Applying Norton’s Theorem, the Norton’s equivalent circuit to the left of the terminals a and b in the below circuit is having equivalent current source (IN ) and equivalent resistance (RN ) as
1
\(\begin{array}{*{20}{c}}
{{{\rm{I}}_{\rm{N}}} = 5{\rm{A}},}&{{{\rm{R}}_{\rm{N}}} = 4{\rm{\Omega }}}
\end{array}\)
2
\(\begin{array}{*{20}{c}}
{{{\rm{I}}_{\rm{N}}} = 4{\rm{A}};}&{{{\rm{R}}_{\rm{N}}} = 6{\rm{\Omega }}}
\end{array}\)
3
\(\begin{array}{*{20}{c}}
{{{\rm{I}}_{\rm{N}}} = 9{\rm{A}};}&{{{\rm{R}}_{\rm{N}}} = 1.6{\rm{\Omega }}}
\end{array}\)
4
\(\begin{array}{*{20}{c}}
{{{\rm{I}}_{\rm{N}}} = 4{\rm{A}};}&{{{\rm{R}}_{\rm{N}}} = 3{\rm{\Omega }}}
\end{array}\)