Question

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:

• A. \(I_N=3.0\) A, \(R_N=24.0\ \Omega\)
• B. \(I_N=12.0\) A, \(R_N=2.0\ \Omega\)
• C. \(I_N=2.0\) A, \(R_N=12.0\ \Omega\)
• D. \(I_N=2.0\) A, \(R_N=24.0\ \Omega\)

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 \(I_N=V_{oc}/R_N\).

Answer 1

The Correct Option is C

Step 1: Check that the terminal relation is linear.
A linear network has a straight-line relationship between terminal voltage \(v_t\) and terminal current \(i_t\), of the form \(v_t = mi_t+c\). Using the first two rows of the table to find the slope:
\[ m = \frac{30-18}{0.5-(-0.5)} = \frac{12}{1.0} = 12 \]

Step 2: Find the intercept.
\[ 18 = 12(-0.5)+c \implies 18 = -6+c \implies c = 24 \]

Step 3: Check the relation against the third row.
\[ v_t = 12i_t+24 \implies v_t(1.0) = 12(1.0)+24 = 36\text{ V} \]
This matches the third row exactly (\(36\) V at \(i_t=1.0\) A), confirming the linear relation \(v_t=12i_t+24\) fits all three data points.

Step 4: Interpret the two constants.
Setting \(i_t=0\) gives the open-circuit voltage:
\[ V_{oc} = 24\text{ V} \]
The coefficient of \(i_t\) is the Thevenin resistance seen from the terminals, since \(i_t\) here is defined flowing into the network at the positive terminal, and the network's driving voltage plus \(i_tR_{th}\) gives the terminal voltage:
\[ R_{th} = 12\ \Omega \]

Step 5: Convert Thevenin to Norton.
\[ I_N = \frac{V_{oc}}{R_{th}} = \frac{24}{12} = 2.0\text{ A},\qquad R_N = R_{th} = 12.0\ \Omega \]

Step 6: Verify against the Norton circuit in Figure (b).
In Figure (b), applying KCL at the top node with the same current convention as \(i_t\):
\[ I_N+i_t = \frac{v_t}{R_N} \implies v_t = R_N I_N+R_N i_t \]
Substituting \(I_N=2.0\) A and \(R_N=12.0\ \Omega\):
\[ v_t = 12(2.0)+12i_t = 24+12i_t \]
which is exactly the fitted line \(v_t=12i_t+24\). This confirms the values.

Step 7: Rule out the other options.
Options (A) and (D) use \(R_N=24\ \Omega\), which is the open-circuit voltage value mistaken for a resistance. Option (B) inverts the roles of the fitted numbers. Only \(I_N=2.0\) A with \(R_N=12.0\ \Omega\) is consistent with all three data points.

Final Answer:
\[ \boxed{I_N=2.0\text{ A},\ R_N=12.0\ \Omega} \]