Question

A linear 2-port network is shown. From the $V_2$–$(-I_2)$ plot: when $V_2 = 5$ V, $I_2 = 0$ mA, and when $V_2 = 4$ V, $I_2 = -4$ mA. Port 2 is now connected to the load in Fig. (c). Find the current $I_2$ (rounded to one decimal place).

Hint:

Any linear 2-port can be replaced by a Thevenin source at the output. Use the $V_2$–$I_2$ line to extract $V_{th}$ and $R_{th}$ directly.

Answer 1

Correct Answer: 3.9

Step 1: Determine the Thevenin equivalent seen at Port 2.
From the graph data: \[ (V_2,\, I_2) = (5\ \text{V}, 0\ \text{mA}), \quad (4\ \text{V}, -4\ \text{mA}) \] Slope of $V_2$–$(-I_2)$ line: \[ \Delta(-I_2) = 4\ \text{mA}, \quad \Delta V_2 = -1\ \text{V} \] Thus: \[ - I_2 = 4\ \text{mA/V}(5 - V_2) \] At $V_2 = 5$ V, $I_2 = 0 \Rightarrow V_{th} = 5$ V. The Thevenin resistance: \[ R_{th} = \frac{\Delta V_2}{\Delta I_2} = \frac{1}{4\ \text{mA}} = 250\ \Omega. \] Thus Port 2 $\Rightarrow$ Thevenin equivalent: \[ V_{th} = 5\ \text{V},\quad R_{th} = 250\ \Omega. \] Step 2: Connect the load of Fig. (c).
Load consists of: \[ 10\ \text{V source in series with }1\ \text{k}\Omega. \] Writing KVL around the loop: \[ V_{th} = I_2 (R_{th} + 1000) + 10 \] Substitute: \[ 5 = I_2 (250 + 1000) + 10 \] \[ 5 = 1250 I_2 + 10 \] \[ -5 = 1250 I_2 \] \[ I_2 = -0.004\ \text{A} = -4.0\ \text{mA} \] Thus the magnitude: \[ \boxed{4.0\ \text{mA}} \]