Question

In the circuit shown below, the network N is described by the following Y matrix: $Y=\left[\begin{array}{cc}0.1S& -0.01S\\ 0.01S& 0.1S\end{array}\right].$ The voltage gain $\frac{V_2}{V_1}$ is

• A. 1/90
• B. -1/90
• C. -1/99
• D. -1/11

Answer 1

The correct option is D -1/11

$I_2=Y_{21}{V}_1+Y_{22}{V}_2$

$\therefore I_2=0.01V_1+0.1V_2$

From given figure,

$V_2=-I_2{R}_L=-100I_2$

$\therefore I_2=-\frac{V_2}{100}$

Putting value of $I_2$ in equation (i),

$-\frac{V_2}{100}=0.01V_1+0.1V_2$

$\Rightarrow -0.01V_2-0.1V_2=0.01V_1$

$\Rightarrow -0.11V_2=0.01V_1$

$\therefore \frac{V_2}{V_1}=-\frac{0.01}{0.11}$

$or,\frac{V_2}{V_1}=-\frac{1}{11}$