Solution to Question 12

Step 1: Formulate the Circuit Equation

We are given:

• Supply Voltage $V_0 = 200\,\text{V}$
• Resistors $R_1 = 20\,\Omega$, $R_2 = 20\,\Omega$
• Device D is in parallel with $R_2$.

Let $V$ be the voltage across the device D (which is also the voltage across $R_2$).

• Current through resistor $R_2$: $I_{R2} = \frac{V}{R_2} = \frac{V}{20}$
• Current through device D: $I_D$ (unknown variable)
• Total current from source: $I_{total} = I_{R2} + I_D = \frac{V}{20} + I_D$

Apply Kirchhoff’s Voltage Law (KVL) to the outer loop:

$$ V_0 = I_{total} R_1 + V $$ $$ 200 = \left( \frac{V}{20} + I_D \right) 20 + V $$ $$ 200 = (V + 20 I_D) + V $$ $$ 200 = 2V + 20 I_D $$

Simplifying by dividing by 2:

$$ 100 = V + 10 I_D $$ $$ 10 I_D = 100 – V $$ $$ I_D = 10 – 0.1 V $$

Step 2: Find the Operating Point (Graphical Solution)

We now have a linear equation for the current through the device: $I_D = 10 – 0.1V$. This is our “Load Line”. We need to find where this line intersects the given V-I characteristic curve of the device.

Let’s test points on the Load Line to see where they land on the graph provided:

• If $V = 50\,\text{V}$, $I_D = 10 – 5 = 5\,\text{A}$. (Graph shows nearly 0 A at 50V)
• If $V = 75\,\text{V}$, $I_D = 10 – 7.5 = 2.5\,\text{A}$.
• If $V = 100\,\text{V}$, $I_D = 10 – 10 = 0\,\text{A}$.

Now look at the provided graph for the device:

• The x-axis grid has major divisions of 25V (0, 25, 50, 75, 100).
• The y-axis grid has major divisions of 1.25A (0, 1.25, 2.5, 3.75, 5.0).
• The curve passes exactly through the point (75 V, 2.5 A).

Since our calculated point $(V=75, I=2.5)$ lies exactly on the device’s characteristic curve, this is the solution.