Discussion
In a DC arc welding operation, the voltage-arc length characteristic was obtained as \(V_{arc}=20+5l\) where the arc length l was varied between 5 mm and 7 mm. Here Varc denotes the arc voltage in Volts. The arc current was varied from 400 A to 500 A. Assuming linear power source characteristic, the open circuit voltage and the short circuit current for the welding operation are
Answer: C
The open circuit voltage is 95 V and the short circuit current 950 A.
A longer arc needs more voltage and so draws less current from a drooping source. The two operating points are:
\[ \begin{aligned} l = 5:\ V &= 45\ \text{V},\ I = 500\ \text{A} \\ l = 7:\ V &= 55\ \text{V},\ I = 400\ \text{A} \end{aligned} \]
A linear source characteristic through both points has slope \(-10/100 = -0.1\) V/A:
\[ \begin{aligned} V &= V_{oc} - 0.1I \\ 45 &= V_{oc} - 50 \Rightarrow V_{oc} = 95\ \text{V} \\ I_{sc} &= 95/0.1 = 950\ \text{A} \end{aligned} \]
Pairing the short arc with the low current instead gives a rising line that has no meaningful open circuit voltage, and 45 V is just the arc voltage at 5 mm. 75 V and 150 V do not pass through both operating points.
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