Given Circuit
The circuit contains:
- Left resistor: \(2\Omega\)
- Right resistor: \(2\Omega\)
- Vertical resistor: \(6\Omega\)
- Dependent current source: \(2V_2\) A, directed upward
- \(I_1\) and \(I_2\) enter the two-port network.
Step 1: Define the Middle Node Voltage
Let the voltage at the middle node be \(V_x\).
Apply KCL at the middle node:
$$
\frac{V_x-V_1}{2}
+
\frac{V_x-V_2}{2}
=
2V_2
$$
Multiplying by 2:
$$
2V_x-V_1-V_2=4V_2
$$
Therefore:
$$
\boxed{
V_x=\frac{V_1}{2}+\frac{5V_2}{2}
}
$$
Step 2: Find \(I_1\)
The current entering port 1 is:
$$
I_1=\frac{V_1-V_x}{2}
$$
Substitute \(V_x\):
$$
I_1=
\frac{
V_1-
\left(
\frac{V_1}{2}+\frac{5V_2}{2}
\right)
}{2}
$$
Therefore:
$$
I_1=
\frac{V_1}{4}
-
\frac{5V_2}{4}
$$
Compare with the Y-parameter equation:
$$
I_1=Y_{11}V_1+Y_{12}V_2
$$
Hence:
$$
\boxed{Y_{11}=\frac14}
\qquad
\boxed{Y_{12}=-\frac54}
$$
Step 3: Find \(I_2\)
The current entering port 2 is:
$$
I_2=\frac{V_2-V_x}{2}
$$
Substitute \(V_x\):
$$
I_2=
\frac{
V_2-
\left(
\frac{V_1}{2}+\frac{5V_2}{2}
\right)
}{2}
$$
Therefore:
$$
I_2=
-\frac{V_1}{4}
-
\frac{3V_2}{4}
$$
Compare with:
$$
I_2=Y_{21}V_1+Y_{22}V_2
$$
Hence:
$$
\boxed{Y_{21}=-\frac14}
\qquad
\boxed{Y_{22}=-\frac34}
$$
Final Y-Matrix
$$
\boxed{
[Y]=
\begin{bmatrix}
\frac14 & -\frac54 \\[2mm]
-\frac14 & -\frac34
\end{bmatrix}
\text{ S}
}
$$
Correct option: (C) (a) & (e) only
\(Y_{21}=-\frac14\) and \(Y_{22}=-\frac34\)
\(Y_{21}=-\frac14\) and \(Y_{22}=-\frac34\)
Further Reading:
GATE EC Previous Year Papers with Solutions