For the given 2-port network, the value of transfer impedance Z_{21} in ohms is _______.

This question was previously asked in

GATE EE 2017 Official Paper: Shift 2

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

Concept:

The Z-parameter for a T-equivalent two-port network is,

\(\left[ Z \right] = \left[ {\begin{array}{*{20}{c}} {{Z_{11}}}&{{Z_{12}}}\\ {{Z_{21}}}&{{Z_{22}}} \end{array}} \right]\)

Where

Z11 is an open circuit impedance

Z12 = Z21 = Transfer impedance

Z22 = Open circuit output impedance

Calculation:

By using delta to star conversion, the given circuit can be reduced.

\({R_1} = \frac{{4 \times 2}}{{4 + 2 + 2}} = \frac{8}{8} = 1\;{\rm{\Omega }}\)

\({R_2} = \frac{{2 \times 2}}{{4 + 2 + 2}} = \frac{4}{8} = \frac{1}{2}\;{\rm{\Omega }}\)

\({R_3} = \frac{{4 \times 2}}{{4 + 2 + 2}} = \frac{8}{8} = 1\;{\rm{\Omega }}\)

The modified circuit is,

Now, by comparing the above circuit with T-equivalent network of Z-parameter matrix,

\(\left[ Z \right] = \left[ {\begin{array}{*{20}{c}} 4&3\\ 3&{3.5} \end{array}} \right]\)

Transfer impedance = Z21 = 3 Ω

__Alternate Method__

Transfer impedance \({z_{21}} = {\left. {\frac{{{V_2}}}{{{I_1}}}} \right|_{{I_2} = 0}}\)

V_{1} = Z_{11} I_{1} + Z_{12} I_{2}

V_{2} = Z_{21} I_{1} + I_{22} I_{2}

At I_{2} = 0,

By applying KVL in the loop 1,

(I_{1} – I_{3}) (2) + (I_{1} – I_{3})(2) – 4(I_{3}) = 0

⇒ 4I_{1} – 8I_{3} = 0 ⇒ I_{1} = 2I_{3}

By applying KVL in the loop 2,

-V_{2} + 2(I_{1} – I_{3}) + 2I_{1} = 0

⇒ V_{2} = 4I_{1} – 2I_{3} = 4(2I_{3}) – 2I_{3} = 6I_{3}

⇒ V_{2} = 6(I_{1}/2)

\(\Rightarrow {Z_{21}} = \frac{{{V_2}}}{{{I_1}}} = 3\)