Figure: Find Thevenin's equivalent ckt:

Solution:

$ \text { Part-1 - To find "V_Th } \\ $

$ V_{\text {Th }}=V_x...(1) $

$ \begin{aligned} \\ &\begin{aligned}\\ &K V L \text { to } 100 p \\\\ &-150+10\left(5+I_{15}\right)-\frac{V_x}{3}+15 I_{15}=0 \\ \end{aligned}\\\\ &\text { But } V_x=15 \cdot I_{15}\\\\ &-150+50+10 I_{15}-{\frac{15 \cdot I_{15}{}}3{+15 I_{15}=0}}\\\\ &-100+10 I_{15}-5 I_{15}+15 I_{15}=0\\\\ &-100=-20 I_{15}\\\\ &I_{15}=5 \mathrm{Amp}\\\\ &\therefore V_x=15(5)=75 \text { Volts }\\\\ &\begin{array}{r} \text { from (I) } \\\\ V_{T H}=75 \text { Volts } \\ \end{array} \\ \end{aligned} \\ $

$ \text { Part-2 - To find "IIN" } \\ $

$ \begin{aligned} \\ &I_N=-a.....(2) \\\\ &a-b=5...(3) \\\\ &V_x=15(b-c)....(4) \\\\ &k V L+100 p(3) \\\\ &-150-10 c-\frac{V x}{3}-15(c-b)=0 \\\\ &-150-25 c+15 b-\frac{15(b-c)}{3}=0 \\\\ &-150-25 c+15 b-5 b+5 c=0 \\\\ &10 b-20 c-150....(5) \\ \end{aligned} \\ $

KVL to super-10op

$ -30 a-15(b-c)=0 \\ $

$ -30 a-15 b+15 c=0....(6) \\ $

from eq-n (3), (5) (6)

$ \begin{gathered} \\ a=-2 \text { Amp } ; b=-7 \text { AmP } ; c=-11 \text { Amp } \\\\ \text { from }(2) \\\\ I_N=2 \text { Amp } \\\\ R_{T H}=\frac{V_{T H}}{I_N}=\frac{75}{2} \quad R_{T H}=37.5 \Omega \\ \end{gathered} \\ $