Question

A two-port network shown below is excited by external dc sources. The voltage and current are measured with voltmeters $V_1$, $V_2$ and ammeters $A_1$, $A_2$ (all assumed to be ideal) as indicated. under following switch conditions, the readings obtained are:

$\left(i\right)S_1-open,S_2-closed{A}_1=0A,V_1=4.5V,V_2=1.5V,A_2=1A$

$\left(ii\right)S_1-closed,S_2-open{A}_1=4A,V_1=6V,V_2=6V,A_2=0A$


The Z-parameter matrix for this network is

• A. $\left[\begin{array}{cc}1.5& 1.5\\ 4.5& 1.5\end{array}\right]$
• B. $\left[\begin{array}{cc}1.5& 4.5\\ 1.5& 4.5\end{array}\right]$
• C. $\left[\begin{array}{cc}1.5& 4.5\\ 1.5& 1.5\end{array}\right]$
• D. $\left[\begin{array}{cc}4.5& 1.5\\ 1.5& 4.5\end{array}\right]$

Answer 1

The correct option is C $\left[\begin{array}{cc}1.5& 4.5\\ 1.5& 1.5\end{array}\right]$

When switch $S_1$is open, current$I_1=0$

then $V_1=4.5V$

$V_2=1.5V$

$I_2=1A$

$V_1=Z_{11}{I}_1+Z_{12}{I}_2$

$V_2=Z_{21}{I}_1+Z_{22}{I}_2$

$Z_{12}={\begin{array}{c}\frac{V_1}{I_2}\end{array}|}_{I_1=0}=\frac{4.5}{1}=4.5\Omega$

$Z_{22}={\begin{array}{c}\frac{V_2}{I_2}\end{array}|}_{I_1=0}=\frac{1.5}{1}=1.5\Omega$

When switch $S_2$ is open, current$I_2=0$

then, $I_1=4A$

$V_1=6V$

$V_2=6V$

$Z_{11}={\begin{array}{c}\frac{V_1}{I_1}\end{array}|}_{I_2=0}=\frac{6}{4}=1.5\Omega$

$Z_{21}={\begin{array}{c}\frac{V_2}{I_1}\end{array}|}_{I_2=0}=\frac{6}{4}=1.5\Omega$

So, Z-parameter matrix is

$\left[\begin{array}{cc}1.5& 4.5\\ 1.5& 1.5\end{array}\right]$