The Z-parameters of a two-port network are Z11 = 2 Ω, Z12 = 1 Ω, Z21 = 10 Ω and Z22 = 11 Ω The corresponding values of hybrid parameters are

This question was previously asked in
UPSC ESE (Prelims) Electronics and Telecommunication Engineering 19 Feb 2023 Official Paper
View all UPSC IES Papers >
  1. \( {\left[\begin{array}{ll} h_{11} & h_{12} \\ h_{21} & h_{22} \end{array}\right]=\left[\begin{array}{rr} \frac{12}{11} & \frac{1}{11} \\ -\frac{10}{11} & \frac{1}{11} \end{array}\right]}\)
  2. \({\left[\begin{array}{ll} h_{11} & h_{12} \\ h_{21} & h_{22} \end{array}\right]=\left[\begin{array}{rr} \frac{1}{11} & \frac{1}{11} \\ -\frac{10}{11} & \frac{12}{11} \end{array}\right]} \)
  3. \({\left[\begin{array}{ll} h_{11} & h_{12} \\ h_{21} & h_{22} \end{array}\right]=\left[\begin{array}{rr} \frac{12}{11} & \frac{10}{11} \\ -\frac{10}{11} & \frac{1}{11} \end{array}\right]}\)
  4. \({\left[\begin{array}{ll} h_{11} & h_{12} \\ h_{21} & h_{22} \end{array}\right]=\left[\begin{array}{rr} \frac{12}{11} & \frac{1}{11} \\ -\frac{10}{11} & \frac{12}{11} \end{array}\right]}\)

Answer (Detailed Solution Below)

Option 1 : \( {\left[\begin{array}{ll} h_{11} & h_{12} \\ h_{21} & h_{22} \end{array}\right]=\left[\begin{array}{rr} \frac{12}{11} & \frac{1}{11} \\ -\frac{10}{11} & \frac{1}{11} \end{array}\right]}\)
Free
ST 1: UPSC ESE (IES) Civil - Building Materials
6.3 K Users
20 Questions 40 Marks 24 Mins

Detailed Solution

Download Solution PDF

Given that, 

Z- parameters.

Z1 = 2\(\Omega\);Z12 = 1\(\Omega\)

Z21 = 10\(\Omega\);Z22 = 11\(\Omega\)

We know that, 

V= Z11 I1 + Z12 I\(\Rightarrow\)V1 = 2I1 + 1I2 ....equation 1.

V2 = Z21 I1 + Z22 I2 \(\Rightarrow\)V2 = 10I1 + 11I2 ....equation 2.

Now, calculating h-parameters.

V1 = h11 I1 + h12 V2 

I2 = h21 + h22 V2 

h11 = \(\frac{V_1}{I_1}|V_2 = 0 \) ; h12 \(\frac{V_1}{V_2}|I_2 = 0 \)

h21 \(\frac{I_2}{I_1}|V_2 = 0 \) ; h22 = \(\frac{I_2}{V_2}|I_1 = 0 \)

When V2 = 0; the equation 2 becomes ; 0 = 10I1 +11I2.

I\(\frac{-10}{11}I_1\) ... equation 3.

\(\frac{I^2}{I^1}=\frac{-10}{11} = \) h21

substitute equation 3 in equation 1; we get

V1 = 2I1 + 1[\(\frac{-10}{11} I_1\)]\(\Rightarrow \frac{V_1}{I_1}=\) h11 = \(\frac{12}{11}\Omega\)

When, I1 = 0; the equation 1 become, V1 = 1I2

 Equation 2 becomes, V2 = 11I2.

h12 = \( \frac{1}{11}\)

h22 =  \(\frac{I_2}{V_2} = \frac{I_2}{I_2} = \frac{1}{11}\Omega\)

h11 = \(\frac{12}{11}\Omega\)

h12 = \(\frac{1}{11}\)

h21 = \(\frac{-10}{11}\)

h22 = \(\frac{1}{11}\)

Here, option 1 is correct.

Latest UPSC IES Updates

Last updated on May 28, 2025

->  UPSC ESE admit card 2025 for the prelims exam has been released. 

-> The UPSC IES Prelims 2025 will be held on 8th June 2025.

-> The selection process includes a Prelims and a Mains Examination, followed by a Personality Test/Interview.

-> Candidates should attempt the UPSC IES mock tests to increase their efficiency. The UPSC IES previous year papers can be downloaded here.

More Two Port Networks Questions

Get Free Access Now
Hot Links: teen patti flush teen patti apk teen patti master golden india