GATE 2016 ME (ME2) – Question 60
In DC arc welding, $V=100+40l$, with arc length l in mm and voltage V in volts. During welding, l varies from 1 to 2 mm and current ranges from 200 to 250 A. Assuming a linear power source, the short-circuit current is ______ A.
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Correct answer: 423 to 428
Explanation
Arc voltage V=100+40l gives140V at l1mm and180V at l2mm. On a drooping linear source these pair with250A and200A, respectively. The voltage–current slope is(180−140)/(200−250)=−0.8V/A, giving source line V=340−0.8I. At short circuit V0, current is **425A**.