The z parameters z11 and z21 for the 2-port network shown in the given figure respectively are :
\(z_{11}=\dfrac{6}{11}\ \Omega;\ z_{21}=\dfrac{-16}{11}\ \Omega\)
Both parameters are open-circuit parameters, so set \(I_{2}=0\) and keep it that way throughout. By definition
\(z_{11}=\left.\dfrac{E_{1}}{I_{1}}\right|_{I_{2}=0}\qquad z_{21}=\left.\dfrac{E_{2}}{I_{1}}\right|_{I_{2}=0}\)
Step 1 — trace the current. With port 2 open, the whole of I1 flows through the 2 Ω series resistor, then down through the 4 Ω shunt resistor, and finally through the dependent source back to the input terminal.
Step 2 — apply KVL round the input loop. Adding the two resistive drops and the controlled source,
\(E_{1}=2I_{1}+4I_{1}-10E_{1}\)
\(E_{1}+10E_{1}=6I_{1}\)
\(11E_{1}=6I_{1}\quad\Rightarrow\quad z_{11}=\dfrac{E_{1}}{I_{1}}=\dfrac{6}{11}\ \Omega\)
The 11 in every option is the signature of this step — it comes from the \(1+10\) produced by the controlled source acting back on its own controlling variable.
Step 3 — write the output voltage. E2 is measured from the shunt node to the reference terminal, so it is the 4 Ω drop plus the source voltage:
\(E_{2}=4I_{1}-10E_{1}\)
Substituting \(E_{1}=\dfrac{6}{11}I_{1}\),
\(E_{2}=4I_{1}-10\left(\dfrac{6}{11}\right)I_{1}=\dfrac{44-60}{11}I_{1}=-\dfrac{16}{11}I_{1}\)
\(z_{21}=-\dfrac{16}{11}\ \Omega\)
which is option 3.
| Parameter | Definition | Value |
|---|---|---|
| z11 | Open-circuit input impedance | +6/11 Ω |
| z21 | Open-circuit forward transfer impedance | −16/11 Ω |
Two checks worth making. First, z11 must be positive here, because the input still looks resistive from the source's point of view — a negative value would mean the network was delivering power back, which this passive-plus-controlled-source combination does not do at the input. Second, z21 may perfectly well be negative: it is a transfer quantity, and the dependent source inverts the sense of the output. That sign is also why the network is non-reciprocal; with a controlled source present \(z_{12}\ne z_{21}\), and reciprocity, which holds for any network of R, L, C and transformers alone, is lost.
Hence, z11 = 6/11 Ω and z21 = −16/11 Ω.
Similar Questions
Read the following statements :
ST 1 : y-parameters can be obtained from Z parameters.
ST 2 : It is not necessary to define y-parameters separately.Match the given lists :
List – I List – II a. Condition of reciprocity i. \(\dfrac{Z_{12}}{Z_{22}}\) b. h12 ii. Z12 = Z21 c. 
\(\begin{bmatrix}R&R\\R&R\end{bmatrix}\)iii. Z d. Condition of symmetry iv. Z11 = Z22 Codes :
The T-parameters for the following cascaded network is :
For the two-port network shown in figure, the z-parameter matrix is given by
The two port network mentioned below can be characterized by four variables V1, V2, I1 and I2, in which only two can be independent.
The h-parameters of the two port network possesses the following :
(a) Linear network should contain no independent sources.
(b) V1 and V2 are taken as independent variables.
(c) V1 and I2 are taken as independent variables.
(d) I1 and V2 are taken as independent variables.Options :
Assertion (A) : In a two port network, with 4 terminals four types of parameters like impedance, admittance, hybrid and transmission are considered and they are related to each other.
Reason (R) : The assumption made for the above statement is that there are no independent sources and non-zero initial conditions within the linear port network.
Select your answer using the codes given below :
The [Y] parameters of the network shown below are given as :
Important Questions from Two Port Networks
A short-circuit admittance matrix of a two-port network is
\(\left[ {\begin{array}{} 0\\ {\frac{1}{2}} \end{array}\begin{array}{} { - \frac{1}{2}}\\ 0 \end{array}} \right]\)
The two-port network is
A two-port network has scattering parameters given \(\left[ s \right] = \left[ {\begin{array}{*{20}{c}} {{s_{11}}}&{{s_{12}}}\\ {{s_{211}}}&{{s_{22}}} \end{array}} \right]\). If the port 2 of the two-port is short-circuited, the s11 parameter for the resultant one-port network is
With 10 V dc connected at port A, the current drawn by 7 Ω connected at port B is
With 6 V dc connected at port A, 1 Ω connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is
In a linear two – port network, when 10 V is applied to Port 1, a current of 4 A flows through Port 2 when it is short-circuited. When 5 V is applied to Port, a current of 1.25 A flows through a 1 Ω resistance connected across Port 2. When 3 V is applied to Port 1, then current (in Ampere) through a 2 Ω resistance connected across Port 2 is __________.
