Question 1.P.49: Determine the Thevenin’s and Norton’s equivalents across a a......

Let V_1=V_{\text {Th }} . Now by KVL in first (only) loop,

V+100-V_1=0

Applying Nodal analysis, we get

\begin{array}{r} -0.01  V_1+\frac{V}{20 \times 10^3}+0=0 \\ -200  V_1+V=0 \\ -200  V_1+V_1-100=0 \\ V_1=\frac{-100}{199}=-502.5  mV =V_{ Th } \end{array}

On short circuiting, we have V_1=0

By KVL,

So,  I_{ sc }=5  mA . \text { Now }

R_{ Th }=\frac{\left|V_{ Th }\right|}{I_{ sc }}=100.5  \Omega

and the equivalent circuit is