The terminal voltage and current of a linear electrical network shown in Figure (a) are given in the table.
\[
\begin{array}{|c|c|}
\hline
\text{Terminal voltage } (v_t) & \text{Terminal current } (i_t) \\
\hline
18\,\text{V} & -0.5\,\text{A} \\
\hline
30\,\text{V} & 0.5\,\text{A} \\
\hline
36\,\text{V} & 1.0\,\text{A} \\
\hline
\end{array}
\]
The correct choice for the parameters ( $\mathrm{I}_{\mathrm{N}}, \mathrm{R}_{\mathrm{N}}$ ) of the Norton equivalent circuit shown in Figure (b) is:
- $\mathrm{I}_{\mathrm{N}}=3.0 \mathrm{~A}, \mathrm{R}_{\mathrm{N}}=24.0 ~\Omega$
- $\mathrm{I}_{\mathrm{N}}=12.0 \mathrm{~A}, \mathrm{R}_{\mathrm{N}}=2.0 ~\Omega$
- $\mathrm{I}_{\mathrm{N}}=2.0 \mathrm{~A}, \mathrm{R}_{\mathrm{N}}=12.0 ~\Omega$
- $\mathrm{I}_{\mathrm{N}}=2.0 \mathrm{~A}, \mathrm{R}_{\mathrm{N}}=24.0 ~\Omega$
