Find the transmission parameter:

Solution:

$ \begin{aligned} \\ &K V L \text { to } 100 p \\\\ &V_1(s)-10 I_1(s)-10\left[I_1(s)-I_3(s)\right]=0 \\\\ &V_1(s)=20 I_1(s)-10 I_3(s)...(1) \\\\ &K V L \text { to } 100 p(2) \\\\ &V_2(s)-10\left[I_2(s)+I_3(s)\right]=0 \\\\ &V_2(s)=10 I_2(s)+10 I_3(s).....(2) \\\\ &K V L \text { to } 100 p(3) \\\\ &-10 I_3(s)-10\left[I_3(s)+I_2(s)\right]-10\left[I_3(s)-I_1(s)\right]=0 \\\\ &-30 I_3(s)+10 I_1(s)-10 I_2(s)=0 \\\\ &30 I_3(s)=10 I_1(s)-10 I_2(s) \\\\ &I_3(s)=\frac{1}{3} I_1(s)-\frac{1}{3} I_2(s).....(3) \\ \end{aligned} \\ $

$ \begin{aligned} \\ &\text { put (3) in (1) }\\\\ &V_1(s)=20 I_1(s)-10\left[\frac{1}{3} I_1(s)-\frac{1}{3} I_2(s)\right]\\\\ &=\frac{50}{3} I_1(S)+\frac{10}{3} I_2(S)....(4)\\\\ &\text { put (3) in (2) }\\\\ &V_2(s)=10 I_2(s)+10\left[\frac{1}{3} I_1(s)-\frac{1}{3} I_2(s)\right]\\\\ &=\frac{10}{3} I_1(s)+\frac{20}{3} I_2(s).....(5)\\\\ &\text { Rearranging eq-n (5) }\\\\ &\frac{10}{3} I_1(S)=V_2(S)-\frac{20}{3} I_2(S)\\\\ &\therefore I_1(s)=\frac{3}{10} V_2(s)-2 I_2(s)...(6)\\ \end{aligned} \\ $

$ \begin{aligned} \\ \text { put } \\ \end{aligned} \begin{aligned} \\ V_1(S) &=\frac{50}{3}\left[\frac{3}{10} V_2(S)-2 I_2(S)\right]+\frac{10}{3} I_2(S) \\\\ &=5 V_2(S)-\frac{100}{3} I_2(S)+\frac{10}{3} I_2(S) \\\\ V_1(S) &=5 V_2(S)-30 I_2(S).....(7) \\ \end{aligned} \\ $

Hence for first network, transmission parameters are

$ \left[\begin{array}{ll}\\ A_1 & B_1 \\\\ C_1 & D_1 \\ \end{array}\right]=\left[\begin{array}{cc} \\ 5 & +30 \\\\ 0.3 & 2 \\ \end{array}\right]\\ $

for resulting network, when similar identical network connected in cascade, then transmission parameter for that network are

$ \left[\begin{array}{ll} A_2 & B_2 \\\\ C_2 & D_2 \\ \end{array}\right]=\left[\begin{array}{ll} A_1 & B_1 \\\\ C_1 & D_1 \\ \end{array}\right]=\left[\begin{array}{ll} 5 & 30 \\\\ 3 / 10 & 2 \\ \end{array}\right] \\ $

Now, by property of cascade connection, transmission parameters of overall resulting network is given by

$ \begin{aligned} \\ {\left[\begin{array}{ll} A & B \\\\ C & D \\ \end{array}\right] } &=\left[\begin{array}{ll} A_1 & B_1 \\\\ C_1 & D_1 \\ \end{array}\right] \times\left[\begin{array}{ll} A_2 & B_2 \\\\ C_2 & D_2 \\ \end{array}\right] \\\\ &=\left[\begin{array}{ll} 5 & 30 \\\\ 0.3 & 2 \\ \end{array}\right]\left[\begin{array}{ll} 5 & 30 \\\\ 0.3 & 2 \\ \end{array}\right] \\ \end{aligned} \\ $

$ \left[\begin{array}{ll} A & B \\\\ C & D \\ \end{array}\right]=\left[\begin{array}{ll} 34 & 210 \\\\ 2.1 & 13 \\ \end{array}\right] \\ $