Step 1: Understanding the Question:
We are given the voltage-current (V-I) characteristic of a DC power source and asked to find the voltage (V) at which the power output (P) is maximum.
Step 2: Key Formula or Approach:
The power (P) delivered by a DC source is given by the formula:
\[ P = V \times I \]
To find the maximum power, we need to express P as a function of a single variable (either V or I) and then find the maximum of that function. This can be done using calculus (finding the derivative and setting it to zero) or by recognizing the function as a parabola.
Step 3: Detailed Explanation:
First, we express the current \(I\) in terms of voltage \(V\) from the given characteristic equation:
\[ 3V + I = 240 \implies I = 240 - 3V \]
Now, substitute this expression for \(I\) into the power equation:
\[ P = V \times I = V(240 - 3V) \]
\[ P(V) = 240V - 3V^2 \]
This is a quadratic equation of the form \(P = aV^2 + bV + c\), with \(a = -3\) and \(b = 240\). Since \(a<0\), the parabola opens downwards, and its vertex represents the maximum point. The voltage at the vertex is given by:
\[ V = -\frac{b}{2a} \]
Substituting the values of a and b:
\[ V = -\frac{240}{2(-3)} = -\frac{240}{-6} = 40 \, \text{V} \]
Step 4: Final Answer:
The voltage that yields the maximum power is 40 V. Therefore, option (C) is the correct answer.