Question

In a DC arc welding process, open-circuit voltage is 100 V and short-circuit current is 1000 A. Arc voltage varies as $V = 10 + 5l$, where $l$ is arc length in mm. The maximum available arc power during the process is ____________ kVA (in integer).

Hint:

Maximum power transfer occurs at the midpoint of the source V–I characteristic.

Answer 1

Correct Answer: 25

The power source has a linear V–I characteristic between:
- Open circuit: \( V = 100\ \text{V},\ I = 0 \)
- Short circuit: \( V = 0,\ I = 1000\ \text{A} \)
The voltage–current relation is a straight line: \[ V = 100 - 0.1 I. \] Arc voltage also obeys: \[ V = 10 + 5l. \] Maximum arc power is obtained from: \[ P = VI = V(1000 - 10V). \] Differentiate: \[ \frac{dP}{dV} = 1000 - 20V = 0 \] \[ V = 50\ \text{V}. \] Then: \[ I = 1000 - 10V = 1000 - 500 = 500\ \text{A}. \] Maximum power: \[ P_{\max} = VI = 50 \times 500 = 25000\ \text{W} = 25\ \text{kVA}. \] Thus, \[ \boxed{25\ \text{kVA}} \]