# GATE Papers >> Mechanical >> 2016 >> Question No 160

Question No. 160

The voltage-length characteristic of a direct current arc in an arc welding process is $V=$ ( 100 + 40 ), where $l$ is the length of the arc in mm and $V$ is arc voltage in volts. During a welding operation, the arc length varies between 1 and 2 mm the welding current is in the range 200-250 A. Assuming a linear power source, the short circuit current is _________ A.

##### Answer : 423 : 428

Solution of Question No 160 of GATE 2016 Mechanical Paper

${\mathcal l}_1=1\;\mathrm{mm}\;\;\;\;\;{\mathcal l}_2=2\;\mathrm{mm}$

More larger arc length more voltage, hence to maintain power current is less

$\begin{array}{l}\left(\mathrm P=\mathrm{VI}\right)\\{\mathrm I}_1=250\mathrm A\;\;\;\;\;{\mathrm I}_2=200\mathrm A\\{\mathrm V}_1=100+4\mathrm\theta{\mathcal l}_1\\{\mathrm V}_1=100+40=140\mathrm V\\{\mathrm V}_2=100+40\times2=180\mathrm V\\\mathrm V=\mathrm{OCV}-\frac{\mathrm{OCV}}{\mathrm{SCC}}\mathrm I\\140=\mathrm{OCV}-\frac{\mathrm{OCV}}{\mathrm{SCC}}\times250.......(\mathrm i)\\180=\mathrm{OCV}-\frac{\mathrm{OCV}}{\mathrm{SCC}}\times200.......(\mathrm{ii})\end{array}$

By solving (i) & (ii)

SCC = 425A

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