Question

For a DC power source for arc welding with characteristics \(3V + I = 240\), what voltage \(V\) yields maximum power?

• A. \(20 \, \text{V}\)
• B. \(30 \, \text{V}\)
• C. \(40 \, \text{V}\)
• D. \(60 \, \text{V}\)

Hint:

For maximum power in quadratic form \(P = aV^2 + bV\), use:
\[ V = \frac{-b}{2a} \]

Answer 1

The Correct Option is C

Concept: Electrical power is given by: \[ P = VI \] To find maximum power, we express power in terms of a single variable and then maximize the function.
Step 1: Express current \(I\) in terms of voltage \(V\).
Given: \[ 3V + I = 240 \Rightarrow I = 240 - 3V \]
Step 2: Substitute into power equation.
\[ P = V \cdot I = V(240 - 3V) \] \[ P = 240V - 3V^2 \]
Step 3: Maximize the power function.
This is a quadratic equation: \[ P = -3V^2 + 240V \] Maximum occurs at: \[ V = \frac{-b}{2a} = \frac{240}{2 \times 3} = 40 \]
Conclusion:
Thus, the voltage that yields maximum power is \(\boxed{40 \, \text{V}}\).