Question

The z-parameters of the two port network shown in the figure are $Z_{11}=40\Omega ,Z_{12}=60\Omega ,Z_{21}=80\Omega$ and $Z_{22}=100\Omega$. The average power delivered to $R_L=20\Omega$, in watts, is

• A. 35.55

Answer 1

The correct option is A 35.55

Given,
$Z_{11}=40\Omega ,Z_{12}=60\Omega$
$Z_{21}=80\Omega ,Z_{22}=100\Omega$

From the figure,

$V_2=-20I_2$ .... (i)

$V_1=40I_1+60I_2$ .... (ii)

$V_2=80I_1+100I_2$ . ..... (iii)

From equation (i) and (iii), we get

So,
$-20I_2=80I_1+100I_2$

$\Rightarrow I_2=\frac{2}{3}{I}_1$ ..... (iv)

Using equation (ii) and (iv), we get

So,

$V_1=40I_1+60I_2$

$=40I_1+60\left(\frac{-2}{3}{I}_1\right)$

$V_1=0$

From the figure,

$20=10I_1+V_1$

$Since,V_1=0$

So,
$I_1=2Aand{I}_2=-\frac{4}{3}A$

Power dissipated in

$P_{R_L}=I_2^2{R}_L={\left(\frac{4}{3}\right)}^2\times 20$

$=\frac{16}{9}\times \left(20\right)=35.55W$