GATE EE 2026 Set 1 — Question 44

For a linear electrical network, the relationship between terminal voltage ($v_t$) and terminal current ($i_t$) can be expressed using Thevenin's equivalent circuit: $v_t = V_{Th} - R_{Th} i_t$, where $i_t$ is the current flowing out of the network. Alternatively, if $i_t$ is the current flowing into the network, the equation is $v_t = V_{Th} + R_{Th} i_t$. The diagram for Figure (a) shows $i_t$ flowing out of the network. However, if we assume the standard convention for a load connected to a source, where $v_t$ is the voltage across the load and $i_t$ is the current flowing into the load (which is equivalent to flowing out of the network), then the equation $v_t = V_{Th} - R_{Th} i_t$ should hold.Let's test this assumption with the given data points:Using the equation $v_t = V_{Th} - R_{Th} i_t$:
From points 1 and 2:
$18 = V_{Th} - R_{Th}(-0.5) \implies 18 = V_{Th} + 0.5 R_{Th}$ (Eq. 1)
$30 = V_{Th} - R_{Th}(0.5) \implies 30 = V_{Th} - 0.5 R_{Th}$ (Eq. 2)Subtract (Eq. 2) from (Eq. 1):
$18 - 30 = (V_{Th} + 0.5 R_{Th}) - (V_{Th} - 0.5 R_{Th})$
$-12 = 1.0 R_{Th} \implies R_{Th} = -12 \Omega$.A negative Thevenin resistance indicates an active network, but it's unusual for a standard problem unless specified. Let's re-evaluate the interpretation of $i_t$.If we assume the current $i_t$ in the table is the current into the network (opposite to the arrow in Figure (a)), then the Thevenin equivalent equation would be $v_t = V_{Th} + R_{Th} i_t$.Let's use this interpretation:Subtract (Eq. 1') from (Eq. 2'):
$30 - 18 = (V_{Th} + 0.5 R_{Th}) - (V_{Th} - 0.5 R_{Th})$
$12 = 1.0 R_{Th} \implies R_{Th} = 12 \Omega$.Now, substitute $R_{Th} = 12 \Omega$ into (Eq. 2') to find $V_{Th}$:
$30 = V_{Th} + 12(0.5) \implies 30 = V_{Th} + 6 \implies V_{Th} = 24 \text{ V}$.Let's verify with (Eq. 3'):
$36 = V_{Th} + 12(1.0) \implies 36 = V_{Th} + 12 \implies V_{Th} = 24 \text{ V}$.The Thevenin parameters are consistent: $V_{Th} = 24 \text{ V}$ and $R_{Th} = 12 \Omega$.For the Norton equivalent circuit, $R_N = R_{Th} = 12 \Omega$.
The Norton current $I_N$ is the short-circuit current, which is $I_N = \frac{V_{Th}}{R_{Th}}$.
$I_N = \frac{24 \text{ V}}{12 \Omega} = 2.0 \text{ A}$.Therefore, the correct parameters for the Norton equivalent circuit are $I_N = 2.0 \text{ A}$ and $R_N = 12.0 \Omega$.The final answer is $\boxed{\text{C}}$