ( t j V j − t i V i ) Substituting
I j ′ = I j t j I_j' = \frac{I_j}{t_j} I j ′ = t j I j :
I j t j = Y ( t j V j − t i V i ) ⟹ I j = − t i t j Y V i + t j 2 Y V j \frac{I_j}{t_j} = Y(t_j V_j - t_i V_i) \implies I_j = -t_i t_j Y V_i + t_j^2 Y V_j t j I j = Y ( t j V j − t i V i ) ⟹ I j = − t i t j Y V i + t j 2 Y V j 6. Bus Admittance Matrix : The Y b u s Y_{bus} Y b u s [ I i I j ] = [ Y i i Y i j Y j i Y j j ] [ V i V j ] \begin{bmatrix} I_i \\ I_j \end{bmatrix} = \begin{bmatrix} Y_{ii} & Y_{ij} \\ Y_{ji} & Y_{jj} \end{bmatrix} \begin{bmatrix} V_i \\ V_j \end{bmatrix} [ I i I j ] = [ Y ii Y j i Y ij Y j j ] [ V i V j ] Y i i = t i 2 Y Y_{ii} = t_i^2 Y Y ii = t i 2 Y Y i j = − t i t j Y Y_{ij} = -t_i t_j Y Y ij = − t i t j Y Y j i = − t i t j Y Y_{ji} = -t_i t_j Y Y j i = − t i t j Y Y j j = t j 2 Y Y_{jj} = t_j^2 Y Y j j = t j 2 Y Thus, the bus admittance matrix is:
Y b u s = [ t i 2 Y − t i t j Y − t i t j Y t j 2 Y ] Y_{bus} = \begin{bmatrix} t_i^2 Y & -t_i t_j Y \\ -t_i t_j Y & t_j^2 Y \end{bmatrix} Y b u s = [ t i 2 Y − t i t j Y − t i t j Y t j 2 Y ] This matches option (C).
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Correct answer (C) [figure]
Solution The system consists of a transmission line with admittance
Y Y Y connected between two ideal transformers.
1. Transformer at bus i i i : It has a turns ratio of 1 : t i 1:t_i 1 : t i i i i V i V_i V i i ′ i' i ′ V i ′ = t i V i V_i' = t_i V_i V i ′ = t i V i i i i I i I_i I i I i ′ = I i t i I_i' = \frac{I_i}{t_i} I i ′ = t i I i
2. Transformer at bus j j j : It has a turns ratio of t j : 1 t_j:1 t j : 1 j j j V j V_j V j j ′ j' j ′ V j ′ = t j V j V_j' = t_j V_j V j ′ = t j V j j j j I j I_j I j I j ′ = I j t j I_j' = \frac{I_j}{t_j} I j ′ = t j I j
3. Line Admittance : The current I i ′ I_i' I i ′ Y Y Y i ′ i' i ′ j ′ j' j ′ I i ′ = Y ( V i ′ − V j ′ ) = Y ( t i V i − t j V j ) I_i' = Y(V_i' - V_j') = Y(t_i V_i - t_j V_j) I i ′ = Y ( V i ′ − V j ′ ) = Y ( t i V i − t j V j ) 4. Relating to Bus Currents : Substituting I i ′ = I i t i I_i' = \frac{I_i}{t_i} I i ′ = t i I i I i t i = Y ( t i V i − t j V j ) ⟹ I i = t i 2 Y V i − t i t j Y V j \frac{I_i}{t_i} = Y(t_i V_i - t_j V_j) \implies I_i = t_i^2 Y V_i - t_i t_j Y V_j t i I i = Y ( t i V i − t j V j ) ⟹ I i = t i 2 Y V i − t i t j Y V j 5. Similarly for Bus j j j : The current I j ′ I_j' I j ′ j ′ j' j ′ i ′ i' i ′ I j ′ = Y ( V j ′ − V i ′ ) = Y ( t j V j − t i V i ) I_j' = Y(V_j' - V_i') = Y(t_j V_j - t_i V_i) I j ′ = Y ( V j ′ − V i ′ ) = Y ( t j V j − t i V i ) Substituting
I j ′ = I j t j I_j' = \frac{I_j}{t_j} I j ′ = t j I j :
I j t j = Y ( t j V j − t i V i ) ⟹ I j = − t i t j Y V i + t j 2 Y V j \frac{I_j}{t_j} = Y(t_j V_j - t_i V_i) \implies I_j = -t_i t_j Y V_i + t_j^2 Y V_j t j I j = Y ( t j V j − t i V i ) ⟹ I j = − t i t j Y V i + t j 2 Y V j 6. Bus Admittance Matrix : The Y b u s Y_{bus} Y b u s [ I i I j ] = [ Y i i Y i j Y j i Y j j ] [ V i V j ] \begin{bmatrix} I_i \\ I_j \end{bmatrix} = \begin{bmatrix} Y_{ii} & Y_{ij} \\ Y_{ji} & Y_{jj} \end{bmatrix} \begin{bmatrix} V_i \\ V_j \end{bmatrix} [ I i I j ] = [ Y ii Y j i Y ij Y j j ] [ V i V j ] Y i i = t i 2 Y Y_{ii} = t_i^2 Y Y ii = t i 2 Y Y i j = − t i t j Y Y_{ij} = -t_i t_j Y Y ij = − t i t j Y Y j i = − t i t j Y Y_{ji} = -t_i t_j Y Y j i = − t i t j Y Y j j = t j 2 Y Y_{jj} = t_j^2 Y Y j j = t j 2 Y Thus, the bus admittance matrix is:
Y b u s = [ t i 2 Y − t i t j Y − t i t j Y t j 2 Y ] Y_{bus} = \begin{bmatrix} t_i^2 Y & -t_i t_j Y \\ -t_i t_j Y & t_j^2 Y \end{bmatrix} Y b u s = [ t i 2 Y − t i t j Y − t i t j Y t j 2 Y ] This matches option (C).
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