The terminal voltage and current of a linear electrical network shown in Figure (a) are given in the table.
Terminal voltage (\(v_t\))
Terminal current (\(i_t\))
18 V
\(-0.5\) A
30 V
0.5 A
36 V
1.0 A
The correct choice for the parameters (\(I_N\), \(R_N\)) of the Norton equivalent circuit shown in Figure (b) is:
Show Hint
Fit the three data points to a straight line \(v_t=mi_t+c\); the intercept gives \(V_{oc}\) and the slope gives \(R_{th}=R_N\), then use \(I_N=V_{oc}/R_N\).
Step 1: Note that only two points are needed to fix a line, and the third checks consistency.
Take any two rows and write two equations for $v_t=mi_t+c$: using $(i_t,v_t)=(0.5,30)$ and $(1.0,36)$:
\[ 30=0.5m+c,\qquad 36=1.0m+c \]
Subtracting the first from the second:
\[ 36-30=(1.0-0.5)m\implies 6=0.5m\implies m=12 \]
Then $c=30-0.5(12)=30-6=24$.
Step 2: Check with the third row.
At $i_t=-0.5$: $v_t=12(-0.5)+24=-6+24=18$ V, matching the table. So the fit $v_t=12i_t+24$ is confirmed by all three rows, not just two.
Step 3: Read off the open-circuit voltage.
Open circuit means $i_t=0$, giving $V_{oc}=24$ V directly from the intercept.
Step 4: Find the short-circuit current.
Short circuit means $v_t=0$:
\[ 0=12i_t+24\implies i_t=-2\text{ A} \]
The magnitude of this short-circuit current gives the Norton current, $I_N=2.0$ A.
Step 5: Get $R_N$ from the slope.
The slope of the $v_t$-$i_t$ line always equals the Thevenin/Norton resistance for a linear one-port, so
\[ R_N=R_{th}=|m|=12.0\ \Omega \]
Step 6: Double check with $V_{oc}=I_NR_N$.
\[ I_NR_N=2.0\times12.0=24\text{ V}=V_{oc} \]
This matches, confirming both values.
\[ \boxed{I_N=2.0\text{ A},\ R_N=12.0\ \Omega} \]
