Question 1.P.48: Norton’s equivalent of the given network is...

Consider the given circuit. The nodal analysis at node 1 gives

\begin{aligned} \frac{V_{ A }-5}{5}+\frac{V_{ A }}{4}+\frac{V_{ A } \times 6}{4}=0 \\ \frac{4 V_{ A }-20+5 V_{ A }+30 V_{ A }}{20}=0 \end{aligned}

or

\begin{array}{r} 39 V_{ A }=20 \Rightarrow V_{ A }=\frac{20}{39}=0.512  V \\ I_{ SC }=\frac{0.512 \times 6}{4}=0.769  A \end{array}

R_{ eq }=\frac{4}{6}+\frac{20}{9}=0.66+2.22=2.88  \Omega