Crey | 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 ratio as shown below. Bus admittance matrix for the system is [ − t i t j Y t j 2 Y t i 2 Y − t i t j Y ] , [ t i t j Y − t j 2 Y − t i 2 Y t i t j Y ] , [ t i 2 Y − t i t j Y − t i t j Y t j 2 Y ] , [ t i t j Y − ( t i − t j ) 2 Y − ( t i − t j ) 2 Y t i t j Y ]

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 ratio as shown below. Bus admittance matrix for the system is

Options

$[\begin{matrix} - t_{i} t_{j} Y & t_{j}^{2} Y \\ t_{i}^{2} Y & - t_{i} t_{j} Y \end{matrix}]$

$[\begin{matrix} t_{i} t_{j} Y & - t_{j}^{2} Y \\ - t_{i}^{2} Y & t_{i} t_{j} Y \end{matrix}]$

$[\begin{matrix} t_{i}^{2} Y & - t_{i} t_{j} Y \\ - t_{i} t_{j} Y & t_{j}^{2} Y \end{matrix}]$

$[\begin{matrix} t_{i} t_{j} Y & - {(t_{i} - t_{j})}^{2} Y \\ - {(t_{i} - t_{j})}^{2} Y & t_{i} t_{j} Y \end{matrix}]$

Correct Answer :

$[\begin{matrix} t_{i}^{2} Y & - t_{i} t_{j} Y \\ - t_{i} t_{j} Y & t_{j}^{2} Y \end{matrix}]$

Solution :

The correct answer is:
$[\begin{matrix} t_{i}^{2} Y & - t_{i} t_{j} Y \\ - t_{i} t_{j} Y & t_{j}^{2} Y \end{matrix}]$

Step-by-Step Derivation and Explanation:

1. Identify the Circuit Parameters from the Image:
From the given system diagram, we observe the following components:
- Bus i has a voltage of _{V_i} and an injected current of _{I_i}.
- An ideal transformer is connected at the end of Bus i with a turns ratio of 1 : t_i.
- A transmission line with series admittance Y connects the two transformer stages.
- Another ideal transformer is connected before Bus j with a turns ratio of t_j : 1.
- Bus j has a voltage of _{V_j} and an injected current of _{I_j}.

2. Calculate the Internal Node Voltages:
Let _{V_i'} be the voltage on the secondary side of the first transformer (facing the transmission line), and _{V_j'} be the voltage on the primary side of the second transformer (facing the transmission line).
Using the turns ratio relationships of ideal transformers:
- For the first transformer with ratio 1 : t_i:
$V_{i}^{'} = t_{i} V_{i}$
- For the second transformer with ratio t_j : 1:
$V_{j}^{'} = t_{j} V_{j}$

3. Determine the Current Flowing Through the Transmission Line:
Let _{I_Y} be the current flowing through the transmission line from left to right (from the first transformer to the second transformer):
$I_{Y} = Y (V_{i}^{'} - V_{j}^{'})$
Substituting the expressions for _{V_i'} and _{V_j'}:
$I_{Y} = Y (t_{i V_{i}} - t_{j V_{j}})$

4. Relate Injected Bus Currents to Line Current:
For ideal transformers, complex power is conserved, meaning the input current and output current scale inversely with the turns ratios:
- At Bus i:
$I_{i} = t_{i} I_{Y}$
Substituting _{I_Y} into the equation:
$I_{i} = t_{i} [Y (t_{i V_{i}} - t_{j V_{j}})] = t_{i}^{2} Y V_{i} - t_{i} t_{j} Y V_{j}$

- At Bus j, the line current _{I_Y} flows into the transformer, so the current flowing to Bus j is related by:
$I_{j} = - t_{j} I_{Y}$
Substituting _{I_Y} into the equation:
$I_{j} = - t_{j [} Y (t_{i V_{i}} - t_{j V_{j}})] = - t_{i t_{j}} Y V_{i} + t_{j}^{2} Y V_{j}$

5. Formulate the Admittance Matrix:
Arranging the linear system of equations in matrix form:

$[\begin{matrix} I_{i} \\ I_{j} \end{matrix}] = [\begin{matrix} t_{i}^{2} Y & - t_{i} t_{j} Y \\ - t_{i} t_{j} Y & t_{j}^{2} Y \end{matrix}] [\begin{matrix} V_{i} \\ V_{j} \end{matrix}]$

Thus, the bus admittance matrix _{Y_bus} is:

$Y_{bus} = [\begin{matrix} t_{i}^{2} Y & - t_{i} t_{j} Y \\ - t_{i} t_{j} Y & t_{j}^{2} Y \end{matrix}]$

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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 ratio as shown below. Bus admittance matrix for the system is

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