Question

Two networks are connected in cascade as shown in the figure. With the usual notations the equivalent A,B,C and D constants are obtained. Given that, $C=0.025\angle {45}^o$, the value of $Z_2$ is

• A. $10\angle {30}^o\Omega$
• B. $40\angle -{45}^o\Omega$
• C. $1\Omega$
• D. $0\Omega$

Answer 1

The correct option is B $40\angle -{45}^o\Omega$


$Z_1=10\angle {30}^o\Omega$

$V_1=A{V}_2+B(-I_2)$

$I_1=C{V}_2+D(-I_2)$

Apply KVL for $V_2$ and $Z_2$ loop,

$V_2=Z_2(I_1+I_2)$ ..... (i)

$C={\begin{array}{c}\frac{I_1}{V_2}\end{array}|}_{I_2=0}$

Putting, $I_2=0$ in equation (i)

$V_2=Z_2{I}_1$

$\Rightarrow Z_2={\begin{array}{c}\frac{V_2}{I_1}\end{array}|}_{I_2=0}=\frac{1}{C}=\frac{1}{0.025\angle {45}^o}$

$Z_2=40\angle -{45}^o\Omega$